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a.) Let U, V be uniformly distributed over the set $\{(u,v): $$0<u<v<1$}.

Let $X$ = $-$$log(U)$, $Y$ = $-$$log(V)$, $Z$ = $max$($X$,$Y$).

a.) Draw the support of the joint distribution ($U$, $V$) and the joint pdf $U$ and $V$.

Here, I use the identity that the joint pdf equals the conditional pdf multiplied by the marginal pdf.

$f_{u,v}$($u$,$v$) = $f_{u|v}$($u|v$)$*f_v(v)$ = $\frac{1}{1-u}$*$I(0<u<v)$, where $I$ is the identity function.

The $\frac{1}{1-u}$ part comes from the PDF of the uniform distribution of $V$ over $(u, 1)$: $f_v(v) = \frac{1}{1-u}$

b.) Find the joint PDF of (X,Y). What is its support?

Really not sure on this one. What are my first steps? I'm guessing they come from part a.) but I think my work for part a.) is wrong.

c.) Find the conditional expectation $E$($Z$|$Y$).

My work so far:

$E$($Z$|$Y$) = $E$($max(X,Y)$|$Y$) = $max(E(X|Y),Y)$, and

$E(X|Y) = \int_0^v f_{X|Y}(x|y)*xdx = \int_0^v \frac{1}{(vu)^2}*xdx = \frac{v^2}{(vu)^2} = \frac{1}{v^2}$.

I calculated $f_{X|Y}(x|y)$ in part b.) (not shown) using the derivative formula for deriving PDFs but I'm almost entirely sure it's wrong.

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In part a, it says uniformly distributed over the set $\{(u,v):0<u<v<1\}$, which means the region between lines $u=0, v=u,v=1$ ($v$ is in y-axis and $u$ is in x-axis). So, he joint PDF is 1/Area of this region.

In part b, you can apply Jacobian technique. Another method is to calculate $F_{XY}(x,y)$ and differentiate wrt $u$ and $v$. It's a good exercise, but I'd highly advise the former. Also, the region of support is very important. Hint: it'll extend to infinity but be under y=x line.

In part c, it seems $\max(X,Y)=X$ because $U<V\rightarrow Y<X$.

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  • $\begingroup$ I'm still having issues on part a.). Drawing the support, I get that the joint PDF is 1/(2*v), because the area between u = 0, v = u and v = 1 is a triangle with height 1 and length v. But this leads to a joint PDF that does not integrate to 1, which violates the definition of the PDF. Do you have any other insights for me? $\endgroup$ – Bo Jack Jun 23 '20 at 21:06
  • $\begingroup$ No, the area is $1/2$. $\endgroup$ – gunes Jun 23 '20 at 21:42
  • $\begingroup$ Thank you. Then I get that the PDF of (U, V) = 2 (a valid PDF) and with the Jacobian technique, I get that the PDF of (X, Y) = 2e^{-x-y} (also a valid PDF). Then, I try to get E[Z|Y] = E[X|Y] by getting the conditional PDF of (X | Y), which I calculate to be -e^{y-x} / {-e^y + 1}. This doesn't seem right to me because my expected value depends on Y even though its conditional on Y. Im not sure where I went wrong here, so if I could get one last piece of advice, that'd be great. $\endgroup$ – Bo Jack Jun 24 '20 at 18:43
  • $\begingroup$ I'm not sure about your algebraical calculations but that is normal, E[X|Y] is a function of $Y$. $\endgroup$ – gunes Jun 24 '20 at 19:09
  • $\begingroup$ Sounds great, thank you! $\endgroup$ – Bo Jack Jun 24 '20 at 19:17

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