Probability Homework Solution: Detailed Answers and Explanations

This document presents a comprehensive solution to a probability assignment, covering a range of concepts from basic probability calculations to more advanced topics such as the gamma and Poisson distributions. The solution includes detailed answers to problems involving normal distributions, expected values, and variance. It also provides step-by-step solutions for calculating probabilities, expectations, and the moment-generating function (mgf) for various distributions. Furthermore, the assignment explores continuous random variables, integration by parts, and proofs related to the properties of distributions. This assignment is a valuable resource for students studying probability and statistics, offering clear explanations and methodologies for solving complex problems. It covers topics from calculating probabilities of radiation exposure to proving the mgf of a gamma distribution and calculating its mean and variance.

Probability
Student Name:
Instructor Name:
Course Number:
7th September 2019

Question 1:
Suppose that the amount of cosmic radiation to which a person is expected while flying from Sydney
to perth is a random variable having a normal distribution with mean μ = 4.35 mrem and standard
deviation σ = 0.59 mrem (here mrem is a unit of radiaton exposure). Calculate the probabilities that
a person will be exposed to:
(a) Less than 3.35 mrem of radiation
Answer
P ( X <3.35 ) =P ( X−μ
σ < 3.35−4.35
0.59 )
¿ P (Z<−1.00
0.59 )
¿ P ( Z ←1.6949 )
¿ 0.0450472
P ( X <3.35 ) =0.0450
(b) More than 5.25 mrem of radiation
Answer
P ( X >5.25 ) =1−P ( X−μ
σ < 5.25−4.35
0.59 )
¿ 1−P (Z < 0.9
0.59 )
¿ 1−P ( Z<1.52542 )
¿ 1−0.9364228
¿ 0.0635772
P ( X >5.25 )=0.0636
(c) Anywhere from 3.5 to 4.5 mrem of radiation
Answer
P ( 3.5<X <4.5 ) =P ( 3.5−4.35
0.59 < X−μ
σ < 4.5−4.35
0.59 )
¿ P ( −0.85
0.59 <Z < 0.15
0.59 )
¿ P (−1.4407< Z< 0.2542 )
¿ P ( Z <0.2542 ) −P ( Z←1.4407 )
¿ 0.6003295−0.0748347
¿ 0.5254948
P ( 3.5< X < 4.5 )=0.5255

Question 2:
A player tosses a coin having the probability 2
5 of coming up heads. The game terminates
when the player tosses a tail. If the first toss is a tail the player wins nothing. Otherwise, the
player wins 2n−1 dollars if n successive heads have been tossed. Let X denote the total
number of dollars won in any particular game. Calculate E(X). How much should a gambler
be willing to pay to play the game?
Answer
P( H )= 2
5
P ( T ) =1− 2
5 =3
5
E ( X ) =∑
i=1
n
xi pi= 2n−1∗2
5
The gambler should be willing to pay 2n−1 dollars to play the game.
Question 3:
Let X have a Poisson distribution with parameter λ. Calculate the expectation of the random
variable
Y = X ( X −1)( X−2)
Answer
E ( Y )=E ( X ( X−1 ) ( X −2 ) )=∑
x=0
∞
(x )(x−1)(x −2) e−λ λx
x !
¿ ∑
x=3
∞
( x)( x−1)(x−2) e− λ λx
x ! because x =0 , x =1∧x=2 terms are themselves 0
¿ ∑
x=3
∞ e− λ λx
( x−3 ) ! divide out by x , x−1∧x−2
¿ ∑
x=3
∞ e− λ λx
( x−3 ) ! divide out by x , x−1∧x−2
¿ λ3 e− λ
∑
x=3
∞ λx−3
( x−3 ) ! factor out e− λ∧λ3

¿ λ3 e− λ
( λ0
0 ! + λ1
1! + λ2
2! + λ3
3 ! +… )
¿ λ3 e− λ eλ
¿ λ3
Question 4:
Let X be a continuous random variable with density function
f X ( x ) = { 3
( 2+ x ) 2 , 0< x <4 ,
0 ,Otherwise .
Calculate E( X)and Var ( X). You may need to use integration by parts
Answer
E ( X ) =∫
0
4
x f ( x ) dx
¿∫
0
4
x 3
( 2+ x ) 2 dx
¿∫
0
4
3 x
( 2+ x )2 dx
¿ 3∫
0
4
x
( 2+ x ) 2 dx
Now lets solve∫
0
4
x
( 2+ x )2 dx first
let u=2+ x → du
dx =1
¿∫
0
4
u−2
u2 du
Expanding yields ;
¿∫
0
4
( 1
u − 2
u2 )du
¿∫
0
4
1
u du−2∫
0
4
1
u2 du

