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6. and 7. Continuous Random Variables and Special Continuous Distributions
β’ Definition 6.1 on P.243:
Let X be a random variable. Suppose that there exists a nonnegative real-values function π: π β 0,β such that for any subset of real numbers A that can be constructed from intervals by a countable number of set operations,
π π β π΄ = π΄
π π₯ ππ₯.
Then X is called absolutely continuous, or for simplicity, continuous. The f is called the probability density function (p.d.f., or pdf) of X. F(t)= ββ
π‘π π₯ ππ₯ is called the (cumulative)
distribution function (c.d.f., or cdf) of X.
6.1 Probability Density Function
6.2 Density Function of a Function of a Random Variable
6.3 Expectations and Variances
1
6. Continuous Random Variables
β’ Properties of p.d.f. (pdf) and c.d.f. (cdf):
The probability density function (p.d.f.) f and its distribution function F of a continuous random variable X has the following properties.
(a) πΉ π‘ = ββπ‘π(π₯) ππ₯. π΄ = (ββ, π‘], πΉ π‘ = π π β€ π‘ = π π β π΄ .
(b) π π β π = βββπ π₯ ππ₯ = 1.
(c) If f is continuous, then πΉβ² π₯ = π π₯ .
(d) For π β€ π, π π β€ π β€ π = πππ π₯ ππ₯.
(e) π π < π < π = π π β€ π < π = π π < π β€ π = π π β€ π β€ π .
2
π π₯ =1
2πβπ₯/2 and πΉ π₯ = 1 β πβπ₯/2 (pdf vs. cdf)
π π₯ =π
2sin ππ₯ vs. πΉ π₯ =
1
2(1 β cos(ππ₯))
Example 6.1 on P.245
Example 6.1 Experience has shown that while walking in a certain park, the time X, in minutes, between seeing two people smoking has a probability density function of the form
β’ f(x)= ππ₯πβπ₯
0,
π₯ > 0ππ‘βπππ€ππ π
(a) Calculate the value π. Ans: π = 1.
(b) Find the distribution function of X. πΉ π‘ = 1 β π‘ + 1 πβπ‘ ππ π‘ β₯ 0.
(c) What is the probability that Jeff, who has just seen a person smoking, will see another person smoking in 2 to 5 minutes? in at least 7 minutes? P(2<X<5)β 0.37, π(π β₯ 7) β 8πβ7 β 0.007.
5
Example 6.2 on P.246
(6.2)(a) Sketch the graph of the function
π π₯ = 1
2β1
4π₯ β 3 , 1 β€ π₯ β€ 5
0 ππ‘βπππ€ππ π
and show that it is the probability density function of a r.v. X.
(b) Find F, the distribution function of X, and sketch its graph.
(c*) Show that F is continuous.
Remark 6.1 on P.249.
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Exercise 3 on P.250
3. The time it takes for a student to finish an aptitude test (in hours) has a probability density function of the form
π π₯ = π π₯ β 1 2 β π₯ ππ 1 < π₯ < 2
(a) Determine the constant c. Ans: 12π π₯ ππ₯ = 1, π π π = 6.
(b) Calculate the distribution function of the time it takes for a randomly selected student to finish the aptitude test.
Ans: πΉ π₯ = 1
π₯π π‘ ππ‘ = β2π₯3 + 9π₯2 β 12π₯ + 5 ππ 1 < π₯ < 2,
πππ πΉ π₯ = 1 ππ π₯ β₯ 2.
(a) What is the probability that a student will finish the aptitude test in less that 75 minutes (i.e., 1.25 hours)? Between 1.5 and 2 hours?
2. πΉ π₯ = 1 β16
π₯2ππ π₯ β₯ 4, πππ 0 ππ‘βπππ€ππ π. Then
(a) π(π₯) =32
π₯3πππ π₯ β₯ 4 π ππππ‘πβ πΉ πππ π πππ π₯ β₯ 4.
π π 5 < π < 7 = πΉ 7 β πΉ(5), d π π β₯ 6 = 1 β πΉ 6 =16
36=
4
9
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Density Function of A Function of A R.V. (P.253)
In probability applications, we may face that the probability density function f of X is known but the p.d.f. h(X) is need. One of the solutions is to find the distribution function G of Y=h(X) first and compute the p.d.f of h(X) by g(y)=Gβ(y).Example (6.3) Let X be a continuous r.v. with the probability density function
π π₯ = 2
π₯2, ππ 1 < π₯ < 2
0, ππ‘βπππ€ππ π.
The distribution function and p.d.f. of π = π2 can be calculated as
πΊ π¦ = π π β€ π¦ = π π2 β€ π¦ = π 1 < π β€ π¦ = 1π¦ 2
π₯2ππ₯ = 2 β
2
π¦,
π π¦ = πΊβ² π¦ =1
π¦ π¦for 1 < π¦ < 4
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Examples 6.4, 6.5 on P.254
(6.4) Let X be a continuous r.v. with p.d.f. f. In terms of f, find the distribution and probability density functions of π = π3.
Hint: πΊ π‘ = π π β€ π‘ = π π3 β€ π‘ = π π β€ 3 π‘ = πΉ(3 π‘), π π‘ = πΊβ² π‘ .
(6.5) The error X of a measurement has the probability density function
π π₯ = 1
2ππ β 1 < π₯ < 1
0 ππ‘βπππ€ππ π
Find the distribution and probability density functions of Y=|X|.
