Compute distribution of determinant Let X, Y, Z be three independent random variables uniformly distributed in (0, 1).
Let's consider the matrix $\begin{bmatrix} X & Z \\ 1 & Y\end{bmatrix}$. I have to determine the distribution of its determinant $D$.
What I've done so far:
The determinant $D$ is $D = XY - Z$.
Thanks to the convolution formula, I know that (if S, T are independent random varibales) the density of S+T is equal to $f_{S+T}(x) = \int f_T(y) f_S(x-y) dy$. Additionally, I know that the density of $XY$ is $-\log u$.
Is this correct until now? How can I proceed?
Thanks for any help.
 A: Hint:
Assume $X$ and $Y$ are independent with corresponding density functions $f_X$, $f_Y$. Then
$$T := Y-X \iff Y = T + X \implies f_{T}(t) = \int_{-\infty}^{\infty} f_X(x) f_Y(t+x) dx$$
$$T := YX \iff Y = T / X \implies f_{T}(t) = \int_{-\infty}^{\infty} f_X(x) f_Y(t/x) dx$$
A: Summarizing, you have:
Two independent random variables, say $U,Z$ where

*

*$f_Z(z)=\mathbb{1}_{(0;1)}(z)$


*$f_U(u)=-\log u\cdot  \mathbb{1}_{(0;1)}(u)$
Thus the joint density is
$$f_{UZ}(u,z)=-\log u$$
over the unit square...
And you have to derive the density of
$$D=U-Z$$

As I showed you in your previous (and related) question, set
$$\begin{cases}
d=u-z\\
w=z
\end{cases}\rightarrow \begin{cases}
u=d+w\\
z=w
\end{cases}$$
This time the jacobian is evidently $|J|=1$ (no calculations are needed) and you have
$$f_{DW}(d,w)=-\log(d+w)$$
to derive $f_D$, as usual, you have to integrate in $dw$
To do that, observe that your vector $(D,W)$ is defined over the following parallelogram (observe that $0<d+w<1$)

and now the integral bounds are evident...

the result of your Determinant's density is
$$f_D(d)=-(d+1)[\log(d+1)-1]\mathbb{1}_{(-1;0)}(d)+(d\log d-d+1)\mathbb{1}_{[0;1)}(d)$$
Here is a drawing of your resulting density



I have to determine the distribution of ...

I do not know if the density is enough for you or if you need to calculate the CDF of your rv. In this second case you can integrate the density or using another method to derive immediately the CDF without passing through the density
