This may sound like a silly question (I may just be blanking right now), but I need to solve the following question:

Let $(X,Y,Z)$ be jointly continuous random variables with joint pdf given by $$ f(x,y,z) = \left\lbrace \begin{array}{cl} cz^{-3} & x^2+y^2 \leq z^2 \text{ and } a\leq z \leq b \\ 0 & \text{otherwise,} \end{array}\right. $$ where $a$ and $b$ are constants satisfying $0<a<b<\infty$.

(a) Find $c$.

(b) Find the marginal joint pdf of $(X,Y)$.

(c) Find $E[X]$ and $E[Y]$.

I am specifically looking for guidance for part (a). I tried solving this myself with the limits of $z$: $a$ to $b$, $y$: $-\sqrt{z^2 - x^2}$ to $\sqrt{z^2 - x^2}$, and $x$: $-\sqrt{z^2 - y^2}$ to $\sqrt{z^2 - y^2}$. Now, my integrals are looking really messy, so I am thinking I do not have the right limits. I was just wondering if anyone could help me? Thanks a lot!

Also some guidance for (c): When I solve for the expectations, do I solve using the joint pdf $f(x, y, z)$ or do I use the marginal pdfs?


The domain of $\Bbb R^3$ defined by $x^2 + y^2 \leq z^2$ and $0<a\leq z \leq b$ is a truncated cone with axis $(Oz)$ and apex angle $\pi/4$. Its volume is given by \begin{aligned} \int_{\Bbb R^3} \mathbf{1}_{x^2 + y^2 \leq z^2}\, \mathbf{1}_{a\leq z\leq b}\, \text{d}x\,\text{d}y\,\text{d}z &= \int_{\Bbb R} \mathbf{1}_{a\leq z\leq b} \int_{\Bbb R^2} \mathbf{1}_{x^2 + y^2 \leq z^2}\, \text{d}x\,\text{d}y\,\text{d}z \\ &= \int_{\Bbb R}\left( \mathbf{1}_{a\leq z\leq b} \int_{0}^{+\infty} \mathbf{1}_{r \leq |z|}\, r\text{d}r \int_{0}^{2\pi}\text{d}\theta\right) \text{d}z \\ &= {\pi} \int_{a}^{b} |z|^2 \,\text{d}z \\ &= \frac{\pi}{3}\left(b^3 - a^3\right) . \end{aligned} This computation can be adapted to the integral of the pdf \begin{aligned} \int_{\Bbb R^3} c z^{-3} \mathbf{1}_{x^2 + y^2 \leq z^2}\, \mathbf{1}_{a\leq z\leq b}\, \text{d}x\,\text{d}y\,\text{d}z &= {\pi}c \int_{a}^{b} z^{-3}\, |z|^2 \,\text{d}z \\ &= {\pi}c\ln\left(b/a\right) \, , \end{aligned} which must be equal to one.

The marginal pdf $g$ of $(X,Y)$ is obtained by integrating the pdf with respect to $z$: \begin{aligned} g(x,y) &= \left\lbrace\begin{array}{ll} 0 &\text {if}\quad \sqrt {x^2+y^2}>b\\ c \int_\sqrt{x^2+y^2}^b z^{-3} \,\text{d}z &\text {if}\quad b>\sqrt{x^2+y^2}>a\\ c \int_a^b z^{-3} \,\text{d}z &\text {if}\quad a>\sqrt{x^2+y^2} \end{array}\right.\\ &= c\int_{\max\left\lbrace a,\sqrt{x^2 + y^2}\right\rbrace}^{\max\left\lbrace b,\sqrt{x^2 + y^2}\right\rbrace} z^{-3} \,\text{d}z \\ &= -\frac{c}{2} \left({\max\left\lbrace b,\sqrt{x^2 + y^2}\right\rbrace}^{-2} - {\max\left\lbrace a,\sqrt{x^2 + y^2}\right\rbrace}^{-2} \right) . \end{aligned}

The marginal pdfs $h$ of $X$ and $Y$ are identical, since $x$ and $y$ play symmetric roles in $g(x,y)$. They write $h(x) = \int_{\Bbb R} g(x,y)\,\text{d}y$. Then, $E[X] = \int_{\Bbb R} x\, h(x) \,\text{d}x = E[Y]$.

  • $\begingroup$ This is really helpful. The only question I have is, when computing the marginal pdf g(x,y), do I sub in both values, and accept the answer that is the largest? Also, if that's the case, why do I take the max between the two? $\endgroup$ – queence Jan 25 '18 at 19:04
  • $\begingroup$ Also, in this case it would be difficult to solve for an absolute answer because I am not given any values right? $\endgroup$ – queence Jan 25 '18 at 19:21
  • $\begingroup$ @queence I added few details. Remind that $(x,y)\mapsto \max\lbrace a, \sqrt{x^2+y^2}\rbrace$ is a function, as well as $x\mapsto \max\lbrace 0, x\rbrace$. $\endgroup$ – Harry49 Jan 25 '18 at 21:39

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.