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Theorem 6.1 Method of Transformations P.255
Let X be a continuous random variable with probability density function ππ and the set of possible values A. For the invertible function
β: π΄ β π , let π = β π be a random variable with the set of possible values B=h(A)= β π : π β π΄ . Suppose that the inverse of π¦ = β π₯ is the function π₯ = ββ1 π¦ , which is differentiable for all values of π¦ β π΅.Then ππ, the probability density function of Y, is given by
ππ π¦ = ππ ββ1 π¦ ββ1 β² π¦ , π¦ β π΅.
See the Proof on P.255
Hint: Compute πΉ π¦ = π π = β π β€ π¦ , π‘βππ ππ π¦ = πΉβ² π¦ .
10
Example 6.6 on P.256
6.6 Let X be a random variable with the probability density function
ππ π₯ = 2πβ2π₯ ππ π₯ > 0
0 ππ‘βπππ€ππ π
The probability density function of π = π is f y = 4yπβ2π¦2,π¦ > 0
Hint: π΄ = 0,β , π΅ = β π΄ = (0,β)
6.7 Let X be a random variable with the probability function
ππ π₯ = 4π₯3 ππ 0 < π₯ < 10 ππ‘βπππ€ππ π
The probability density function of π = 1 β 3π2 is f y =2
91 β π¦ , π¦ β (β2,1)
Hint: A=(0,1), and B=h(A)=(-2,1).
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6.3 Expectations and Variances on P.259-264
Definition 6.2 If X is a continuous random variable with probability density function f, the expected value of X is defined as
πΈ π = βββπ₯π π₯ ππ₯. π or E[X] is an alternative notation.
Definition 6.3 If X is a continuous random variable with πΈ π = π, then Var(X) and ππ, called variance and standard deviation of X, respectively, are defined by
Var π = πΈ[ π β π)2 = βββ
π₯ β π 2π π₯ ππ₯. ππ = πππ π .
12
Examples of Expectation and Variance
(6.8) In a group of adult males, the difference between uric acid and 6, the standard value, is a random variable X with the following p.d.f.
π π₯ =27
4903π₯2 β 2π₯ ππ
2
3< π₯ < 3, π π₯ = 0 ππ‘βπππ€ππ π.
Find the mean of these differences for the group, that is, E(X). Ans: 2.36
(6.9) A random variable X with the following p.d.f. is called a Cauchy r.v.
π π₯ =π
1+π₯2, ββ < π₯ < β
(a) Find c. c=1/π.
(b) Show that E(X) does not exist.
13
Theorems 6.2, 6.3, Corollary, P.261-264 (Skip)
Theorem 6.2 πΈ π = 0β[1 β πΉ π‘ ] ππ‘ β 0
βπΉ βπ‘ ππ‘
Remark 6.4
Theorem 6.3 πΈ β π = ββββ π₯ π(π₯) ππ₯
Example 6.10 A point X is selected from (0,π/4) randomly. Calculate
E(cos2X)=2
π, and E(πππ 2π)=
1
2+
1
π.
Remarks 6.5, 6.6., 6.7: About πΈ(ππ+1)
14
πΈ(ππ): The ππ‘β ππππππ‘ ππ π
β’ Let X be a discrete r.v. with probability mass function f π₯ , then
β’ πΈ ππ = βββπ₯ππ(π₯) ππ₯ is called the ππ‘β moment of X if it exists.
β’ When π = 1, πΈ π ππ π‘βπ ππ₯ππππ‘ππ π£πππ’π ππ ππ₯ππππ‘ππ‘πππ.
β’ When π = 2, πΈ π2 ππ ππππππ π‘βπ π πππππ ππππππ‘.
β’ Note that πππ π = πΈ π2 β (πΈ[π])2
β’ Remark: πΈ(ππ+1) exits implies that πΈ ππ . See proof on p.184.
A6. Let X be a discrete r.v. with the set of possible values {π₯1, π₯2, β― , π₯π}; X is called uniform if π π = π₯π =
1
π, 1 β€ π β€ π.
E(X)=(π + 1)/2 and Var(X)=(π2 β 1)/12 for the case that π₯π = π, 1 β€ π β€ π.
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Exercises A1,A6,A7,A10, B13,B14 on P.267-270
(A1) πΉ π₯ = 1 β 16/π₯2 if π₯ β₯ 4, πΉ π₯ = 0 ππ π₯ < 4. Then
(a) E(X)=8, (b) Show that V(X) does not exist.
(A6) π π₯ = 3πβ3π₯ ππ 0 β€ π₯ < β. Then πΈ ππ =3
2.
(A7) π π₯ =1
π 1βπ₯2ππ β 1 < π₯ < 1. Then E(X)=0.
(A10) π π₯ =2
π₯2ππ 1 < π₯ < 2. Then E lnX = 1 β ππ2.
(B13) π π₯ =1
π(1+π₯2), ππ β β < π₯ < β.
Prove that πΈ(|π|πΌ) converges if 0 < πΌ < 1 and diverges if πΌ β₯ 1.
(B14) π π > π‘ = πΌπβππ‘ + π½πβππ‘ , π‘ β₯ 0, πΌ + π½ = 1. Then E(X)=πΌ
π+
π½
π.
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