Reconsider the minicomputer component lifetimes X and Y as described in Exercise 12. Determine E(XY). What can be said about Cov(X, Y) and ρ?

EXERCISE 12
Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y:

f(x, y)= { ^xe^-x(1+y) X>=0 and y>= 0 _0 otherwise

To determine \( E(XY) \), we need to calculate the expectation of the product of the random variables \( X \) and \( Y \) with the given joint probability density function (pdf):

\[
f(x, y) = \begin{cases} 
x e^{-x(1+y)} & \text{if } x \geq 0 \text{ and } y \geq 0, \\
0 & \text{otherwise}.
\end{cases}
\]

### Step 1: Calculate \( E(XY) \)

The expectation \( E(XY) \) is given by the double integral:

\[
E(XY) = \int_0^{\infty} \int_0^{\infty} x y f(x, y) \, dy \, dx.
\]

Substitute \( f(x, y) = x e^{-x(1+y)} \):

\[
E(XY) = \int_0^{\infty} \int_0^{\infty} x y \cdot x e^{-x(1+y)} \, dy \, dx.
\]

Simplifying the integrand:

\[
E(XY) = \int_0^{\infty} \int_0^{\infty} x^2 y e^{-x(1+y)} \, dy \, dx.
\]

### Step 2: Integrate with respect to \( y \)

Let's focus on the inner integral with respect to \( y \):

\[
\int_0^{\infty} y e^{-x(1+y)} \, dy.
\]

To solve this, we can use integration by parts or look up the known integral formula for moments of the exponential function. 

Since the steps involve some complex integrations, I will confirm and compute the final values in a detailed step-by-step solution. 

Once we find \( E(XY) \), we will consider \( \text{Cov}(X, Y) \) and \( \rho(X, Y) \) using

Reconsider the minicomputer component lifetimes X and Y as described in Exercise 12. Determine E(XY). What can be said about Cov(X, Y) and ρ?

Explore how to determine E(XY), Cov(X, Y), and correlation ρ for minicomputer component lifetimes with a given joint PDF. This solution walks through calculating the expected product of component lifetimes and the statistical relationships between them. Suitable for undergraduate students in probability and statistics.

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