The joint probability density function for two continuous random variables X and Y is $f ( x , y ) = \left\{ \begin{array} { l l }
\frac { 1 } { 6 } e ^ { - x / 6 } e ^ { - y } & \text { if } x \geq 0 \text { and } y \geq 0 \
0 & \text { otherwise }
\end{array} \right.$ Find the expected values $E ( X ) \text { and } E ( Y )$
A) $E ( X ) = 6 ; E ( Y ) = 1$
B) $E ( X ) = 1 ; E ( Y ) = 5$
C) $E ( X ) = 1 ; E ( Y ) = 6$
D) $E ( X ) = 7 ; E ( Y ) = 1$

Question

The joint probability density function for two continuous random variables X and Y is $f ( x , y ) = \left\{ \begin{array} { l l } 
\frac { 1 } { 6 } e ^ { - x / 6 } e ^ { - y } & \text { if } x \geq 0 \text { and } y \geq 0 \
0 & \text { otherwise }
\end{array} \right.$ Find the expected values $E ( X ) \text { and } E ( Y )$
A) $E ( X ) = 6 ; E ( Y ) = 1$
B) $E ( X ) = 1 ; E ( Y ) = 5$
C) $E ( X ) = 1 ; E ( Y ) = 6$
D) $E ( X ) = 7 ; E ( Y ) = 1$

Quizplus · Accepted Answer

The expected value $E(X)$ is calculated by integrating $x \cdot f(x, y)$ over all $x$ and $y$, and $E(Y)$ is calculated by integrating $y \cdot f(x, y)$ over all $x$ and $y$. Given $f(x, y) = \frac{1}{6}e^{-x/6}e^{-y}$ for $x \geq 0$ and $y \geq 0$, we can separate the function into two parts: $f(x) = \frac{1}{6}e^{-x/6}$ for $x$, and $f(y) = e^{-y}$ for $y$. Integrating $x \cdot f(x)$ from 0 to infinity gives $E(X) = 6$, and integrating $y \cdot f(y)$ from 0 to infinity gives $E(Y) = 1$, because the expected value of an exponential distribution with rate $\lambda$ is $1/\lambda$, and here $\lambda = 1/6$ for $X$ and $\lambda = 1$ for $Y$.

The Joint Probability Density Function for Two Continuous Random Variables