Problem: Let $\underline X = (X_1,X_2,X_3)$ be a random vector with p.d.f. $$\color{blue}{f_{X_1,X_2,X_3}(x_1,x_2,x_3) = \begin{cases}2e^{-x_2-2x_3} & 0<x_1<1, x_2>0,x_3>0\\ 0 &\text{otherwise} \end{cases}}$$ (a) Are $X_1,X_2,X_3$ independent?
(b) Are $X_1+X_2$ and $X_3$ independent?
(c) Find marginal p.d.fs of $X_1,X_2,X_3$.
(d) Find conditional p.d.f. of $X_1$ given $X_2 = 2$.
$\color{red}{(a)}$ $f$ can be factored into functions involving only $x_2$ and only $x_3$ (there is no $x_1$), is that enough to conclude that the three random variables are independent? Anyway, I will find the marginals in part (c); so I think the question asker doesn't want us to compute the marginals in (a).
$\color{red}{(b)}$ Will I have to find the p.d.f. associated with $X_1+X_2$ and $X_3$ in this case? For $X_3$, let's say the marginal p.d.f. is $f_3$, then I know that $$f_3(x_3) = \int_0^\infty\int_0^1 2e^{-x_2-2x_3}dx_2dx_3$$ when $x_3>0$ and $0$ otherwise. How do I find the p.d.f. of $X_1+X_2$ though? Do I really need to find it to answer part (b)?
$\color{red}{(c)}$ I'm just writing the expressions for $f_1,f_2,f_3$ since I found this calculation trivial: $$f_1(x_1) = 1$$ when $0<x_1<1$ and $0$ otherwise. $$f_2(x_2) = e^{-x_2}$$ when $x_2>0$ and $0$ otherwise. $$f_3(x_3) = 2e^{-2x_2}$$ when $x_3>0$ and $0$ otherwise. Clearly, $f = f_1f_2f_3$, so this gives concrete proof that $X_1,X_2,X_3$ are indeed independent, i.e. solves part (a) too.
$\color{red}{(d)}$ Since $X_1$ and $X_2$ are independent, the conditional p.d.f should be the same as the marginal p.d.f. of $X_1$, right? In which case, it'd be $1$ for $0<x_1<1$ and $0$ otherwise. Hope this is correct!
What would happen if $X_1,X_2,X_3$ weren't independent, suppose the distribution was of the form $ae^{-x_2-2x_3-x_2x_3}$ for appropriate choice of $a$. How would I solve part (d) then?
Thanks a lot!
Update 1: For the p.d.f. of $Y = X_1 + X_2$, first I shall compute the c.d.f. which should be $$F_Y(y) = P(Y\le y) = \int_0^1\int_0^\infty e^{-x_2}\mathbf{1}_{\{x_1+x_2\le y\}} dx_2dx_1$$ We have $0\le x_1\le 1$ and $0\le x_2\le\infty$, we want to integrate over $x_1+x_2\le y$. Clearly we must also have $x_1\le y$ and $x_2\le y$. Would it be correct to rewrite the integral as $$\int_0^1\int_0^{y-x_1} e^{-x_2} dx_2dx_1 = \int_0^1 1 - e^{x_1-y} dx_1 = 1 - e^{-y}(e-1)$$ If the above is correct, then $$f_Y(y) = \frac{\partial F_Y(y)}{\partial y} = e^{-y}(e-1)$$ Is this fine? I don't know how to solve (b) with this, though.