Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x,y)=\begin{cases}xe^{-x(1+y)} & x\geq0\ and\ \geq

Armorikam

Armorikam

Answered question

2021-05-03

Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y:
f(x,y)={xex(1+y)x0 and 00otherwise
a) What is the probability that the lifetime X of the first component exceeds 3?
b) What are the marginal pdfs

Answer & Explanation

Adnaan Franks

Adnaan Franks

Skilled2021-05-04Added 92 answers

Step 1
Consider two random variables X and Y which indicates the lifetime of the two components of the minicomputers.
The joint probability density function is provided as:
f(x,y)={xex(1+y)x0 and 00otherwise
The required probability does not depend on Y values. Hence consider all the possible values of Y and the values of the X ranges from 3 to as the first component exceeds 3.
So, the required probability is obtained as:
P(x>3)=30xex(1+y)dydx
=3([ex(1+y)]0)dx
=3exdx
=[ex1]3
=0.049787
0.05
The probability indicates that there is 5% chance that the lifetime X of the first component exceeds 3.
Step 2
Substitute the provided probability density function and integrates with respect to y to obtain the marginal probability density function of X.
Hence, the marginal distribution of X can be obtained as:
fx(x)=f(x,y)dy for 
=0xex(1+y)dy
=[ex(1+y)]0
Substitute the given probability density function and integrates with respect to x to obtain the marginal probability density function of Y.
Hence, the marginal distribution of Y can be obtained as:
fy(y)=0xex(1+y)dx
=[xex(1+y)(1+y)]00ex(1+y)(1+y)dx
=[((y1)x+1)eyxzy2+2y1]0
=1(1+y)2
The marginal probability distribution of the random variable X and Y indicates the individual probability of occurrence of the random variable without considering the occurrence of the other variable.
Substitute the obtained marginal probability density functions and the given joint probability density function in the condition for independence to find whether the two lifetimes are independent.
f(x,y)=fx(x)×fy(y)
xex(1+y)(ex)(1(1+y)2)
Since the product of marginal probability density function is not same as the given joint probability density function, the two lifetimes are not independent.
As the joint probability distribution of the random variables does not re[presented in the form of the product of two marginal distribution so the random variable are not independent of each other that is the occurrence of one event effects the occurrence of other.
Step 3
The probability that the lifetime of at least one component exceeds 3 is given by:
P(X>3 or Y>3)=1P(X,Y3)
=10303xex(1+y)dydx
=103([ex(1+y)]03)dx
=103(e4x(e3x1))dx
The probability can be simplified as:
P(X>3 or Y>3)=1[e4x4e$x]03
=10.700
=0.3000
The probability indicates that there is 30% chance that the lifetime of at least one of the components will exceed 3.

Jeffrey Jordon

Jeffrey Jordon

Expert2023-04-30Added 2605 answers

a) The probability that the lifetime X of the first component exceeds 3 is given by:
P(X>3)=30f(x,y)dydx
=30xex(1+y)dydx
=3[xex(1+y)x]y=0y=dx
=30(xex)dx
=3xexdx
We can use integration by parts to solve the integral:
3xexdx=[xex]3+31exdx
=limt[xex]3t+[ex]3
=3e3+e3
0.021
Therefore, the probability that the lifetime X of the first component exceeds 3 is approximately 0.021.
b) To find the marginal pdfs, we integrate the joint pdf f(x,y) over the other variable. First, we integrate over y to get the marginal pdf of X:
fX(x)=0f(x,y)dy
=0xex(1+y)dy
=[ex(1+y)x]y=0y=
=1x2[ex+1]
=1x21x2ex
Next, we integrate over x to get the marginal pdf of Y:
fY(y)=0f(x,y)dx
=0xex(1+y)dx
=1(1+y)2
Therefore, the marginal pdf of X is fX(x)=1x21x2ex, and the marginal pdf of Y is fY(y)=1(1+y)2.
RizerMix

RizerMix

Expert2023-04-30Added 656 answers

a) To find the probability that the lifetime X of the first component exceeds 3, we need to integrate the joint pdf over all values of Y for which X is greater than 3:
P(X>3)=30f(x,y) dy dx
=30xex(1+y) dy dx
=3[xex(1+y)x]y=0y= dx
=3(0xex) dx
=[xex2]x=3x=
=3e32
Therefore, the probability that the lifetime X of the first component exceeds 3 is 3e32.
b) To find the marginal pdf of X, we need to integrate the joint pdf over all values of Y:
fX(x)=0f(x,y) dy
=0xex(1+y) dy
=[ex(1+y)x]y=0y=
=1x
Therefore, the marginal pdf of X is fX(x)=1x for x0.
To find the marginal pdf of Y, we need to integrate the joint pdf over all values of X:
fY(y)=0f(x,y) dx
=0xex(1+y) dx
=1(1+y)2
Therefore, the marginal pdf of Y is fY(y)=1(1+y)2 for y0.

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