Conditionate expectation and characteristic function

Question

Let $X,Y$ two random variables such that $X\sim \mathcal{E}(2)$ and $Y_{|X=x}\sim \mathcal{E}(x)$ for all $x>0$.

1. Find the density of $Y$.

2. Calculate $\mathbb{P}(Y\leq X)$.

3. Calculate the characteristic function of $XY$ and deduce its density.

My attempt is the following:

1. I find that:

$$f_{(X,Y)}(x,y)=f_{Y|X}(y|x)\cdot f_X(x)=$$

$$=xe^{-xy}\cdot \mathbf{1}_{(0,+\infty)}(y)\cdot 2e^{-2x}\cdot \mathbf{1}_{(0,+\infty)}(x)= $$

$$=2xe^{-x(y+2)}\mathbf{1}_{(0,+\infty)\times (0,+\infty)}(x,y)$$

So

$$f_Y(y)=\int_{0}^{+\infty}2xe^{-x(y+2)}\ dx=\frac{2}{(y+2)^2}\mathbf{1}_{(0,+\infty)}(y)$$

2. I have no idea, I tried to use conditionate probability but doesn't work much.

3. $\varphi_{XY}(\theta)=\mathbb{E}(e^{i\theta XY})=\mathbb{E}\left(\mathbb{E}(e^{i\theta xY}|X=x)\right)=e^{i\theta x }\mathbb{E}(e^{i\theta Y})=e^{i\theta x}\frac{x}{x-i\theta} $

and now I don't know how to do...Maybe, it's not the right way in this case

What's correct? Any suggestions for 2 and 3?

Answer 1

The first one is correct. For the second, you can notice that you have the joint distribution of $X,Y$, so it is simply integrating \begin{align*} P(Y \leq X) & = \int_{0}^{\infty} \int_{0}^{x} 2xe^{-x(y+2)} dydx \\ & = \int_{0}^{\infty} 2 e^{-2x} - 2e^{-x^{2} - 2x} dx \\ & = 1 - 2\int_{0}^{\infty}e^{-x^{2} - 2x} dx \\ & = 1- 2e \int_{0}^{\infty}e^{-x^{2} - 2x-1} dx \\ & = 1 - 2e \int_{0}^{\infty}e^{-(x+1)^{2}} dx \end{align*} I'll leave it at that, and you can easily continue calculating that (Normal dist.). Now, for the 3, you can calculate directly $$ \varphi_{XY}(t) = \iint_{(0,\infty)^{2}} e^{itxy} 2xe^{-x(y+2)} dydx $$ Maybe there is another easier or different way, but this way everything comes out "directly"