Find the joint density function $f(x_1, x_2, x_3)$ of the random vector $(X_1, X_2, X_3)$.

Solution 1:

There are a few issues with your notation. Also, you're asked to compute a joint density, not a cumulative distribution function.

You're given that $$X_3|X_1,X_2\sim\mathcal{U}[0,X_2]$$ $$X_2|X_1\sim\mathcal{U}[0,X_1]$$ $$X_1\sim \mathcal{U}[0,1]$$ So, $$\begin{eqnarray*}f_{X_1,X_2,X_3}(x_1,x_2,x_3) &=& f_{X_3|X_1,X_2}(x_3|x_1,x_2)f_{X_2|X_1}(x_2|x_1)f_{X_1}(x_1) \\ &=& \frac{1}{x_2} \cdot \frac{1}{x_1} \cdot 1 \cdot 1_{\{0\leq x_3 \leq x_2 \leq x_1\leq 1\}}\end{eqnarray*}$$

Solution 2:

Previous answer was to different problem. For this problem $f(x_1,x_2,x_3)=\frac{1}{x_1x_2}$ for $0\le x_3\le x_3\le x_1\le 1$ and $=0$ otherwise. To get marginal distribution for any variable, need to integrate other two. $f_{X_1}(x_1)=\frac{1}{x_1}\int_0^{x_1}\frac{dx_2}{x_2}\int_0^{x_2}dx_3=1$ $f_{X_2}(x_2)=\frac{1}{x_2}\int_{x_2}^1\frac{dx_1}{x_1}\int_0^{x_2}dx_3=-ln(x_2)$ $f_{X_3}(x_3)=\int_{x_3}^1\frac{dx_2}{x_2}\int_{x_2}^1\frac{dx_1}{x_1}=\frac{ln^2(x_3)}{2}$