# Let $(X, Y, Z)$ be jointly continuous random variables over a truncated cone

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, $E[X] = \int_{\Bbb R} x\, h(x) \,\text{d}x = E[Y]$.
• @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 $x\mapsto \max\lbrace 0, x\rbrace$. – Harry49 Jan 25 '18 at 21:39