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Premium equations
Actuarial mathematics 3280Benefit premiums
Edward Furman
Department of Mathematics and StatisticsYork University
January 14, 2010
Edward Furman Actuarial mathematics 3280 1 / 46
Premium equations
Insurance businessAn insurance system is a mechanism for reducing the adversefinancial impact of random events that prevent the fulfillment ofreasonable expectations. (see, Bowers et al., 1989)
Definition 1.1Conditional state independence Let X be a set of insurancerisks X1,X2, . . ., then a premium calculation principle (pcp) isdefined as the map
π : X → [0,∞],
such thatπ[X ] ≥ E[X ],
for every X ∈ X .
Note that π implies ordering for each X ,Y ∈ X .
Edward Furman Actuarial mathematics 3280 2 / 46
Premium equations
Insurance businessAn insurance system is a mechanism for reducing the adversefinancial impact of random events that prevent the fulfillment ofreasonable expectations. (see, Bowers et al., 1989)
Definition 1.1Conditional state independence Let X be a set of insurancerisks X1,X2, . . ., then a premium calculation principle (pcp) isdefined as the map
π : X → [0,∞],
such that
π[X ] ≥ E[X ],
for every X ∈ X .
Note that π implies ordering for each X ,Y ∈ X .
Edward Furman Actuarial mathematics 3280 2 / 46
Premium equations
Insurance businessAn insurance system is a mechanism for reducing the adversefinancial impact of random events that prevent the fulfillment ofreasonable expectations. (see, Bowers et al., 1989)
Definition 1.1Conditional state independence Let X be a set of insurancerisks X1,X2, . . ., then a premium calculation principle (pcp) isdefined as the map
π : X → [0,∞],
such thatπ[X ] ≥ E[X ],
for every X ∈ X .
Note that π implies ordering for each X ,Y ∈ X .
Edward Furman Actuarial mathematics 3280 2 / 46
Premium equations
Insurance businessAn insurance system is a mechanism for reducing the adversefinancial impact of random events that prevent the fulfillment ofreasonable expectations. (see, Bowers et al., 1989)
Definition 1.1Conditional state independence Let X be a set of insurancerisks X1,X2, . . ., then a premium calculation principle (pcp) isdefined as the map
π : X → [0,∞],
such thatπ[X ] ≥ E[X ],
for every X ∈ X .
Note that π implies ordering for each X ,Y ∈ X .
Edward Furman Actuarial mathematics 3280 2 / 46
Premium equations
What you pay is what you get
Letw1 denote the capital an insurance buyer has;X be the risk the person faces;π represent the premium the buyer is ready to pay.
Then π may be determined from the equation
w1 − π = E[w1 − X ] = w1 − E[X ],
thus yieldingπ = E[X ].
Edward Furman Actuarial mathematics 3280 3 / 46
Premium equations
What you pay is what you get
Letw1 denote the capital an insurance buyer has;X be the risk the person faces;π represent the premium the buyer is ready to pay.
Then π may be determined from the equation
w1 − π = E[w1 − X ] = w1 − E[X ],
thus yielding
π = E[X ].
Edward Furman Actuarial mathematics 3280 3 / 46
Premium equations
What you pay is what you get
Letw1 denote the capital an insurance buyer has;X be the risk the person faces;π represent the premium the buyer is ready to pay.
Then π may be determined from the equation
w1 − π = E[w1 − X ] = w1 − E[X ],
thus yieldingπ = E[X ].
Edward Furman Actuarial mathematics 3280 3 / 46
Premium equations
Insured side
Let us now assume the above mentioned person has autility function that describes his/her preferences. Moreprecisely, u : Y → R, a function from e.g. a consumptionset to reals.
The premium defining equation from the buyer’s point ofview becomes:
u(w1 − π) = E[u(w1 − X )]
We will further assume that u : R→ R and the person1 prefers more to less, and2 enjoys much more when is poor and much less when is
rich.
Edward Furman Actuarial mathematics 3280 4 / 46
Premium equations
Insured side
Let us now assume the above mentioned person has autility function that describes his/her preferences. Moreprecisely, u : Y → R, a function from e.g. a consumptionset to reals.The premium defining equation from the buyer’s point ofview becomes:
u(w1 − π) = E[u(w1 − X )]
We will further assume that u : R→ R and the person1 prefers more to less, and2 enjoys much more when is poor and much less when is
rich.
Edward Furman Actuarial mathematics 3280 4 / 46
Premium equations
Insured side
Let us now assume the above mentioned person has autility function that describes his/her preferences. Moreprecisely, u : Y → R, a function from e.g. a consumptionset to reals.The premium defining equation from the buyer’s point ofview becomes:
u(w1 − π) = E[u(w1 − X )]
We will further assume that u : R→ R and the person1 prefers more to less, and2 enjoys much more when is poor and much less when is
rich.
Edward Furman Actuarial mathematics 3280 4 / 46
Premium equations
The latter two assumption are reformulated as
u′(x) > 0
and
u′′(x) < 0.
Using the above and Jensen’s inequality, we arrive at
u(w1 − π) = E[u(w1 − X )] ≤ u(w1 − E[X ]),
leading to the requirement
π ≥ E[X ],
or alternatively0 ≥ E[X ]− π.
Edward Furman Actuarial mathematics 3280 5 / 46
Premium equations
The latter two assumption are reformulated as
u′(x) > 0
andu′′(x) < 0.
Using the above and Jensen’s inequality, we arrive at
u(w1 − π) = E[u(w1 − X )] ≤ u(w1 − E[X ]),
leading to the requirement
π ≥ E[X ],
or alternatively0 ≥ E[X ]− π.
Edward Furman Actuarial mathematics 3280 5 / 46
Premium equations
The latter two assumption are reformulated as
u′(x) > 0
andu′′(x) < 0.
Using the above and Jensen’s inequality, we arrive at
u(w1 − π) = E[u(w1 − X )] ≤
u(w1 − E[X ]),
leading to the requirement
π ≥ E[X ],
or alternatively0 ≥ E[X ]− π.
Edward Furman Actuarial mathematics 3280 5 / 46
Premium equations
The latter two assumption are reformulated as
u′(x) > 0
andu′′(x) < 0.
Using the above and Jensen’s inequality, we arrive at
u(w1 − π) = E[u(w1 − X )] ≤ u(w1 − E[X ]),
leading to the requirement
π ≥ E[X ],
or alternatively0 ≥ E[X ]− π.
Edward Furman Actuarial mathematics 3280 5 / 46
Premium equations
The latter two assumption are reformulated as
u′(x) > 0
andu′′(x) < 0.
Using the above and Jensen’s inequality, we arrive at
u(w1 − π) = E[u(w1 − X )] ≤ u(w1 − E[X ]),
leading to the requirement
π ≥ E[X ],
or alternatively0 ≥ E[X ]− π.
Edward Furman Actuarial mathematics 3280 5 / 46
Premium equations
Insurer’s side
Let us now look at the insurer side. Letw2 denote the capital the insurer has;X be the risk the insurer takes (transferred to him by thebuyer) ;θ represent the premium the insurer intends to charge.
Then the premium defining equation from the insurer’s side is:
u(w2) = E[u(w2 + θ − X )] ≤ u(E[w2 + θ − X ])
thus yieldingE[θ − X ] ≥ 0⇔ θ ≥ E[X ],
or alternatively demanding non-positive loss, i.e.,
E[X ]− θ ≤ 0.
Edward Furman Actuarial mathematics 3280 6 / 46
Premium equations
Insurer’s side
Let us now look at the insurer side. Letw2 denote the capital the insurer has;X be the risk the insurer takes (transferred to him by thebuyer) ;θ represent the premium the insurer intends to charge.
Then the premium defining equation from the insurer’s side is:
u(w2) = E[u(w2 + θ − X )] ≤ u(E[w2 + θ − X ])
thus yieldingE[θ − X ] ≥ 0⇔ θ ≥ E[X ],
or alternatively demanding non-positive loss, i.e.,
E[X ]− θ ≤ 0.
Edward Furman Actuarial mathematics 3280 6 / 46
Premium equations
Insurer’s side
Let us now look at the insurer side. Letw2 denote the capital the insurer has;X be the risk the insurer takes (transferred to him by thebuyer) ;θ represent the premium the insurer intends to charge.
Then the premium defining equation from the insurer’s side is:
u(w2) = E[u(w2 + θ − X )] ≤
u(E[w2 + θ − X ])
thus yieldingE[θ − X ] ≥ 0⇔ θ ≥ E[X ],
or alternatively demanding non-positive loss, i.e.,
E[X ]− θ ≤ 0.
Edward Furman Actuarial mathematics 3280 6 / 46
Premium equations
Insurer’s side
Let us now look at the insurer side. Letw2 denote the capital the insurer has;X be the risk the insurer takes (transferred to him by thebuyer) ;θ represent the premium the insurer intends to charge.
Then the premium defining equation from the insurer’s side is:
u(w2) = E[u(w2 + θ − X )] ≤ u(E[w2 + θ − X ])
thus yielding
E[θ − X ] ≥ 0⇔ θ ≥ E[X ],
or alternatively demanding non-positive loss, i.e.,
E[X ]− θ ≤ 0.
Edward Furman Actuarial mathematics 3280 6 / 46
Premium equations
Insurer’s side
Let us now look at the insurer side. Letw2 denote the capital the insurer has;X be the risk the insurer takes (transferred to him by thebuyer) ;θ represent the premium the insurer intends to charge.
Then the premium defining equation from the insurer’s side is:
u(w2) = E[u(w2 + θ − X )] ≤ u(E[w2 + θ − X ])
thus yieldingE[θ − X ] ≥ 0⇔ θ ≥ E[X ],
or alternatively demanding non-positive loss, i.e.,
E[X ]− θ ≤ 0.
Edward Furman Actuarial mathematics 3280 6 / 46
Premium equations
Feasibility
Definition 1.2An insurance contract is called feasible whenever
π ≥ θ ≥ E[X ].
