I am practising some exam questions and am failing to understand the problem at hand. I believe I am supposed to take the double integral of the joint PDF that can be calculated by noting that $f_X,_Y(x,y) = f_{X|Y}(x)*f_Y(y)$.

The problem is as such:

The conditional pdf for X given Y = y, is

$ f_{X|Y=y}(x) = \begin{cases} 1/y^2, & \text{for 0 $\leq$ x $\leq$ $y^2$} \\ 0, & \text{otherwise,} \end{cases} $

While the marginal density of Y is

$ f_Y(y) = \begin{cases} 4y^3, & \text{for 0 $\leq$ y $\leq$ 1} \\ 0, & \text{otherwise.} \end{cases} $

Now I think that X and Y are not independent, this is because looking at the limits of $f_{X|Y}(x)$ it is clear that if y = 0 then x must be 0. Hence, I need to double integrate over the joint pdf to find E(XY), I assume. The problem is how do I determine the limits of my integral?

Thanks for your patience, help and time! It is much appreciated!


The limits of the integral are in fact given to you.   They are: $0\leq y\leq 1$ and $0\leq x\leq y^2$.

$$\begin{align} \mathsf E(XY) & = \int_{y=0}^1 \int_{x=0}^{y^2} x\,y\; f_Y(y)\;f_{X\mid Y=y}(x)\operatorname d x\operatorname d y \\[1ex] & = \int_{y=0}^1 y\cdot 4y^3/y^2 \int_{x=0}^{y^2} x\operatorname d x\operatorname d y \end{align}$$

  • $\begingroup$ Thanks Graham, may I ask why we need take the integral of $xy*f_{X|Y}(x)*f_Y(y)$ ? More specifically the $xy$. Thanks again. $\endgroup$ – orwellian Jun 11 '15 at 8:27
  • $\begingroup$ @Orwellian By definition of expectation: $$\begin{align}\mathsf E(g(X,Y)) & = \iint_{\mathcal {Y\times X}} g(x,y)\; f_{X,Y}(x,y)\operatorname d x \operatorname d y \\[1ex] & = \iint_{\mathcal {Y\times X}} g(x,y)\; f_{X\mid Y=y}(x)\;f_Y(y)\operatorname d x \operatorname d y \end{align}$$ $\endgroup$ – Graham Kemp Jun 11 '15 at 8:33
  • $\begingroup$ @Orwellian As $\int x\operatorname d x = \frac 1 2 x^2 +\text{constant}$ then $$\begin{align}\int _0^{y^2} x \operatorname d x & = \frac {(y^2)^2}2-\frac{0^2}{2} \\[1ex] & = \tfrac 1 2 y^4\end{align}$$ $\endgroup$ – Graham Kemp Jun 11 '15 at 8:50
  • $\begingroup$ Once again thanks, @Graham! I have a complete solution now but to ensure I fully understand, if I were to find the marginal distribution of X for instance I would simply be doing the following: $\int_{0}^{1} 4y dy$ where 4y is the joint pdf here. Have I got the right idea and limits for this? $\endgroup$ – orwellian Jun 11 '15 at 9:10
  • $\begingroup$ @Orwellian The marginal distribution of $x$ is the integral of the joint distribution with respect to $y\in[\sqrt x; 1]$. $$\begin{align} f_X(x) & = \int_{\sqrt{x\;}}^1 f_Y(y)\,f_{X\mid Y=y}(x) \operatorname d y \\[1ex] & = \int_{\sqrt{x\;}}^1 4 y^3\,/y^2 \operatorname d y \\[1ex] & = 4 \int_{\sqrt{x\;}}^1 y \operatorname d y \\[1ex] & = 2 -2 x \end{align}$$ $\endgroup$ – Graham Kemp Jun 11 '15 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.