Covariance of two random variable, with one uniformly distributed and the other dependent on it.

Question

Problem from actuarial Exam P:

Let $X$ and $Y$ denote the values of two stocks at the end of a five-year period. $X$ is uniformly distributed on the interval $(0,12)$. Given $X=x$, $Y$ is uniformly distributed on the interval $(0,x)$. Determine $\operatorname{Cov}(X,Y)$ according to this model.

I know that $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]$, and $E[X]=6$.

So ultimately my question is that when I am finding $E[Y]$, I don't understand how $f(x,y)$ becomes $1/12x$. Is it $1/12 \cdot 1/x$? If correct, why?

Answer 1

It's because $f(x,y)= f(y|x) f(x)$. But $f(y|x)= \frac{1}{x}$ and $f(x)=\frac{1}{12}$.

Make sure you understand this! You'll need it for P, and you'll need it again for C.

Note: This seems the wrong forum for this question. You should ask it over at Actuarial Outpost.

Answer 2

Note that $Y=ZX$ where $Z$ is uniform on $(0,1)$ and independent of $X$, hence $XY=ZX^2$ and $\mathrm{Cov}(X,Y)$ is $$E(ZX^2)-E(X)E(ZX)=E(Z)E(X^2)-E(Z)E(X)^2=E(Z)\mathrm{Var}(X).$$ Can you finish this?