There are a lot of various pricing principles in the literature.In this course we will mostly assume linear utility(u′′(x) = 0) that is
π = θ = E[X ],
and thus the insurance contract is always feasible.Moreover, the loss has then to be zero (equivalenceprinciple), i.e., e.g.,
E[X ]− π = 0.
Edward Furman Actuarial mathematics 3280 7 / 46
Premium equations
Feasibility
Definition 1.2An insurance contract is called feasible whenever
π ≥ θ ≥ E[X ].
There are a lot of various pricing principles in the literature.
In this course we will mostly assume linear utility(u′′(x) = 0) that is
π = θ = E[X ],
and thus the insurance contract is always feasible.Moreover, the loss has then to be zero (equivalenceprinciple), i.e., e.g.,
E[X ]− π = 0.
Edward Furman Actuarial mathematics 3280 7 / 46
Premium equations
Feasibility
Definition 1.2An insurance contract is called feasible whenever
π ≥ θ ≥ E[X ].
There are a lot of various pricing principles in the literature.In this course we will mostly assume linear utility(u′′(x) = 0) that is
π = θ = E[X ],
and thus the insurance contract is always feasible.Moreover, the loss has then to be zero (equivalenceprinciple), i.e., e.g.,
E[X ]− π = 0.
Edward Furman Actuarial mathematics 3280 7 / 46
Premium equations
Example 1.1
A person age (x) enters a contract that pays $1 at the end ofthe year of death of the buyer. In return, the buyer is required topay premiums at the beginning of every year while he is alive.State the formula for the premium using the equivalenceprinciple.
SolutionWe observe that
E[π] = E[P..aK+1 ] = P
..ax
andE[X ] = E[vK+1] = Ax .
Edward Furman Actuarial mathematics 3280 8 / 46
Premium equations
Example 1.1
A person age (x) enters a contract that pays $1 at the end ofthe year of death of the buyer. In return, the buyer is required topay premiums at the beginning of every year while he is alive.State the formula for the premium using the equivalenceprinciple.
SolutionWe observe that
E[π] = E[P..aK+1 ] = P
..ax
andE[X ] = E[vK+1] = Ax .
Edward Furman Actuarial mathematics 3280 8 / 46
Premium equations
Example 1.1
A person age (x) enters a contract that pays $1 at the end ofthe year of death of the buyer. In return, the buyer is required topay premiums at the beginning of every year while he is alive.State the formula for the premium using the equivalenceprinciple.
SolutionWe observe that
E[π] = E[P..aK+1 ] =
P..ax
andE[X ] = E[vK+1] = Ax .
Edward Furman Actuarial mathematics 3280 8 / 46
Premium equations
Example 1.1
A person age (x) enters a contract that pays $1 at the end ofthe year of death of the buyer. In return, the buyer is required topay premiums at the beginning of every year while he is alive.State the formula for the premium using the equivalenceprinciple.
SolutionWe observe that
E[π] = E[P..aK+1 ] = P
..ax
andE[X ] = E[vK+1] =
Ax .
Edward Furman Actuarial mathematics 3280 8 / 46
Premium equations
Example 1.1
A person age (x) enters a contract that pays $1 at the end ofthe year of death of the buyer. In return, the buyer is required topay premiums at the beginning of every year while he is alive.State the formula for the premium using the equivalenceprinciple.
SolutionWe observe that
E[π] = E[P..aK+1 ] = P
..ax
andE[X ] = E[vK+1] = Ax .
Edward Furman Actuarial mathematics 3280 8 / 46
Premium equations
SolutionThus, due to the equivalence principle, we require that
Ax − P..ax = 0⇔ P = Ax/
..ax .
The loss random variable is given by
L = vK+1 − P..aK+1 ,
we can find its variance:
Var[L] = Var[vK+1 − P..aK+1 ]
= Var[vK+1 − P
1− vK+1
d
]= Var
[vK+1 − P
d+ P
vK+1
d
]= Var
[vK+1
(1 +
Pd
)− P
d
].
Edward Furman Actuarial mathematics 3280 9 / 46
Premium equations
SolutionThus, due to the equivalence principle, we require that
Ax − P..ax = 0⇔ P = Ax/
..ax .
The loss random variable is given by
L = vK+1 − P..aK+1 ,
we can find its variance:
Var[L] = Var[vK+1 − P..aK+1 ]
= Var[vK+1 − P
1− vK+1
d
]= Var
[vK+1 − P
d+ P
vK+1
d
]= Var
[vK+1
(1 +
Pd
)− P
d
].
Edward Furman Actuarial mathematics 3280 9 / 46
Premium equations
SolutionThus, due to the equivalence principle, we require that
Ax − P..ax = 0⇔ P = Ax/
..ax .
The loss random variable is given by
L = vK+1 − P..aK+1 ,
we can find its variance:
Var[L] =
Var[vK+1 − P..aK+1 ]
= Var[vK+1 − P
1− vK+1
d
]= Var
[vK+1 − P
d+ P
vK+1
d
]= Var
[vK+1
(1 +
Pd
)− P
d
].
Edward Furman Actuarial mathematics 3280 9 / 46
Premium equations
SolutionThus, due to the equivalence principle, we require that
Ax − P..ax = 0⇔ P = Ax/
..ax .
The loss random variable is given by
L = vK+1 − P..aK+1 ,
we can find its variance:
Var[L] = Var[vK+1 − P..aK+1 ]
=
Var[vK+1 − P
1− vK+1
d
]= Var
[vK+1 − P
d+ P
vK+1
d
]= Var
[vK+1
(1 +
Pd
)− P
d
].
Edward Furman Actuarial mathematics 3280 9 / 46
Premium equations
SolutionThus, due to the equivalence principle, we require that
Ax − P..ax = 0⇔ P = Ax/
..ax .
The loss random variable is given by
L = vK+1 − P..aK+1 ,
we can find its variance:
Var[L] = Var[vK+1 − P..aK+1 ]
= Var[vK+1 − P
1− vK+1
d
]=
Var[vK+1 − P
d+ P
vK+1
d
]= Var
[vK+1
(1 +
Pd
)− P
d
].
Edward Furman Actuarial mathematics 3280 9 / 46
Premium equations
SolutionThus, due to the equivalence principle, we require that
Ax − P..ax = 0⇔ P = Ax/
..ax .
The loss random variable is given by
L = vK+1 − P..aK+1 ,
we can find its variance:
Var[L] = Var[vK+1 − P..aK+1 ]
= Var[vK+1 − P
1− vK+1
d
]= Var
[vK+1 − P
d+ P
vK+1
d
]=
Var[vK+1
(1 +
Pd
)− P
d
].
Edward Furman Actuarial mathematics 3280 9 / 46
Premium equations
SolutionThus, due to the equivalence principle, we require that
Ax − P..ax = 0⇔ P = Ax/
..ax .
The loss random variable is given by
L = vK+1 − P..aK+1 ,
we can find its variance:
Var[L] = Var[vK+1 − P..aK+1 ]
= Var[vK+1 − P
1− vK+1
d
]= Var
[vK+1 − P
d+ P
vK+1
d
]= Var
[vK+1
(1 +
Pd
)− P
d
].
Edward Furman Actuarial mathematics 3280 9 / 46
Premium equations
SolutionThen,
Var[L] =
Var[vK+1
(1 +
Pd
)]= Var
[vK+1
](1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Ax..axd
)2
=2Ax − (Ax )2
(..axd)2
,
since..axd + Ax = 1.
Edward Furman Actuarial mathematics 3280 10 / 46
Premium equations
SolutionThen,
Var[L] = Var[vK+1
(1 +
Pd
)]=
Var[vK+1
](1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Ax..axd
)2
=2Ax − (Ax )2
(..axd)2
,
since..axd + Ax = 1.
Edward Furman Actuarial mathematics 3280 10 / 46
Premium equations
SolutionThen,
Var[L] = Var[vK+1
(1 +
Pd
)]= Var
[vK+1
](1 +
Pd
)2
=
(2Ax − (Ax )2)
(1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Ax..axd
)2
=2Ax − (Ax )2
(..axd)2
,
since..axd + Ax = 1.
Edward Furman Actuarial mathematics 3280 10 / 46
Premium equations
SolutionThen,
Var[L] = Var[vK+1
(1 +
Pd
)]= Var
[vK+1
](1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Pd
)2
=
(2Ax − (Ax )2)
(1 +
Ax..axd
)2
=2Ax − (Ax )2
(..axd)2
,
since..axd + Ax = 1.
Edward Furman Actuarial mathematics 3280 10 / 46
Premium equations
SolutionThen,
Var[L] = Var[vK+1
(1 +
Pd
)]= Var
[vK+1
](1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Ax..axd
)2
=
2Ax − (Ax )2
(..axd)2
,
since..axd + Ax = 1.
Edward Furman Actuarial mathematics 3280 10 / 46
Premium equations
SolutionThen,
Var[L] = Var[vK+1
(1 +
Pd
)]= Var
[vK+1
](1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Pd
)2
= (2Ax − (Ax )2)
(1 +
Ax..axd
)2
=2Ax − (Ax )2
(..axd)2
,
since..axd + Ax = 1.
Edward Furman Actuarial mathematics 3280 10 / 46
Premium equations
We can easily generalize the ideas in Example 1. Indeed,denote by
b(K + 1) the benefit function,
v(K + 1) the discount function,P the general symbol for the premium,Z the discrete annuity random variable (e.g.
..aK+1 ).
Then using the already familiar ‘equivalence principle’, wearrive at:
P =E[b(K + 1)v(K + 1)]
E[Z ].
Putting b(K + 1) ≡ 1, v(K + 1) = vk+1 and Z =..aK+1 , we
arrive at the case of Example 1.
Edward Furman Actuarial mathematics 3280 11 / 46
Premium equations
We can easily generalize the ideas in Example 1. Indeed,denote by
b(K + 1) the benefit function,v(K + 1) the discount function,
P the general symbol for the premium,Z the discrete annuity random variable (e.g.
