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Random Variable (RV)
A function that assigns a numerical value to each outcome of an experiment.
Notation: X, Y, Z, etc
Observed values: x, y, z, etc
Example 1
Two fair coins tossed. Let X = No of Heads
Outcomes Probability Value of X
HH ¼ 2
HT ¼ 1
TH ¼ 1
TT ¼ 0
Random Variable
Discrete RV – finite or countable number of values
Continuous RV – taking values in an interval
Probability Distribution
Probability distribution of a discrete RV described by what is known as a Probability Mass Function (PMF).
Probability distribution of a continuous RV described by what is known as a Probability Density Function (PDF).
Probability Mass Function (PMF)
p(x) = Pr (X = x) satisfying
p(x) ≥ 0 for all x
∑p(x) = 1
Example 1 (Contd)
X = No of Heads. The PMF of X is:
x Pr (X = x)
0 1/4
1 2/4=1/2
2 1/4
Example 1 (Contd)
Probability Density Function (PDF)
f(x) satisfying f(x) ≥ 0 for all x ∫f(x) dx = 1 ∫a
b f(x) dx = Pr (a < X < b) Pr (X = a) = 0
Example 2
Example 3
Expectation
E (X) = ∑x p (x) for a Discrete RV
E (X) = ∫x f (x) dx for a Continuous RV
Expectation
E (X2)= ∑x2 p (x) for a Discrete RV
E (X2) = ∫x2 f (x) dx for a Continuous RV
Expectation
E (g(X)) = ∑g(x) p (x) for a Discrete RV
E (g(X)) = ∫g(x) f (x) dx for a Continuous RV
Variance
Var (X) = E (X2) – (E(X))2
Standard Deviation
SD (X) = √Var (X)
Properties of Expectation
E (c) = c for a constant c
E (c X) = c E (X) for a constant c
E (c X + d) = c E (X) + d for constants c & d
Properties of Variance
Var (c) = 0 for a constant c
Var (c X) = c2 Var (X) for a constant c
Var (c X + d) = c2 Var (X) for constants c & d
Example 1 (Contd)
X = No of Heads. Find the following:
(a) E (X) Ans: 1
(b) E (X2) Ans: 1.5
(c) E ((X+10)2) Ans: 121.5
(d) E (2X) Ans: 2.25
(e) Var (X) Ans: 0.5
(f) SD (X) Ans: 1/√2
Example 4
If X is a random variable with the probability density function
f (x) = 2 (1 - x) for 0 < x < 1
find the following:
(a) E (X) Ans: 1/3(b) E (X2) Ans: 1/6(c) E ((X+10)2) Ans: 106.8333(d) Var (X) Ans: 1/18(e) SD (X) Ans: 1/(3√2)
Example 5
An urn contains 4 balls numbered 1, 2, 3 & 4. Let X denote the number that occurs if one ball is drawn at random from the urn. What is the PMF of X?
Example 5 (Contd)
Two balls are drawn from the urn without replacement. Let X be the sum of the two numbers that occur. Derive the PMF of X.
Example 6
The church lottery is going to give away a £3,000 car and 10,000 tickets at £1 a piece.
(a) If you buy 1 ticket, what is your expected gain. (Ans: -0.7)
(b) What is your expected gain if you buy 100 tickets? (Ans: -70)
(c) Compute the variance of your gain in these two instances. (Ans: 899.91 & 89100)
Example 7
A box contains 20 items, 4 of them are defective. Two items are chosen without replacement. Let X = No of defective items chosen. Find the PMF of X.
Example 8
You throw two fair dice, one green and one red.Find the PMF of X if X is defined as:
A) Sum of the two numbersB) Difference of the two numbersC) Minimum of the two numbersD) Maximum of the two numbers
Example 9
If X has the PMF p (x) = ¼ for x = 2, 4, 8, 16
compute the following:
(a) E (X) Ans: 7.5
(b) E (X2) Ans: 85
(c) E (1/X) Ans: 15/64
(d) E (2X/2) Ans: 139/2
(e) Var (X) Ans: 115/4
(f) SD (X) Ans: √115/2
Example 10
If X is a random variable with the probability density function
f (x) = 10 exp (-10 x) for x > 0
find the following:
(a) E (X) Ans: 0.1(b) E (X2) Ans: 0.02(c) E ((X+10)2) Ans: 102.02(d) Var (X) Ans: 0.01(e) SD (X) Ans: 0.1
Example 11
If X is a random variable with the probability density function
f (x) = (1/√(2)) exp (-0.5 x2) for - < x <
find the following:
(a) E (X) Ans: 0(b) E (X2) Ans: 1(c) E ((X+10)2) Ans: 101(d) Var (X) Ans: 1(e) SD (X) Ans: 1
Example 12
A game is played where a person pays to roll two fair six-sided dice. If exactly one six is shown uppermost, the player wins £5. If exactly 2 sixes are shown uppermost, then the player wins £20. How much should be charged to play this game is the player is to break-even?
Example 13
Mr. Smith buys a £4000 insurance policy on his son’s violin. The premium is £50 per year. If the probability that the violin will need to be replaced is 0.8%, what is the insurance company’s gain (if any) on this policy?