A machine has two components and fails when both components fail. The
number of years from now until the first component fails, X, and the
number of years from now until the machine fails, Y , are random
variables with joint density function
$$f(x,y)=\begin{cases}\frac1{18}e^{-(x+y)/6}&\text{if }0<x<y\\
0&\text{otherwise}\end{cases}$$
Calculate $\operatorname{Var}(Y |X = 2)$.
My answer: I can notice that the ftp is a product of two exponentials rv, with $\lambda= \frac16$. And the variance of an exponential is $\frac{1}{\lambda^2}$. Then the variance of this pdf is $1/(1/6)^2=36$. Is that right? The book said it is $36$, but I'm not sure if my argument is correct.
My question: How to calculate $\operatorname{Var}(Y |X = 2)$?
Best Answer
I don´t know if your argument is right. I have used the definition of the conditional variance.
$Var(Y|X)=\int_{-\infty}^{\infty} y^2\cdot h(y|x)\ dy- \left[ \int_{-\infty}^{\infty} y\cdot h(y|x)\ dy\right] ^2$
$h(y|x)=\frac{f(x,y)}{f_X(x)}$
$f_X(x)=\int_x^{\infty} \frac{1}{18}\cdot e^{-(x+y)/6} \ dy=\frac{1}{3}e^{-x/3}$
$h(y|x)=\frac{\frac{1}{18} e^{-(x+y)/6}}{\frac{1}{3}e^{-x/3}}=\frac{1}{6}e^{(x-y)/6}$
$h(y|2)=\frac{1}{6}e^{1/3-y/6}$
$\int_{2}^{\infty} y^2\cdot h(y|2) \ dy=\int_2^{\infty} y^2\cdot 1/6\cdot e^{1/3-y/6} \ dy=100$
$\int_{2}^{\infty} y\cdot h(y|2) \ dy=\int_2^{\infty} y \cdot 1/6\cdot e^{1/3-y/6} \ dy=8$