..aK+1 ).
Then using the already familiar ‘equivalence principle’, wearrive at:
P =E[b(K + 1)v(K + 1)]
E[Z ].
Putting b(K + 1) ≡ 1, v(K + 1) = vk+1 and Z =..aK+1 , we
arrive at the case of Example 1.
Edward Furman Actuarial mathematics 3280 11 / 46
Premium equations
We can easily generalize the ideas in Example 1. Indeed,denote by
b(K + 1) the benefit function,v(K + 1) the discount function,P the general symbol for the premium,
Z the discrete annuity random variable (e.g...aK+1 ).
Then using the already familiar ‘equivalence principle’, wearrive at:
P =E[b(K + 1)v(K + 1)]
E[Z ].
Putting b(K + 1) ≡ 1, v(K + 1) = vk+1 and Z =..aK+1 , we
arrive at the case of Example 1.
Edward Furman Actuarial mathematics 3280 11 / 46
Premium equations
We can easily generalize the ideas in Example 1. Indeed,denote by
b(K + 1) the benefit function,v(K + 1) the discount function,P the general symbol for the premium,Z the discrete annuity random variable (e.g.
..aK+1 ).
Then using the already familiar ‘equivalence principle’, wearrive at:
P =E[b(K + 1)v(K + 1)]
E[Z ].
Putting b(K + 1) ≡ 1, v(K + 1) = vk+1 and Z =..aK+1 , we
arrive at the case of Example 1.
Edward Furman Actuarial mathematics 3280 11 / 46
Premium equations
We can easily generalize the ideas in Example 1. Indeed,denote by
b(K + 1) the benefit function,v(K + 1) the discount function,P the general symbol for the premium,Z the discrete annuity random variable (e.g.
..aK+1 ).
Then using the already familiar ‘equivalence principle’, wearrive at:
P =E[b(K + 1)v(K + 1)]
E[Z ].
Putting b(K + 1) ≡ 1, v(K + 1) = vk+1 and Z =..aK+1 , we
arrive at the case of Example 1.
Edward Furman Actuarial mathematics 3280 11 / 46
Premium equations
Example 1.2Let us now consider the case of the n-year term insurance withpremiums payable at the beginning of each year until end of thecontract. Derive the equation for the premium and calculate thevariance of the loss.
Solution
We have b(k + 1) ≡ 1, v(k + 1) = vk+1 and for the premiums
Z =
{ ..aK+1 , 0 ≤ K < n
..an , K ≥ n
.
As to the loss, let
Z ∗ =
{vK+1, K = 0,1,2, . . . ,n − 1vn, elsewhere
.
Edward Furman Actuarial mathematics 3280 12 / 46
Premium equations
Example 1.2Let us now consider the case of the n-year term insurance withpremiums payable at the beginning of each year until end of thecontract. Derive the equation for the premium and calculate thevariance of the loss.
Solution
We have b(k + 1) ≡ 1, v(k + 1) = vk+1 and for the premiums
Z =
{ ..aK+1 , 0 ≤ K < n
..an , K ≥ n
.
As to the loss, let
Z ∗ =
{vK+1, K = 0,1,2, . . . ,n − 1vn, elsewhere
.
Edward Furman Actuarial mathematics 3280 12 / 46
Premium equations
Example 1.2Let us now consider the case of the n-year term insurance withpremiums payable at the beginning of each year until end of thecontract. Derive the equation for the premium and calculate thevariance of the loss.
Solution
We have b(k + 1) ≡ 1, v(k + 1) = vk+1 and for the premiums
Z =
{ ..aK+1 , 0 ≤ K < n
..an , K ≥ n
.
As to the loss, let
Z ∗ =
{vK+1, K = 0,1,2, . . . ,n − 1vn, elsewhere
.
Edward Furman Actuarial mathematics 3280 12 / 46
Premium equations
Example 1.2Let us now consider the case of the n-year term insurance withpremiums payable at the beginning of each year until end of thecontract. Derive the equation for the premium and calculate thevariance of the loss.
Solution
We have b(k + 1) ≡ 1, v(k + 1) = vk+1 and for the premiums
Z =
{ ..aK+1 , 0 ≤ K < n
..an , K ≥ n
.
As to the loss, let
Z ∗ =
{vK+1, K = 0,1,2, . . . ,n − 1vn, elsewhere
.
Edward Furman Actuarial mathematics 3280 12 / 46
Premium equations
Example 1.2Let us now consider the case of the n-year term insurance withpremiums payable at the beginning of each year until end of thecontract. Derive the equation for the premium and calculate thevariance of the loss.
Solution
We have b(k + 1) ≡ 1, v(k + 1) = vk+1 and for the premiums
Z =
{ ..aK+1 , 0 ≤ K < n
..an , K ≥ n
.
As to the loss, let
Z ∗ =
{vK+1, K = 0,1,2, . . . ,n − 1
vn, elsewhere.
Edward Furman Actuarial mathematics 3280 12 / 46
Premium equations
Example 1.2Let us now consider the case of the n-year term insurance withpremiums payable at the beginning of each year until end of thecontract. Derive the equation for the premium and calculate thevariance of the loss.
Solution
We have b(k + 1) ≡ 1, v(k + 1) = vk+1 and for the premiums
Z =
{ ..aK+1 , 0 ≤ K < n
..an , K ≥ n
.
As to the loss, let
Z ∗ =
{vK+1, K = 0,1,2, . . . ,n − 1vn, elsewhere
.
Edward Furman Actuarial mathematics 3280 12 / 46
Premium equations
SolutionThus the premium is formulated as
Px :n =E[Z ∗]E[Z ]
=
Ax :n..ax :n
.
The loss random variable is
L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗
d.
Thus the variance is
Var[L] = Var[Z ∗ − Px :n
1− Z ∗
d
]= Var
[Z ∗(
1 +Px :n
d
)− Px :n
d
]= Var
[Z ∗(
1 +Px :n
d
)].
Edward Furman Actuarial mathematics 3280 13 / 46
Premium equations
SolutionThus the premium is formulated as
Px :n =E[Z ∗]E[Z ]
=Ax :n..ax :n
.
The loss random variable is
L =
Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗
d.
Thus the variance is
Var[L] = Var[Z ∗ − Px :n
1− Z ∗
d
]= Var
[Z ∗(
1 +Px :n
d
)− Px :n
d
]= Var
[Z ∗(
1 +Px :n
d
)].
Edward Furman Actuarial mathematics 3280 13 / 46
Premium equations
SolutionThus the premium is formulated as
Px :n =E[Z ∗]E[Z ]
=Ax :n..ax :n
.
The loss random variable is
L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗
d.
Thus the variance is
Var[L] =
Var[Z ∗ − Px :n
1− Z ∗
d
]= Var
[Z ∗(
1 +Px :n
d
)− Px :n
d
]= Var
[Z ∗(
1 +Px :n
d
)].
Edward Furman Actuarial mathematics 3280 13 / 46
Premium equations
SolutionThus the premium is formulated as
Px :n =E[Z ∗]E[Z ]
=Ax :n..ax :n
.
The loss random variable is
L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗
d.
Thus the variance is
Var[L] = Var[Z ∗ − Px :n
1− Z ∗
d
]=
Var[Z ∗(
1 +Px :n
d
)− Px :n
d
]= Var
[Z ∗(
1 +Px :n
d
)].
Edward Furman Actuarial mathematics 3280 13 / 46
Premium equations
SolutionThus the premium is formulated as
Px :n =E[Z ∗]E[Z ]
=Ax :n..ax :n
.
The loss random variable is
L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗
d.
Thus the variance is
Var[L] = Var[Z ∗ − Px :n
1− Z ∗
d
]= Var
[Z ∗(
1 +Px :n
d
)− Px :n
d
]=
Var[Z ∗(
1 +Px :n
d
)].
Edward Furman Actuarial mathematics 3280 13 / 46
Premium equations
SolutionThus the premium is formulated as
Px :n =E[Z ∗]E[Z ]
=Ax :n..ax :n
.
The loss random variable is
L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗
d.
Thus the variance is
Var[L] = Var[Z ∗ − Px :n
1− Z ∗
d
]= Var
[Z ∗(
1 +Px :n
d
)− Px :n
d
]= Var
[Z ∗(
1 +Px :n
d
)].
Edward Furman Actuarial mathematics 3280 13 / 46
Premium equations
SolutionFurther because of the equivalence principle, we have that
Var[L] =
(1 +
Px :n
d
)2 (2Ax :n − (Ax :n )2
)=
(2Ax :n − (Ax :n )2)(d
..ax :n )2
,
recalling that1 = d
..ax :n + Ax :n
that leads to
1 + Px :n /d = 1 +Ax :n
d..ax :n
=1
d..ax :n
.
Edward Furman Actuarial mathematics 3280 14 / 46
Premium equations
SolutionFurther because of the equivalence principle, we have that
Var[L] =
(1 +
Px :n
d
)2 (2Ax :n − (Ax :n )2
)=
(2Ax :n − (Ax :n )2)(d
..ax :n )2
,
recalling that1 = d
..ax :n + Ax :n
that leads to
1 + Px :n /d = 1 +Ax :n
d..ax :n
=1
d..ax :n
.
Edward Furman Actuarial mathematics 3280 14 / 46
Premium equations
SolutionFurther because of the equivalence principle, we have that
Var[L] =
(1 +
Px :n
d
)2 (2Ax :n − (Ax :n )2
)=
(2Ax :n − (Ax :n )2)(d
..ax :n )2
,
recalling that1 = d
..ax :n + Ax :n
that leads to
1 + Px :n /d = 1 +Ax :n
d..ax :n
=1
d..ax :n
.
Edward Furman Actuarial mathematics 3280 14 / 46
Premium equations
Quantile premium
Besides the equivalence principle, there are a lot of other usefulways of pricing insurance contracts. We shall now define the socalled quantile based premium.
