The joint probability density for $X$ and $Y$ is
$$ f_{XY}(x,y)=\begin{cases}
2e^{-(x+2y)}, & x>0,y>0\\
0, & \text{otherwise}\\
\end{cases}$$
Calculate the variance of $Y$ given that $X>3$ and $Y>3$.
Correct answer: 0.25
My work:$\def\Var{\operatorname{Var}}$
$\Var(Y\mid Y>3, X>3) = E[Y^2\mid Y>3, X>3] – E[Y\mid Y>3, X>3]^2$
We know $$E[Y\mid Y>3, X>3] = \int_\infty^\infty yf_{X\mid Y}(y\mid Y>3, X>3)dy$$
$\int_3^\infty y\frac{f_{X,Y}(y, X>3)}{f_{X,Y}(X>3, Y>3)}dy$?
That denominator is usually just the marginal of one RV, but in this case it is a function of $X$ and $Y$ so s it a joint? Doesn't it cancel out with joint pdf above when integrated?
Best Answer
Just use the definition for conditioning expectation of a random variable $Z$ over an event $\mathcal A$. $~\mathsf E(Z\mid\mathcal A)=\left.{\mathsf E(Z~\mathbf 1_\mathcal A)}\middle/{\mathsf E(\mathbf 1_\mathcal A)}\right.$ (Where the indicator random variable, $\mathbf 1_\mathcal A$, equals $1$ where the event occurs and $0$ otherwise.)
Thus, use a double integral...
$$\begin{align}\mathsf E(Y^n\mid X>3, Y>4) &= \dfrac{\mathsf E(Y^n\mathbf 1_{X>3, Y>3)})}{\mathsf E(\mathbf 1_{X>3, Y>3)})}\\[1ex]&=\dfrac{\displaystyle\int_3^\infty\int_3^\infty 2y^n\,\mathrm e^{-(x+2y)}\,\mathrm d x\,\mathrm d y}{\displaystyle\int_3^\infty\int_3^\infty 2\,\mathrm e^{-(x+2y)}\,\mathrm d x\,\mathrm d y}\end{align}$$