Joint pdf of $Y$ and $Z$ given $X_1$ and $X_2$

Question

Let $X_1$ ~ uniform(0,1) and $X_2$ ~ uniform(0,1). Given that $X_1$ and $X_2$ are independently distributed:

A) Determine the joint probability density function of $X_1$ and $X_2$.

B) Let $Y = X_1 + X_2$ and $Z = X_2$. Determine the joint pdf of Y and Z.

C) Also, determine that marginal density of Y.

I found A to be

$f(x_1, x_2) = 1$ for 0 < x < 1 and 0 < y < 1, 0 otherwise

I found B to be

$f(y, z) = 1$ for 0 < y-z < 1 and 0 < z < 1, 0 otherwise

Am I on the right track? Any tips on how to approach part C?

Answer 1

1. You've got the right joint density. As $X_1$ and $X_2$ are independent, you just need to multiply the two densities to form the joint density.
2. You've noticed that $X_1 = Y-Z$, so you substituted $x$ in "$0 < x < 1$" in the previous part with $y - z$.
3. You need to work on $0 < y - z < 1$ to obtain the appropriate range of $y$ : $z < y < 1 + z$. $$f_Y(y) = \int_0^1 f_{Y,Z}(y,z)\,\mathrm{d}z = \int_{\max(y-1,0)}^{\min(y,1)} 1 \,\mathrm{d}z = \begin{cases} y &\text{ if } y \in [0,1] \\ 2-y &\text{ if } y \in (1,2] \\ 0 &\text{ otherwise} \end{cases} = (1-|1-y|)_+$$ The marginal density of $Y$ is $f_Y(y) = (1-|1-y|)_+$, i.e. the positive part of $1 - |1 - y|$.