Definition 1.3 (Quantile or Value-at-Risk premium)The quantile based premium for a continuous random variableL is the solution in π of the following equation:
P(L(π) ≥ 0) = q,
where L represents the loss random variable and q ∈ (0,1).
It is useful to recall the following definition of the q-th quantile ofX v F ,
xq = inf{x : F (x) ≥ q} = sup{x : F (x) ≤ q}.
Edward Furman Actuarial mathematics 3280 15 / 46
Premium equations
Example 1.3Consider a 10000 fully discrete whole life insurance. Use ELTand interest rate equal to 6% and issue age 35. Calculate therequired premium such that
the expectation of the distribution of loss is zero,the probability is less than 0.5 that the loss is positive(approximate the smallest premium).the probability of a positive total loss is 0.05 using thenormal. approximation
Find the variance of the loss distribution in these three cases.
Edward Furman Actuarial mathematics 3280 16 / 46
Premium equations
Solutiona.
Due to the equivalence principle
P = 10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.
The variance is:
Var[L] = 1000022A35 − (A35)2
d..a35
= 100002 0.0348843− 0.12871942
((0.06/1.06)(15.39262))2
= 2412713
and √Var[L] = 1553.3.
Edward Furman Actuarial mathematics 3280 17 / 46
Premium equations
Solutiona. Due to the equivalence principle
P = 10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.
The variance is:
Var[L] = 1000022A35 − (A35)2
d..a35
= 100002 0.0348843− 0.12871942
((0.06/1.06)(15.39262))2
= 2412713
and √Var[L] = 1553.3.
Edward Furman Actuarial mathematics 3280 17 / 46
Premium equations
Solutiona. Due to the equivalence principle
P =
10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.
The variance is:
Var[L] = 1000022A35 − (A35)2
d..a35
= 100002 0.0348843− 0.12871942
((0.06/1.06)(15.39262))2
= 2412713
and √Var[L] = 1553.3.
Edward Furman Actuarial mathematics 3280 17 / 46
Premium equations
Solutiona. Due to the equivalence principle
P = 10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.
The variance is:
Var[L] = 1000022A35 − (A35)2
d..a35
= 100002 0.0348843− 0.12871942
((0.06/1.06)(15.39262))2
= 2412713
and √Var[L] = 1553.3.
Edward Furman Actuarial mathematics 3280 17 / 46
Premium equations
Solutiona. Due to the equivalence principle
P = 10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.
The variance is:
Var[L] = 1000022A35 − (A35)2
d..a35
= 100002 0.0348843− 0.12871942
((0.06/1.06)(15.39262))2
= 2412713
and √Var[L] = 1553.3.
Edward Furman Actuarial mathematics 3280 17 / 46
Premium equations
Solutionb. We are interested in
P(L(π) > 0) < 0.5⇔ P(
10000vK+1 − π1− vK+1
d> 0
)< 0.5.
Thus we arrive at
P(
10000vK+1 − π
d+ π
vK+1
d> 0
)< 0.5
⇔ P(
vK+1(
10000 +π
d
)− π
d> 0
)< 0.5
⇔ P(
vK+1 >π
d10000 + π
)< 0.5
⇔ P(
(K + 1) log(v) > log(
π
d10000 + π
))< 0.5.
Edward Furman Actuarial mathematics 3280 18 / 46
Premium equations
Solutionb. We are interested in
P(L(π) > 0) < 0.5⇔
P(
10000vK+1 − π1− vK+1
d> 0
)< 0.5.
Thus we arrive at
P(
10000vK+1 − π
d+ π
vK+1
d> 0
)< 0.5
⇔ P(
vK+1(
10000 +π
d
)− π
d> 0
)< 0.5
⇔ P(
vK+1 >π
d10000 + π
)< 0.5
⇔ P(
(K + 1) log(v) > log(
π
d10000 + π
))< 0.5.
Edward Furman Actuarial mathematics 3280 18 / 46
Premium equations
Solutionb. We are interested in
P(L(π) > 0) < 0.5⇔ P(
10000vK+1 − π1− vK+1
d> 0
)< 0.5.
Thus we arrive at
P(
10000vK+1 − π
d+ π
vK+1
d> 0
)< 0.5
⇔ P(
vK+1(
10000 +π
d
)− π
d> 0
)< 0.5
⇔ P(
vK+1 >π
d10000 + π
)< 0.5
⇔ P(
(K + 1) log(v) > log(
π
d10000 + π
))< 0.5.
Edward Furman Actuarial mathematics 3280 18 / 46
Premium equations
Solutionb. We are interested in
P(L(π) > 0) < 0.5⇔ P(
10000vK+1 − π1− vK+1
d> 0
)< 0.5.
Thus we arrive at
P(
10000vK+1 − π
d+ π
vK+1
d> 0
)< 0.5
⇔ P(
vK+1(
10000 +π
d
)− π
d> 0
)< 0.5
⇔ P(
vK+1 >π
d10000 + π
)< 0.5
⇔ P(
(K + 1) log(v) > log(
π
d10000 + π
))< 0.5.
Edward Furman Actuarial mathematics 3280 18 / 46
Premium equations
Solutionb. We are interested in
P(L(π) > 0) < 0.5⇔ P(
10000vK+1 − π1− vK+1
d> 0
)< 0.5.
Thus we arrive at
P(
10000vK+1 − π
d+ π
vK+1
d> 0
)< 0.5
⇔ P(
vK+1(
10000 +π
d
)− π
d> 0
)< 0.5
⇔ P(
vK+1 >π
d10000 + π
)< 0.5
⇔ P(
(K + 1) log(v) > log(
π
d10000 + π
))< 0.5.
Edward Furman Actuarial mathematics 3280 18 / 46
Premium equations
Solutionb. We are interested in
P(L(π) > 0) < 0.5⇔ P(
10000vK+1 − π1− vK+1
d> 0
)< 0.5.
Thus we arrive at
P(
10000vK+1 − π
d+ π
vK+1
d> 0
)< 0.5
⇔ P(
vK+1(
10000 +π
d
)− π
d> 0
)< 0.5
⇔ P(
vK+1 >π
d10000 + π
)< 0.5
⇔ P(
(K + 1) log(v) > log(
π
d10000 + π
))< 0.5.
Edward Furman Actuarial mathematics 3280 18 / 46
Premium equations
SolutionAs log(v) < 0, we have that
⇔ P(
K + 1 < log(
π
d10000 + π
)/ log(v)
)< 0.5
⇔ P(
K < log(
π
d10000 + π
)/ log(v)− 1
)< 0.5
⇔ P(
K (x) ≥ log(
π
d10000 + π
)/ log(v)− 1
)> 0.5
We thus look for the highest value k for which P(K (x) > k − 1)=k px > 0.5. Take k − 1 = 41, then 42p35 = 0.5125. Go further,take k − 1 = 42, then 43p35 = 0.4808 - less than 0.5. Thus thenecessary age is k − 1 = 41. Solving
log(
π
d10000 + π
)/ log(v)− 1 = 42
Edward Furman Actuarial mathematics 3280 19 / 46
Premium equations
SolutionAs log(v) < 0, we have that
⇔ P(
K + 1 < log(
π
d10000 + π
)/ log(v)
)< 0.5
⇔ P(
K < log(
π
d10000 + π
)/ log(v)− 1
)< 0.5
⇔ P(
K (x) ≥ log(
π
d10000 + π
)/ log(v)− 1
)> 0.5
We thus look for the highest value k for which P(K (x) > k − 1)=k px > 0.5. Take k − 1 = 41, then 42p35 = 0.5125. Go further,take k − 1 = 42, then 43p35 = 0.4808 - less than 0.5. Thus thenecessary age is k − 1 = 41. Solving
log(
π
d10000 + π
)/ log(v)− 1 = 42
Edward Furman Actuarial mathematics 3280 19 / 46
Premium equations
SolutionAs log(v) < 0, we have that
⇔ P(
K + 1 < log(
π
d10000 + π
)/ log(v)
)< 0.5
⇔ P(
K < log(
π
d10000 + π
)/ log(v)− 1
)< 0.5
⇔ P(
K (x) ≥ log(
π
d10000 + π
)/ log(v)− 1
)> 0.5
We thus look for the highest value k for which P(K (x) > k − 1)=k px > 0.5. Take k − 1 = 41, then
42p35 = 0.5125. Go further,take k − 1 = 42, then 43p35 = 0.4808 - less than 0.5. Thus thenecessary age is k − 1 = 41. Solving
log(
π
d10000 + π
)/ log(v)− 1 = 42
Edward Furman Actuarial mathematics 3280 19 / 46
Premium equations
SolutionAs log(v) < 0, we have that
⇔ P(
K + 1 < log(
π
d10000 + π
)/ log(v)
)< 0.5
⇔ P(
K < log(
π
d10000 + π
)/ log(v)− 1
)< 0.5
⇔ P(
K (x) ≥ log(
π
d10000 + π
)/ log(v)− 1
)> 0.5
We thus look for the highest value k for which P(K (x) > k − 1)=k px > 0.5. Take k − 1 = 41, then 42p35 = 0.5125. Go further,take k − 1 = 42, then 43p35 = 0.4808 - less than 0.5. Thus thenecessary age is k − 1 = 41. Solving
log(
π
d10000 + π
)/ log(v)− 1 = 42
Edward Furman Actuarial mathematics 3280 19 / 46
Premium equations
SolutionAs log(v) < 0, we have that
⇔ P(
K + 1 < log(
π
d10000 + π
)/ log(v)
)< 0.5
⇔ P(
K < log(
π
d10000 + π
)/ log(v)− 1
)< 0.5
⇔ P(
K (x) ≥ log(
π
d10000 + π
)/ log(v)− 1
)> 0.5
We thus look for the highest value k for which P(K (x) > k − 1)=k px > 0.5. Take k − 1 = 41, then 42p35 = 0.5125. Go further,take k − 1 = 42, then 43p35 = 0.4808 - less than 0.5. Thus thenecessary age is k − 1 = 41. Solving
log(
π
d10000 + π
)/ log(v)− 1 = 42
Edward Furman Actuarial mathematics 3280 19 / 46
Premium equations
SolutionOr equivalently
log(
π
d10000 + π
)= 43 log(v),
that is
π
d10000 + π= v43
leads to
π =d10000v43
1− v43 =46.20240.9184
= 50.31.