∫
0
4
1
u du=ln u
2∫
0
4
1
u2 = 2
u
So we have;
ln u+ 2
u
Substituting yields;
ln (x+2)+ 2
x+2
So we have;
[3 ln ( x+2 ) + 6
x+ 2 ]0
4
¿ 3 (ln ( 6 )−ln (2 )− 2
3 )
¿ 3 ln ( 6 ) −3 ln ( 2 ) −2
≈ 1.2958
Var ( X ) =∫
0
4
x2 f ( x ) dx− [ E(X ) ] 2
∫
0
4
x2 f ( x ) dx
¿∫
0
4
x2 3
( 2+ x )2 dx
¿ 3∫
0
4
x2
( 2+ x ) 2 dx
let u=2+ x → du
dx =1
¿∫
0
4 ( u−2 ) 2
u2 du
Solvingthe integral yields ;
¿−12 ln ( x +2 )− 12
x +2 +3 x
¿ [−12 ln ( x +2 )− 12
x+2 +3 x ]0
4

¿ 3 (−4 ln ( 6 ) +4 ln ( 2 )+ 16
3 )
¿−12 ln (6 ) +4 ln ( 2 )+ 16
≈ 2.8167
Var ( X )=2.8167−1.29582
¿ 2.8167−1.6792
¿ 1.1375
Question 5:
Let X be a continuous random variable whose pdf f X ( x ) is a gamma distribution with
parameters α and λ.
(a) Prove that the mgf of X is
ψ X ( t )=( λ
λ−t )α
,−∞<t <λ
Answer
mgf =M X ( t ) =E ( etX ) =∫
0
∞
etX f X ( x)dx
We 1st take t <λ
Now we have;
M X ( t ) = λα
Γ ( α ) ∫
0
∞
xα −1 e− ( λ−t ) x dx
¿ λα
Γ ( α ) ∫
0
∞
( u
λ−t )α −1
e−u du
λ−t substituting ( λ−t ) x=u
¿ λα
Γ ( α ) ( λ−t )α ∫
0
∞
uα−1 e−u du
¿ λα
Γ ( α ) ( λ−t )α Γ ( α ) defn of Gamma distribution
¿ ( λ
λ−t )
α
(b) Use ψ X ( t ) to find the formula for E ( X k ).
Answer
Since

ψ X ( t ) =( λ
λ−t )
α
But we know that ψX ( t ) =E ( etX )
SoE ( X k ) would be;
E ( X k )=1− ( λ
λ−t ) α
(c) Calculate the mean and variance of X.
Answer
E ( X ) =∫
−∞
∞
x f ( x ) dx
¿∫
0
∞
x xα −1 e
− x
λ
Γ ( α ) λα dx
¿ Γ ( α+1 ) λα+1
Γ ( α ) λα ∫
0
∞
x(α +1)−1 e
− x
λ
Γ ( α+1 ) λα+ 1 dx
¿ α Γ ( α ) λα +1
Γ ( α ) λα
¿ αλ
To obtain the variance, Var ( X ), we know that;
M ( t )= ( λ
λ−t )α
= ( 1−λ t )−α
It therefore follows that
M ' ( t ) = ( −α ) ( 1−λt )−α −1 ( − λ ) =αλ ( 1− λt ) −α−1
SoM ' ( t )=αλ
M ' ' ( t )=αλ (−α−1 ) (1−λt )−α−2 (−λ )
¿ α λ2 (α +1) ( 1−λt )−α −2
E ( X2 ) =M ' ' ( 0 )
M ' ' ( 0 )=α λ2 ( α +1 )=α 2 λ2+α λ2
V ar ( X )=E ( X2 )− [ E ( X ) ]2
¿ α 2 λ2+ α λ2−α2 λ2
¿ α λ2