I'm asked to solve the following exercises:
Let $X$ be a random variable with exponential law of parameter $1$, $X \sim \Gamma(1,1)$. Determine the density of $Y:=X^2$ and compute its expected value.
Given $f\colon \mathbb{R}^2\to \mathbb{R}_+$, $f(x,y)=\alpha(x+y)1_{(0,2)^2}(x,y)$ do the following:
1) Choose $\alpha$ such that $f$ is the density of a random vector variable $(X,Y)$;
2) Find the marginal densities of $X$ and $Y$ and tell if they are independent;
3) Compute the probability that $X>Y$.
That's what I did:
$F_Y(t)=F_{X^2}(t)=\mathbb{P}(X^2 \leq t) = \mathbb{P}(-\sqrt{t} \leq X \leq \sqrt{t})=\mathbb{P}(X \leq \sqrt{t}) - \mathbb{P}(X <-\sqrt{t})=F_X(\sqrt{t})-F_X(-\sqrt{t})=F_X(\sqrt{t})$ and the last equality follows by $$f_X(t)=\begin{cases}e^{-t} \quad t\geq 0 \\ 0 \qquad t < 0 \end{cases}$$ So now we have $f_Y(t)=(F_Y(t))'=\frac{1}{2\sqrt{t}}e^{-\sqrt{t}}$ and then $$\mathbb{E}[Y]=\int_0^{\infty}f_Y(t)dt = 2$$
For $f$ to be a density we need $$\int_{\mathbb{R}^2}f(x,y)dxdy = 1 = \alpha \int_0^2 \int_0^2 (x+y)dxdy$$ so $\alpha= \frac{1}{8}$. To find marginal densities we remind $\mathbb{P}(X \leq t) = \mathbb{P}(X \leq t, Y \in \mathbb{R})$ and so $$\mathbb{P}(X \leq t)=\alpha \int_0^t \left(\int_0^2 (x+y)dy \right)dx=\alpha(t^2+2t)=\mathbb{P}(Y \leq t)$$ but $$\mathbb{P}(X\leq t)\mathbb{P}(Y \leq t)= f_Xf_Y \neq f_{X,Y}$$ so the variables are not independent each other. For the last point: $$\mathbb{P}(X>Y)=\mathbb{P}((X,Y) \in A) \qquad A=\{(x,y) \in \mathbb{R}^2 | x>y\}$$ $$\mathbb{P}(X>Y)=\int_0^2 \int_y^2 \alpha(x+y)= \frac{1}{2}$$
Is it alright?
Mostly. Remember to include the supports so$$\Bbb P(X\leq x)=\alpha(x^2+2x)\mathbf 1_{\small(0..2)}(x)+\mathbf 1_{\small[2..\infty)}(x)$$
And independence is then denied since (generally): $\Bbb P(X\leq x)\Bbb P(Y\leq y)\neq \Bbb P(X\leq x,Y\leq y)$ $$8^{-2}(x^2+2x)(y^2+2y)\neq 8^{-1}(x+y)\qquad\text{for }\langle x,y\rangle\in(0..2)^2$$