The variance is given by
Var[L(π)] = 100002(2A35 − (A35)2)(
1 +π
d10000
)2
Thus√
Var[L(π)] = 1473.65.
Edward Furman Actuarial mathematics 3280 20 / 46
Premium equations
SolutionOr equivalently
log(
π
d10000 + π
)= 43 log(v),
that isπ
d10000 + π= v43
leads to
π =d10000v43
1− v43 =46.20240.9184
= 50.31.
The variance is given by
Var[L(π)] = 100002(2A35 − (A35)2)(
1 +π
d10000
)2
Thus√
Var[L(π)] = 1473.65.
Edward Furman Actuarial mathematics 3280 20 / 46
Premium equations
SolutionOr equivalently
log(
π
d10000 + π
)= 43 log(v),
that isπ
d10000 + π= v43
leads to
π =d10000v43
1− v43 =46.20240.9184
= 50.31.
The variance is given by
Var[L(π)] = 100002(2A35 − (A35)2)(
1 +π
d10000
)2
Thus√
Var[L(π)] = 1473.65.
Edward Furman Actuarial mathematics 3280 20 / 46
Premium equations
Solutionc. This time we are interested in
P(L(π) > 0) = 0.05
using normal approximation. Thus,
P((L(π)−E[L(π)])/√
Var[L(π)] > −E[L(π)]/√
Var[L(π)]) = 0.05,
leading to
P((L(π)−E[L(π)])/√
Var[L(π)] ≤ −E[L(π)]/√
Var[L(π)]) = 0.95.
We know that in the normal case, Φ−1(0.95) = 1.6449, thus tofind the premium the following equation in π has to be solved:
−E[L(π)]/√
Var[L(π)] = 1.6449
Edward Furman Actuarial mathematics 3280 21 / 46
Premium equations
Solutionc. This time we are interested in
P(L(π) > 0) = 0.05
using normal approximation. Thus,
P((L(π)−E[L(π)])/√
Var[L(π)] > −E[L(π)]/√
Var[L(π)]) = 0.05,
leading to
P((L(π)−E[L(π)])/√
Var[L(π)] ≤ −E[L(π)]/√
Var[L(π)]) = 0.95.
We know that in the normal case, Φ−1(0.95) = 1.6449, thus tofind the premium the following equation in π has to be solved:
−E[L(π)]/√
Var[L(π)] = 1.6449
Edward Furman Actuarial mathematics 3280 21 / 46
Premium equations
Solutionc. This time we are interested in
P(L(π) > 0) = 0.05
using normal approximation. Thus,
P((L(π)−E[L(π)])/√
Var[L(π)] > −E[L(π)]/√
Var[L(π)]) = 0.05,
leading to
P((L(π)−E[L(π)])/√
Var[L(π)] ≤ −E[L(π)]/√
Var[L(π)]) = 0.95.
We know that in the normal case, Φ−1(0.95) = 1.6449, thus tofind the premium the following equation in π has to be solved:
−E[L(π)]/√
Var[L(π)] = 1.6449
Edward Furman Actuarial mathematics 3280 21 / 46
Premium equations
Solutionc. This time we are interested in
P(L(π) > 0) = 0.05
using normal approximation. Thus,
P((L(π)−E[L(π)])/√
Var[L(π)] > −E[L(π)]/√
Var[L(π)]) = 0.05,
leading to
P((L(π)−E[L(π)])/√
Var[L(π)] ≤ −E[L(π)]/√
Var[L(π)]) = 0.95.
We know that in the normal case, Φ−1(0.95) = 1.6449, thus tofind the premium the following equation in π has to be solved:
−E[L(π)]/√
Var[L(π)] = 1.6449
Edward Furman Actuarial mathematics 3280 21 / 46
Premium equations
Solutionc. This time we are interested in
P(L(π) > 0) = 0.05
using normal approximation. Thus,
P((L(π)−E[L(π)])/√
Var[L(π)] > −E[L(π)]/√
Var[L(π)]) = 0.05,
leading to
P((L(π)−E[L(π)])/√
Var[L(π)] ≤ −E[L(π)]/√
Var[L(π)]) = 0.95.
We know that in the normal case, Φ−1(0.95) = 1.6449, thus tofind the premium the following equation in π has to be solved:
−E[L(π)]/√
Var[L(π)] = 1.6449
Edward Furman Actuarial mathematics 3280 21 / 46
Premium equations
SolutionNote that
E[L(π)] =
(10000− π
d
)A35 −
π
d.
andVar[L(π)] =
(10000− π
d
)2(2A35 − (A35)2).
Taking into account the above, the premium is 100.66.
Under the equivalence principle we have that
..a−1x = πx + d and πx =
dAx
1− Ax.
Check the above for term insurances.
Edward Furman Actuarial mathematics 3280 22 / 46
Premium equations
SolutionNote that
E[L(π)] =(
10000− π
d
)A35 −
π
d.
andVar[L(π)] =
(10000− π
d
)2(2A35 − (A35)2).
Taking into account the above, the premium is 100.66.
Under the equivalence principle we have that
..a−1x = πx + d and πx =
dAx
1− Ax.
Check the above for term insurances.
Edward Furman Actuarial mathematics 3280 22 / 46
Premium equations
SolutionNote that
E[L(π)] =(
10000− π
d
)A35 −
π
d.
andVar[L(π)] =
(10000− π
d
)2(2A35 − (A35)2).
Taking into account the above, the premium is 100.66.
Under the equivalence principle we have that
..a−1x = πx + d and πx =
dAx
1− Ax.
Check the above for term insurances.
Edward Furman Actuarial mathematics 3280 22 / 46
Premium equations
It is interesting to note that both sides of
..a−1x = πx + d
stand for the annual amount paid by a life annuityproduced by one monetary unit at time zero. Indeed
1 =..a−1x
..ax = (πx + d)
..ax .
Edward Furman Actuarial mathematics 3280 23 / 46
Premium equations
It is interesting to note that both sides of
..a−1x = πx + d
stand for the annual amount paid by a life annuityproduced by one monetary unit at time zero. Indeed
1 =..a−1x
..ax = (πx + d)
..ax .
Edward Furman Actuarial mathematics 3280 23 / 46
Premium equations
Problem 1.1A person age (x) borrows an amount of money equal to Ax inorder to buy a whole life insurance with unit benefit and a singlepremium payed at the beginning of the contract. He repays theinterest on Ax in advance at the beginning of each year ifsurvives and agrees to repay the loan itself from the unit deathbenefit at the end of the year of death. What is the annualpremium for a full unit of this insurance?
SolutionThe equation is actually
dAx..ax + Ax · Ax = Ax
ordAx
..ax = (1− Ax )Ax .
Edward Furman Actuarial mathematics 3280 24 / 46
Premium equations
Problem 1.1A person age (x) borrows an amount of money equal to Ax inorder to buy a whole life insurance with unit benefit and a singlepremium payed at the beginning of the contract. He repays theinterest on Ax in advance at the beginning of each year ifsurvives and agrees to repay the loan itself from the unit deathbenefit at the end of the year of death. What is the annualpremium for a full unit of this insurance?
SolutionThe equation is actually
dAx..ax
+ Ax · Ax = Ax
ordAx
..ax = (1− Ax )Ax .
Edward Furman Actuarial mathematics 3280 24 / 46
Premium equations
Problem 1.1A person age (x) borrows an amount of money equal to Ax inorder to buy a whole life insurance with unit benefit and a singlepremium payed at the beginning of the contract. He repays theinterest on Ax in advance at the beginning of each year ifsurvives and agrees to repay the loan itself from the unit deathbenefit at the end of the year of death. What is the annualpremium for a full unit of this insurance?
SolutionThe equation is actually
dAx..ax + Ax
· Ax = Ax
ordAx
..ax = (1− Ax )Ax .
Edward Furman Actuarial mathematics 3280 24 / 46
Premium equations
Problem 1.1A person age (x) borrows an amount of money equal to Ax inorder to buy a whole life insurance with unit benefit and a singlepremium payed at the beginning of the contract. He repays theinterest on Ax in advance at the beginning of each year ifsurvives and agrees to repay the loan itself from the unit deathbenefit at the end of the year of death. What is the annualpremium for a full unit of this insurance?
SolutionThe equation is actually
dAx..ax + Ax ·
Ax = Ax
ordAx
..ax = (1− Ax )Ax .
Edward Furman Actuarial mathematics 3280 24 / 46
Premium equations
Problem 1.1A person age (x) borrows an amount of money equal to Ax inorder to buy a whole life insurance with unit benefit and a singlepremium payed at the beginning of the contract. He repays theinterest on Ax in advance at the beginning of each year ifsurvives and agrees to repay the loan itself from the unit deathbenefit at the end of the year of death. What is the annualpremium for a full unit of this insurance?
SolutionThe equation is actually
dAx..ax + Ax · Ax =
Ax
ordAx
..ax = (1− Ax )Ax .
Edward Furman Actuarial mathematics 3280 24 / 46
Premium equations
Problem 1.1A person age (x) borrows an amount of money equal to Ax inorder to buy a whole life insurance with unit benefit and a singlepremium payed at the beginning of the contract. He repays theinterest on Ax in advance at the beginning of each year ifsurvives and agrees to repay the loan itself from the unit deathbenefit at the end of the year of death. What is the annualpremium for a full unit of this insurance?
