Let the joint density $ f_{X,Y}(x,y)=\begin{cases} c(x^3+2xy),\ 0\le x,y\le 2\\ 0, \text{ else}\end{cases}$

be given. I want to compute $Var(Y|X=1)=\int^\infty_{-\infty} (y-E(Y|X=1))^2f_{Y|X=1}(y)\,\mathrm{d}y$.

I computed $E(Y|X=1)=11/6$ and $f_{Y|X=1}(y)=1/6(1+2y)$

Then $Var(Y|X=1)=\int_{0}^2(y-11/6)^2 1/6(1+2y)\,\mathrm{d}y$

Is this correct so far?


1 Answer 1


The expectation is wrong because $11/6$ is too close to $2$. The PDF has a trapezoidal shape, increasing as $y$ increase; therefore the mean could have been $2/3$ at max. Specifically, it'll be $11/9$:

$$E[Y|X=1]=\int_0^2y(1+2y)/6 dy=1/6(y^2/2+2y^3/3)|_0^2=11/9$$ The rest can be solved by your way, but I think using $E[Y^2|X=1]$ will be a bit simpler to do.

  • $\begingroup$ I got $23/81$ for the conditional variance. What do you mean by using $E[Y^2|X=1]$? In what way is this related to $E[Y|X=1]$? $\endgroup$ Oct 13, 2019 at 13:53
  • 1
    $\begingroup$ It seems correct. Integrating for $Y^2$ is easier than $(Y-11/9)^2$. Then, you can find the variance by $E[Y^2]-E[Y]^2$ (ignored conditionals for notational simplicity). $\endgroup$
    – gunes
    Oct 13, 2019 at 14:08

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