Let $X \sim \mathrm{Unif }[0,1]$ and $Y\sim \mathrm{Gamma}(2,r)$, with $r>0$, be two independent random variables.
I have to find $\mathrm{Cov}(U,V)$ where $U = X$ and $V = XY$.
From previous points, I have calculated that $$f_{U,V}(u,v) = \frac{r^{2}ve^{-r\frac{v}{u}}}{u^{2}} $$ with $0 < u < 1 $ and $ v > 0 $,
and$$f_{V}(v) = re^{-rv}$$ with $v > 0$, hence, $V \sim \textit{Exp}(r)$.
Thus,
\begin{align*}
\mathrm{Cov}(U,V) &= \mathbb{E}(U\cdot V) – \mathbb{E}U \cdot\mathbb{E}V \\
&= \mathbb{E}(X\cdot XY) – \mathbb{E}X\cdot\mathbb{E}(XY) \\
&\overset{indep.}{=} \mathbb{E}(X^{2}\cdot Y) – (\mathbb{E}X)^{2}\cdot\mathbb{E}Y
\end{align*}
- If $X^{2}$ and $Y$ are independent, then $$ = \mathbb{E}Y\cdot \textit{Var}(X)$$ which would be pretty convenient. But also how should I prove this?
- If they're not independent, then I'm wondering if there is an easy way to evaluate it, without having to find the density function of $UV$ which I predict would end up having the same "problem". I know that there is a way to approximate $\lvert\mathbb{E}(UV)\rvert$ (Cauchy-Schwarz inequality), that is,
$$\lvert\mathbb{E}(UV)\rvert \leq \sqrt{\mathbb{E}(U^{2})\cdot\mathbb{E}(V^{2})}$$ but I'm not sure this is the way to evaluate this.
(I've searched around for an answer, but they all involve Measure Theory which I have no knowledge of).
Any suggestion/help would be appreciated.
Best Answer
For any continuous functions $f$ and $g$ (and, in fact, for any Borel measurable functions $f$ and $g$) the variables $f(X)$ and $g(Y)$ are independent. So it is certainly true that $X^{2}$ and $Y$ are independent.