SolutionThe equation is actually
dAx..ax + Ax · Ax = Ax
or
dAx..ax = (1− Ax )Ax .
Edward Furman Actuarial mathematics 3280 24 / 46
Premium equations
Problem 1.1A person age (x) borrows an amount of money equal to Ax inorder to buy a whole life insurance with unit benefit and a singlepremium payed at the beginning of the contract. He repays theinterest on Ax in advance at the beginning of each year ifsurvives and agrees to repay the loan itself from the unit deathbenefit at the end of the year of death. What is the annualpremium for a full unit of this insurance?
SolutionThe equation is actually
dAx..ax + Ax · Ax = Ax
ordAx
..ax = (1− Ax )Ax .
Edward Furman Actuarial mathematics 3280 24 / 46
Premium equations
Figure: Discrete premiums
Edward Furman Actuarial mathematics 3280 25 / 46
Premium equations
Continuous premiums
The same reasoning applies to continuous premiums. Forinstance, when we have a whole life continuous insurance Axand premiums are payable continuously then due to theequivalence principle, we obtain that:
P(Ax ) =Ax
ax
Also
Var[L] = Var[vT − P
1− vT
δ
]=
2Ax − (Ax )2
(δax )2 .
Edward Furman Actuarial mathematics 3280 26 / 46
Premium equations
Continuous premiums
The same reasoning applies to continuous premiums. Forinstance, when we have a whole life continuous insurance Axand premiums are payable continuously then due to theequivalence principle, we obtain that:
P(Ax ) =Ax
ax
Also
Var[L] =
Var[vT − P
1− vT
δ
]=
2Ax − (Ax )2
(δax )2 .
Edward Furman Actuarial mathematics 3280 26 / 46
Premium equations
Continuous premiums
The same reasoning applies to continuous premiums. Forinstance, when we have a whole life continuous insurance Axand premiums are payable continuously then due to theequivalence principle, we obtain that:
P(Ax ) =Ax
ax
Also
Var[L] = Var[vT − P
1− vT
δ
]=
2Ax − (Ax )2
(δax )2 .
Edward Furman Actuarial mathematics 3280 26 / 46
Premium equations
Example 1.4Calculate the premium when the force of mortality is constantµ, and the force of interest is δ.
SolutionWith the above assumptions
Ax =
∫ ∞0
e−δtµe−µtdt
=µ
µ+ δ
∫ ∞0
(µ+ δ)e−(δ+µ)tdt
=µ
µ+ δ.
In our caseAx = 0.04/0.1 = 0.4.
Also,
Edward Furman Actuarial mathematics 3280 27 / 46
Premium equations
Example 1.4Calculate the premium when the force of mortality is constantµ, and the force of interest is δ.
SolutionWith the above assumptions
Ax =
∫ ∞0
e−δtµe−µtdt
=
µ
µ+ δ
∫ ∞0
(µ+ δ)e−(δ+µ)tdt
=µ
µ+ δ.
In our caseAx = 0.04/0.1 = 0.4.
Also,
Edward Furman Actuarial mathematics 3280 27 / 46
Premium equations
Example 1.4Calculate the premium when the force of mortality is constantµ, and the force of interest is δ.
SolutionWith the above assumptions
Ax =
∫ ∞0
e−δtµe−µtdt
=µ
µ+ δ
∫ ∞0
(µ+ δ)e−(δ+µ)tdt
=
µ
µ+ δ.
In our caseAx = 0.04/0.1 = 0.4.
Also,
Edward Furman Actuarial mathematics 3280 27 / 46
Premium equations
Example 1.4Calculate the premium when the force of mortality is constantµ, and the force of interest is δ.
SolutionWith the above assumptions
Ax =
∫ ∞0
e−δtµe−µtdt
=µ
µ+ δ
∫ ∞0
(µ+ δ)e−(δ+µ)tdt
=µ
µ+ δ.
In our caseAx = 0.04/0.1 = 0.4.
Also,
Edward Furman Actuarial mathematics 3280 27 / 46
Premium equations
Solution
ax =
1− Ax
δ=
1− 0.40.06
= 10.
The premium is then:
P(Ax ) = 0.4/10 = 0.04 = µ,
a coincidence? As to the variance, we have that:
2Ax =
∫ ∞0
e−2δtµe−µtdt
=µ
2δ + µ= 0.25.
Thus
Var[L] =25− 0.42
(0.06 · 10)2 = 0.25.
Edward Furman Actuarial mathematics 3280 28 / 46
Premium equations
Solution
ax =1− Ax
δ=
1− 0.40.06
= 10.
The premium is then:
P(Ax ) = 0.4/10 = 0.04 = µ,
a coincidence? As to the variance, we have that:
2Ax =
∫ ∞0
e−2δtµe−µtdt
=µ
2δ + µ= 0.25.
Thus
Var[L] =25− 0.42
(0.06 · 10)2 = 0.25.
Edward Furman Actuarial mathematics 3280 28 / 46
Premium equations
Solution
ax =1− Ax
δ=
1− 0.40.06
= 10.
The premium is then:
P(Ax ) = 0.4/10 = 0.04 = µ,
a coincidence? As to the variance, we have that:
2Ax =
∫ ∞0
e−2δtµe−µtdt
=µ
2δ + µ= 0.25.
Thus
Var[L] =25− 0.42
(0.06 · 10)2 = 0.25.
Edward Furman Actuarial mathematics 3280 28 / 46
Premium equations
Solution
ax =1− Ax
δ=
1− 0.40.06
= 10.
The premium is then:
P(Ax ) = 0.4/10 = 0.04 = µ,
a coincidence? As to the variance, we have that:
2Ax =
∫ ∞0
e−2δtµe−µtdt
=µ
2δ + µ= 0.25.
Thus
Var[L] =25− 0.42
(0.06 · 10)2 = 0.25.
Edward Furman Actuarial mathematics 3280 28 / 46
Premium equations
Solution
ax =1− Ax
δ=
1− 0.40.06
= 10.
The premium is then:
P(Ax ) = 0.4/10 = 0.04 = µ,
a coincidence? As to the variance, we have that:
2Ax =
∫ ∞0
e−2δtµe−µtdt
=
µ
2δ + µ= 0.25.
Thus
Var[L] =25− 0.42
(0.06 · 10)2 = 0.25.
Edward Furman Actuarial mathematics 3280 28 / 46
Premium equations
Solution
ax =1− Ax
δ=
1− 0.40.06
= 10.
The premium is then:
P(Ax ) = 0.4/10 = 0.04 = µ,
a coincidence? As to the variance, we have that:
2Ax =
∫ ∞0
e−2δtµe−µtdt
=µ
2δ + µ= 0.25.
Thus
Var[L] =
25− 0.42
(0.06 · 10)2 = 0.25.
Edward Furman Actuarial mathematics 3280 28 / 46
Premium equations
Solution
ax =1− Ax
δ=
1− 0.40.06
= 10.
The premium is then:
P(Ax ) = 0.4/10 = 0.04 = µ,
a coincidence? As to the variance, we have that:
2Ax =
∫ ∞0
e−2δtµe−µtdt
=µ
2δ + µ= 0.25.
Thus
Var[L] =25− 0.42
(0.06 · 10)2 = 0.25.
Edward Furman Actuarial mathematics 3280 28 / 46
Premium equations
Let us have a closer look at the loss random variable forthe whole insurance contract.
Figure: Loss rv L is a non-increasing function since log(v) < 0
Note that the events {L > 0} and {T < c} are equivalent.
Edward Furman Actuarial mathematics 3280 29 / 46
Premium equations
Let us have a closer look at the loss random variable forthe whole insurance contract.
Figure: Loss rv L is a non-increasing function since log(v) < 0
Note that the events {L > 0} and {T < c} are equivalent.
Edward Furman Actuarial mathematics 3280 29 / 46
Premium equations
Let us have a closer look at the loss random variable forthe whole insurance contract.
Figure: Loss rv L is a non-increasing function since log(v) < 0
Note that the events {L > 0} and {T < c} are
equivalent.
Edward Furman Actuarial mathematics 3280 29 / 46
Premium equations
Let us have a closer look at the loss random variable forthe whole insurance contract.
Figure: Loss rv L is a non-increasing function since log(v) < 0
Note that the events {L > 0} and {T < c} are equivalent.
Edward Furman Actuarial mathematics 3280 29 / 46
Premium equations
If the premiums is chosen such that P[L > 0] = q, thenbecause of the equivalence of {L > 0} and {T < c}, itholds that P[T < c] = q, where c = xq of T .
Thusvxq − P
..axq = 0,
from which
P =vxq
..axq
=1
..axq (1 + i)xq
=1..sxq
.
Intuitive?Of course! With probability q, we have that
..sT /
..sxq < 1,
and with probability 1− q, we have that..sT /
..sxq > 1
Edward Furman Actuarial mathematics 3280 30 / 46
Premium equations
If the premiums is chosen such that P[L > 0] = q, thenbecause of the equivalence of {L > 0} and {T < c}, itholds that P[T < c] = q, where c = xq of T .Thus
vxq − P..axq = 0,
from which
P =vxq
..axq
=
1..axq (1 + i)xq
=1..sxq
.
Intuitive?Of course! With probability q, we have that
..sT /
..sxq < 1,
and with probability 1− q, we have that..sT /
..sxq > 1
Edward Furman Actuarial mathematics 3280 30 / 46
Premium equations
If the premiums is chosen such that P[L > 0] = q, thenbecause of the equivalence of {L > 0} and {T < c}, itholds that P[T < c] = q, where c = xq of T .Thus
vxq − P..axq = 0,
from which
P =vxq
..axq
=1
..axq (1 + i)xq
=
1..sxq
.
Intuitive?Of course! With probability q, we have that
..sT /
..sxq < 1,
and with probability 1− q, we have that..sT /
..sxq > 1
Edward Furman Actuarial mathematics 3280 30 / 46
Premium equations
If the premiums is chosen such that P[L > 0] = q, thenbecause of the equivalence of {L > 0} and {T < c}, itholds that P[T < c] = q, where c = xq of T .Thus
vxq − P..axq = 0,
from which
P =vxq
..axq
=1
..axq (1 + i)xq
=1..sxq
.
