Question

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

Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y:
\(f(x,y)=\begin{cases}xe^{-x(1+y)} & x\geq0\ and\ \geq0\\ 0 & otherwise \end{cases}\)
a) What is the probability that the lifetime X of the first component exceeds 3?
b) What are the marginal pdf's of X and Y? Are the two lifetimes independent? Explain.
c) What is the probability that the lifetime of at least one component exceeds 3?

Answers (1)

2021-05-04

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:
\(f(x,y)=\begin{cases}xe^{-x(1+y)} & x\geq0\ and\ \geq0\\ 0 & otherwise \end{cases}\)
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 \(\infty\) as the first component exceeds 3.
So, the required probability is obtained as:
\(P(x>3)=\int_{3}^{\infty}\int_{0}^{\infty}xe^{-x(1+y)}dydx\)
\(=\int_{3}^{\infty}([-e^{-x(1+y)}]_{0}^{\infty})dx\)
\(=\int_{3}^{\infty}e^{x}dx\)
\(=\left[\frac{e^{-x}}{-1}\right]_{3}^{\infty}\)
\(=0.049787\)
\(\approx0.05\)
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:
\(f_{x}(x)=\int_{\infty}^{\infty}f(x,y)dy \ for \ -\infty\)
\(=\int_{0}^{\infty}xe^{-x(1+y)}dy\)
\(=[-e^{-x(1+y)}]_{0}^{\infty}\)
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:
\(f_{y}(y)=\int_{0}^{\infty}xe^{-x(1+y)}dx\)
\(=\left[\frac{xe^{-x(1+y)}}{-(1+y)}\right]_{0}^{\infty}-\int_{0}^{\infty}\frac{e^{-x(1+y)}}{-(1+y)}dx\)
\(=\left[\frac{((y-1)x+1)e^{-yx-z}}{y^{2}+2y-1}\right]_{0}^{\infty}\)
\(=\frac{1}{(1+y)^{2}}\)
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.
\(f(x,y)=f_{x}(x)\times f_{y}(y)\)
\(xe^{-x(1+y)}\neq(e^{-x})\left(\frac{1}{(1+y)^{2}}\right)\)
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:
\(P(X>3\ or\ Y>3)=1-P(X,Y\leq3)\)
\(=1-\int_{0}^{3}\int_{0}^{3}xe^{-x(1+y)}dydx\)
\(=1-\int_{0}^{3}([-e^{-x(1+y)}]_{0}^{3})dx\)
\(=1\int_{0}^{3}(e^{-4x}(e^{3x}-1))dx\)
The probability can be simplified as:
\(P(X>3\ or\ Y>3)=1-\left[\frac{e^{-4x}}{4}-e${-x}\right]_{0}^{3}\)
\(=1-0.700\)
\(=0.3000\)
The probability indicates that there is \(30\%\) chance that the lifetime of at least one of the components will exceed 3.

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