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?


2 Answers 2


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...

$$\begin{align}\mathsf E(Y^n\mid X>3, Y>4) &= \dfrac{\mathsf E(Y^n\mathbf 1_{X>3, Y>3)})}{\mathsf E(\mathbf 1_{X>3, Y>3)})}\\[1ex]&=\dfrac{\displaystyle\int_3^\infty\int_3^\infty 2y^n\,\mathrm e^{-(x+2y)}\,\mathrm d x\,\mathrm d y}{\displaystyle\int_3^\infty\int_3^\infty 2\,\mathrm e^{-(x+2y)}\,\mathrm d x\,\mathrm d y}\end{align}$$


by definition $$E(Y|A)=\frac{E(Y1_{A})}{p(A)}$$ Conditional_expectation_with_respect_to_an_event

let $A=\{Y>3 ,X>3 \}$ so

$$E(Y|Y>3,X>3)=E(Y|A)=\frac{E(Y1_{A})}{p(A)}=\frac{E(Y1_{(Y>3,X>3)})}{p(Y>3,X>3)}=\frac{\int_{y>3} \int_{x>3} y f_{(X,Y)}(x,y) dx \ dy}{\int_{y>3} \int_{x>3} f_{(X,Y)}(x,y) dy \ dx}$$

$$E(Y^2|Y>3,X>3)=E(Y^2|A)=\frac{E(Y^21_{A})}{p(A)}=\frac{E(Y^21_{(Y>3,X>3)})}{p(Y>3,X>3)}=\frac{\int_{y>3} \int_{x>3} y^2 f_{(X,Y)}(x,y) dx \ dy}{\int_{y>3} \int_{x>3} f_{(X,Y)}(x,y) dy \ dx}$$

  • 1
    $\begingroup$ thank you again! $\endgroup$ Mar 13, 2020 at 8:51

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