Intuitive?
Of course! With probability q, we have that..sT /
..sxq < 1,
and with probability 1− q, we have that..sT /
..sxq > 1
Edward Furman Actuarial mathematics 3280 30 / 46
Premium equations
If the premiums is chosen such that P[L > 0] = q, thenbecause of the equivalence of {L > 0} and {T < c}, itholds that P[T < c] = q, where c = xq of T .Thus
vxq − P..axq = 0,
from which
P =vxq
..axq
=1
..axq (1 + i)xq
=1..sxq
.
Intuitive?Of course!
With probability q, we have that..sT /
..sxq < 1,
and with probability 1− q, we have that..sT /
..sxq > 1
Edward Furman Actuarial mathematics 3280 30 / 46
Premium equations
If the premiums is chosen such that P[L > 0] = q, thenbecause of the equivalence of {L > 0} and {T < c}, itholds that P[T < c] = q, where c = xq of T .Thus
vxq − P..axq = 0,
from which
P =vxq
..axq
=1
..axq (1 + i)xq
=1..sxq
.
Intuitive?Of course! With probability q, we have that
..sT /
..sxq < 1,
and with probability 1− q, we have that..sT /
..sxq > 1
Edward Furman Actuarial mathematics 3280 30 / 46
Premium equations
Figure: The cdf of the loss rv associated with a whole life insurance.The cdf is a function of two variables, i.e., the premium and the loss.
Edward Furman Actuarial mathematics 3280 31 / 46
Premium equations
Example 1.5
Let fT (55)(t) =t p55µ(55 + t) = 1/45, 0 < t < 45. Display thecdf of L and determine the amount P as the smallestnon-negative number for which P[L(P) > 0] ≤ 0.25, in thecontext of
a. 20 year endowment insuranceb. 20 year term insurancec. 10 year term insurance.
Also the force of interest is 0.06.
Solution
We find FL first. Note that FL(u) = 0 for all u < v20 − Pa20 . Theloss is thus bounded from below, i.e., the profit cannot exceedcertain value.
Further for v20 − Pa20 ≤ u ≤ 1, we have that
Edward Furman Actuarial mathematics 3280 32 / 46
Premium equations
Example 1.5
Let fT (55)(t) =t p55µ(55 + t) = 1/45, 0 < t < 45. Display thecdf of L and determine the amount P as the smallestnon-negative number for which P[L(P) > 0] ≤ 0.25, in thecontext of
a. 20 year endowment insuranceb. 20 year term insurancec. 10 year term insurance.
Also the force of interest is 0.06.
Solution
We find FL first. Note that FL(u) = 0 for all u < v20 − Pa20 . Theloss is thus bounded from below, i.e., the profit cannot exceedcertain value. Further for v20 − Pa20 ≤ u ≤ 1, we have that
Edward Furman Actuarial mathematics 3280 32 / 46
Premium equations
Solution
FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]
= P[vT − P
1− vT
δ≤ u
]= P
[vT ≤ δu + P
δ + P
]
= P
[T ≥ −1
δlog
(δu + Pδ + P
)]
= 1− FT
(−1δ
log
(δu + Pδ + P
)),
for all u > −P/δ. Surprised?
Edward Furman Actuarial mathematics 3280 33 / 46
Premium equations
Solution
FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]
= P[vT − P
1− vT
δ≤ u
]
= P
[vT ≤ δu + P
δ + P
]
= P
[T ≥ −1
δlog
(δu + Pδ + P
)]
= 1− FT
(−1δ
log
(δu + Pδ + P
)),
for all u > −P/δ. Surprised?
Edward Furman Actuarial mathematics 3280 33 / 46
Premium equations
Solution
FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]
= P[vT − P
1− vT
δ≤ u
]= P
[vT ≤ δu + P
δ + P
]
= P
[T ≥ −1
δlog
(δu + Pδ + P
)]
= 1− FT
(−1δ
log
(δu + Pδ + P
)),
for all u > −P/δ. Surprised?
Edward Furman Actuarial mathematics 3280 33 / 46
Premium equations
Solution
FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]
= P[vT − P
1− vT
δ≤ u
]= P
[vT ≤ δu + P
δ + P
]
= P
[T ≥ −1
δlog
(δu + Pδ + P
)]
= 1− FT
(−1δ
log
(δu + Pδ + P
)),
for all u > −P/δ. Surprised?
Edward Furman Actuarial mathematics 3280 33 / 46
Premium equations
Solution
FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]
= P[vT − P
1− vT
δ≤ u
]= P
[vT ≤ δu + P
δ + P
]
= P
[T ≥ −1
δlog
(δu + Pδ + P
)]
= 1− FT
(−1δ
log
(δu + Pδ + P
)),
for all u > −P/δ. Surprised?
Edward Furman Actuarial mathematics 3280 33 / 46
Premium equations
Solution
FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]
= P[vT − P
1− vT
δ≤ u
]= P
[vT ≤ δu + P
δ + P
]
= P
[T ≥ −1
δlog
(δu + Pδ + P
)]
= 1− FT
(−1δ
log
(δu + Pδ + P
)),
for all u > −P/δ. Surprised?
Edward Furman Actuarial mathematics 3280 33 / 46
Premium equations
SolutionWe know the cdf of T , i.e., we know that
FT (u) =145
u, 0 < u < 45.
Hence
P[L > 0] =
1− P[L ≤ 0] = FT
(−1δ
log
(P
δ + P
))= 0.25.
Equivalently
1 +1
45
(1
0.06log
(P
0.06 + P
))= 0.75.
Solving for P, we find thatP = 0.06224.
Edward Furman Actuarial mathematics 3280 34 / 46
Premium equations
SolutionWe know the cdf of T , i.e., we know that
FT (u) =145
u, 0 < u < 45.
Hence
P[L > 0] = 1− P[L ≤ 0] =
FT
(−1δ
log
(P
δ + P
))= 0.25.
Equivalently
1 +1
45
(1
0.06log
(P
0.06 + P
))= 0.75.
Solving for P, we find thatP = 0.06224.
Edward Furman Actuarial mathematics 3280 34 / 46
Premium equations
SolutionWe know the cdf of T , i.e., we know that
FT (u) =145
u, 0 < u < 45.
Hence
P[L > 0] = 1− P[L ≤ 0] = FT
(−1δ
log
(P
δ + P
))= 0.25.
Equivalently
1 +1
45
(1
0.06log
(P
0.06 + P
))= 0.75.
Solving for P, we find thatP = 0.06224.
Edward Furman Actuarial mathematics 3280 34 / 46
Premium equations
SolutionWe know the cdf of T , i.e., we know that
FT (u) =145
u, 0 < u < 45.
Hence
P[L > 0] = 1− P[L ≤ 0] = FT
(−1δ
log
(P
δ + P
))= 0.25.
Equivalently
1 +1
45
(1
0.06log
(P
0.06 + P
))= 0.75.
Solving for P, we find thatP = 0.06224.
Edward Furman Actuarial mathematics 3280 34 / 46
Premium equations
SolutionAnother way would be to recall that {L > 0} is equivalent to{T < c}. Thus the premium is
P =1
..sx0.25
,
wherex0.25 = F−1
T (0.25) = 0.25 · 45 = 11.25.
The premium is then again
P = 0.06224.
Going back to the cdf of L, we note that FL(u) for u > 1 is 1.
Edward Furman Actuarial mathematics 3280 35 / 46
Premium equations
SolutionAnother way would be to recall that {L > 0} is equivalent to{T < c}. Thus the premium is
P =1
..sx0.25
,
wherex0.25 = F−1
T (0.25) = 0.25 · 45 = 11.25.
The premium is then again
P = 0.06224.
Going back to the cdf of L, we note that FL(u) for u > 1 is 1.
Edward Furman Actuarial mathematics 3280 35 / 46
Premium equations
Solution
Figure: The cdf of L when the insurance is a 20 year endowment one.Edward Furman Actuarial mathematics 3280 36 / 46
Premium equations
Solutionb. Now we have a 20 year term insurance. This means that theloss rv L can take on values in the interval
[−Pa20 ,1
].
Thus FL(u) = 0 for all u < −Pa20 .We also know that for u > 1 we have that FL(u) = 1.Another interesting point is v20 − Pa20 . However in theinterval
[v20 − Pa20 ,1
]the cdf behaves as the cdf of a
loss of the whole life insurance. So
FL(u) = 1 +145
(1
0.06log
(0.06u + P0.06 + P
))
for u ∈[v20 − Pa20 ,1
].
Edward Furman Actuarial mathematics 3280 37 / 46
Premium equations
Solutionb. Now we have a 20 year term insurance. This means that theloss rv L can take on values in the interval
[−Pa20 ,1
].
Thus FL(u) = 0 for all u
< −Pa20 .We also know that for u > 1 we have that FL(u) = 1.Another interesting point is v20 − Pa20 . However in theinterval
[v20 − Pa20 ,1
]the cdf behaves as the cdf of a
loss of the whole life insurance. So
FL(u) = 1 +145
(1
0.06log
(0.06u + P0.06 + P
))
for u ∈[v20 − Pa20 ,1
].
Edward Furman Actuarial mathematics 3280 37 / 46
Premium equations
Solutionb. Now we have a 20 year term insurance. This means that theloss rv L can take on values in the interval
[−Pa20 ,1
].
Thus FL(u) = 0 for all u < −Pa20 .We also know that for u > 1 we have that FL(u) =
1.Another interesting point is v20 − Pa20 . However in theinterval
[v20 − Pa20 ,1
]the cdf behaves as the cdf of a
loss of the whole life insurance. So
FL(u) = 1 +145
(1
0.06log
(0.06u + P0.06 + P
))
for u ∈[v20 − Pa20 ,1
].
