# 3-part question on joint PDFs

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

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.

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

In part a, it says uniformly distributed over the set $$\{(u,v):0, which means the region between lines $$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.
• No, the area is $1/2$. – gunes Jun 23 '20 at 21:42
• I'm not sure about your algebraical calculations but that is normal, E[X|Y] is a function of $Y$. – gunes Jun 24 '20 at 19:09