How to calculate a joint pdf by convolution

Question

Suppose $A, B, C$ are independent and uniformly distributed on $[0,1]$

Let $X = A + B$ and $Y = A + C$.

Can anyone show me how to calculate the PDF of the joint distribution of $(X,Y)$?

---

I can get the PDF of $X$ (or $Y$) by convolution, e.g., if $0 \leq x \leq 1$, then

\begin{align} f_X(x) = \int_{-\infty}^{\infty} f_A(t)f_B(x-t)dt = \int_0^x 1 dt = x \end{align}

and if $1 < x \leq 2$, then

\begin{align} f_X(x) = \int_{-\infty}^{\infty} f_A(t)f_B(x-t)dt = \int_{x-1}^1 1 dt = 2 - x \end{align}

but I don't see how to use this approach for a joint distribution

Answer 1

Hint: $$f_{X,Y}(x,y) = f_{Y\mid X}(y\mid x)f_X(x) = f_{X\mid Y}(x\mid y)f_Y(y)$$

Answer 2

Without using convolutions, since $X, Y, Z$ are independent, the density on the unit cube is 1. And $Pr(A \leq a, B \leq b)$ is the volume of the intersection of two wedges defined by $A \leq a$ and $B \leq b$.

Suppose that $a \leq b \leq 1$. Then calculating the volume with integrals shows that

\begin{equation} Pr(A \leq a, B \leq b) = \frac{1}{2} (b - a) a^2 + \frac{a^3}{3} \end{equation}

If $b > 1$ the CDF is more complicated