Edward Furman Actuarial mathematics 3280 37 / 46
Premium equations
Solutionb. Now we have a 20 year term insurance. This means that theloss rv L can take on values in the interval
[−Pa20 ,1
].
Thus FL(u) = 0 for all u < −Pa20 .We also know that for u > 1 we have that FL(u) = 1.Another interesting point is v20 − Pa20 . However in theinterval
[v20 − Pa20 ,1
]the cdf behaves as the cdf of a
loss of the whole life insurance. So
FL(u) = 1 +145
(1
0.06log
(0.06u + P0.06 + P
))
for u ∈[v20 − Pa20 ,1
].
Edward Furman Actuarial mathematics 3280 37 / 46
Premium equations
Solution
In the interval u ∈[−Pa20 , v
20 − Pa20
], we have that
FL(u) = P[T > 20] = 1− 20/45 = 25/45.
The premium is the same as in a. since we again have tofind the 0.25 quantile of L, and it again ‘sits’ on theincreasing part of FL(u).
Edward Furman Actuarial mathematics 3280 38 / 46
Premium equations
Solution
In the interval u ∈[−Pa20 , v
20 − Pa20
], we have that
FL(u) = P[T > 20] = 1− 20/45 = 25/45.The premium is the same as in a.
since we again have tofind the 0.25 quantile of L, and it again ‘sits’ on theincreasing part of FL(u).
Edward Furman Actuarial mathematics 3280 38 / 46
Premium equations
Solution
In the interval u ∈[−Pa20 , v
20 − Pa20
], we have that
FL(u) = P[T > 20] = 1− 20/45 = 25/45.The premium is the same as in a. since we again have tofind the 0.25 quantile of L, and it again ‘sits’ on theincreasing part of FL(u).
Edward Furman Actuarial mathematics 3280 38 / 46
Premium equations
Solution
Figure: The cdf of L when the insurance is a 20 year term one.
Edward Furman Actuarial mathematics 3280 39 / 46
Premium equations
Solutionc. We now have a 10 year term insurance. So
FL(u) = 0 for all u < −Pa10 ,
Also, FL(u) = 1 for all u > 1,
In the interval u ∈[−Pa10 , v
10 − Pa10
], we can calculate
the cdf from the probabilityP[T > 10] = 1− P[T ≤ 10] = 1− 10/45 = 35/45 > 0.75.So?As in part b., we have that the cdf is zero first and then wehave a jump to 35/45. This time however the premium isnot P = 0.06224 anymore.
Edward Furman Actuarial mathematics 3280 40 / 46
Premium equations
Solutionc. We now have a 10 year term insurance. So
FL(u) = 0 for all u < −Pa10 ,Also, FL(u) = 1 for all u > 1,
In the interval u ∈[−Pa10 , v
10 − Pa10
],
we can calculatethe cdf from the probabilityP[T > 10] = 1− P[T ≤ 10] = 1− 10/45 = 35/45 > 0.75.So?As in part b., we have that the cdf is zero first and then wehave a jump to 35/45. This time however the premium isnot P = 0.06224 anymore.
Edward Furman Actuarial mathematics 3280 40 / 46
Premium equations
Solutionc. We now have a 10 year term insurance. So
FL(u) = 0 for all u < −Pa10 ,Also, FL(u) = 1 for all u > 1,
In the interval u ∈[−Pa10 , v
10 − Pa10
], we can calculate
the cdf from the probabilityP[T > 10] = 1− P[T ≤ 10] = 1− 10/45 = 35/45 > 0.75.So?
As in part b., we have that the cdf is zero first and then wehave a jump to 35/45. This time however the premium isnot P = 0.06224 anymore.
Edward Furman Actuarial mathematics 3280 40 / 46
Premium equations
Solutionc. We now have a 10 year term insurance. So
FL(u) = 0 for all u < −Pa10 ,Also, FL(u) = 1 for all u > 1,
In the interval u ∈[−Pa10 , v
10 − Pa10
], we can calculate
the cdf from the probabilityP[T > 10] = 1− P[T ≤ 10] = 1− 10/45 = 35/45 > 0.75.So?As in part b., we have that the cdf is zero first and then wehave a jump to 35/45. This time however the premium isnot P = 0.06224 anymore.
Edward Furman Actuarial mathematics 3280 40 / 46
Premium equations
Solution
Remember that we choose P as the smallest value suchthat P[L(P) > 0] ≤ 0.25.
Try the smallest possible premiumP = 0. Then
P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.
Zero premium and non zero benefit. Reasonable? No.Does it mean that the quantile based premium is “bad”?Not at all. Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.
Edward Furman Actuarial mathematics 3280 41 / 46
Premium equations
Solution
Remember that we choose P as the smallest value suchthat P[L(P) > 0] ≤ 0.25. Try the smallest possible premiumP = 0. Then
P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.
Zero premium and non zero benefit. Reasonable?
No.Does it mean that the quantile based premium is “bad”?Not at all. Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.
Edward Furman Actuarial mathematics 3280 41 / 46
Premium equations
Solution
Remember that we choose P as the smallest value suchthat P[L(P) > 0] ≤ 0.25. Try the smallest possible premiumP = 0. Then
P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.
Zero premium and non zero benefit. Reasonable? No.Does it mean that the quantile based premium is “bad”?
Not at all. Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.
Edward Furman Actuarial mathematics 3280 41 / 46
Premium equations
Solution
Remember that we choose P as the smallest value suchthat P[L(P) > 0] ≤ 0.25. Try the smallest possible premiumP = 0. Then
P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.
Zero premium and non zero benefit. Reasonable? No.Does it mean that the quantile based premium is “bad”?Not at all.
Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.
Edward Furman Actuarial mathematics 3280 41 / 46
Premium equations
Solution
Remember that we choose P as the smallest value suchthat P[L(P) > 0] ≤ 0.25. Try the smallest possible premiumP = 0. Then
P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.
Zero premium and non zero benefit. Reasonable? No.Does it mean that the quantile based premium is “bad”?Not at all. Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.
Edward Furman Actuarial mathematics 3280 41 / 46
Premium equations
Solution
Figure: The cdf of L when the insurance is a 10 year term one.
Edward Furman Actuarial mathematics 3280 42 / 46
Premium equations
Figure: Continuous premiums
Edward Furman Actuarial mathematics 3280 43 / 46
Premium equations
Example 1.6Let the utility function of an insurer be
u(x) = x − 0.01x2.
Also, the insurer is known to have the initial wealth w .Determine the annual premium that maximizes the utility above.
Solution
We must find π such that
π = argmaxP
E[u(w + P − X )].
Thus we have a closer look at
∂
∂PE[u(w + P − X )] = E
[∂
∂Pu(w + P − X )
]= 0
Edward Furman Actuarial mathematics 3280 44 / 46
Premium equations
Example 1.6Let the utility function of an insurer be
u(x) = x − 0.01x2.
Also, the insurer is known to have the initial wealth w .Determine the annual premium that maximizes the utility above.
SolutionWe must find π such that
π = argmaxP
E[u(w + P − X )].
Thus we have a closer look at
∂
∂PE[u(w + P − X )] = E
[∂
∂Pu(w + P − X )
]= 0
Edward Furman Actuarial mathematics 3280 44 / 46
Premium equations
Example 1.6Let the utility function of an insurer be
u(x) = x − 0.01x2.
Also, the insurer is known to have the initial wealth w .Determine the annual premium that maximizes the utility above.
SolutionWe must find π such that
π = argmaxP
E[u(w + P − X )].
Thus we have a closer look at
∂
∂PE[u(w + P − X )]
= E[∂
∂Pu(w + P − X )
]= 0
Edward Furman Actuarial mathematics 3280 44 / 46
Premium equations
Example 1.6Let the utility function of an insurer be
u(x) = x − 0.01x2.
Also, the insurer is known to have the initial wealth w .Determine the annual premium that maximizes the utility above.
SolutionWe must find π such that
π = argmaxP
E[u(w + P − X )].
Thus we have a closer look at
∂
∂PE[u(w + P − X )] = E
[∂
∂Pu(w + P − X )
]= 0
Edward Furman Actuarial mathematics 3280 44 / 46
Premium equations
SolutionSubstituting u as in the question leads to
E[∂
∂Pu(w + P − X )
]
= E[∂
∂P
((w + P − X )− 0.01(w + P − X )2
)]= E [1− 0.02(w + P − X )] = 0.
which is equivalent to
0.02E [(w + P − X )] = 1
orE [(w + P − X )] = 50
Edward Furman Actuarial mathematics 3280 45 / 46
Premium equations
SolutionSubstituting u as in the question leads to
E[∂
∂Pu(w + P − X )
]= E
[∂
∂P
((w + P − X )− 0.01(w + P − X )2
)]
= E [1− 0.02(w + P − X )] = 0.
which is equivalent to
0.02E [(w + P − X )] = 1
orE [(w + P − X )] = 50
Edward Furman Actuarial mathematics 3280 45 / 46
Premium equations
SolutionSubstituting u as in the question leads to
E[∂
∂Pu(w + P − X )
]= E
[∂
∂P
((w + P − X )− 0.01(w + P − X )2
)]= E [1− 0.02(w + P − X )] = 0.
which is equivalent to
0.02E [(w + P − X )] = 1
orE [(w + P − X )] = 50
Edward Furman Actuarial mathematics 3280 45 / 46
Premium equations
SolutionSubstituting u as in the question leads to
E[∂
∂Pu(w + P − X )
]= E
[∂
∂P
((w + P − X )− 0.01(w + P − X )2
)]= E [1− 0.02(w + P − X )] = 0.
which is equivalent to
0.02E [(w + P − X )] = 1
orE [(w + P − X )] = 50
Edward Furman Actuarial mathematics 3280 45 / 46
Premium equations
SolutionIn other words
π = 50− w + E[X ].
Edward Furman Actuarial mathematics 3280 46 / 46