Composite of uniform distributions

Question

If $X_1$ is uniform $(0,1)$ and $X_2$ is uniform $(0, X_1+1)$, what is $X_3$, which is characterized as uniform $(0, X_2+1)$?

I simulated $X_3$ and got the following graph but found it hard to compute the precise distribution:

Answer 1

$P[X_3\leq x]=\int^1_0\frac{1}{1+x_1}\int^{x_1+1}_0\frac{1}{1+x_2}\int^{x\wedge (x_2+1)}_0\,dx_3\,dx_2\,dx_1$

To see this, notice that $$P[X_3\leq z|X_2]=\frac{1}{1+X_2}\int^{x\wedge(X_2+1)}_0\,du$$ and $$ P[X_2\leq y|X_1]=\frac{1}{1+X_1}\int^{y\wedge(X_1+1)}_0\,dt$$ Hence,for any bounded measurable function $f$ on $\mathbb{R}$ $$E[f(X_2)|X_1]=\frac{1}{1+X_1}\int^{1+X_1}_0f(t)\,dt$$