Let $A$ be a $2 \times 2$ matrix, i.e. \begin{equation} A = \begin{pmatrix} X_1 & X_2 \\ X_3 & X_4 \\ \end{pmatrix} \end{equation}

where the $X_{i}$'s entries i.i.d normal random variable with mean $0$ and variance $1$. The problem was to determine the ditribution of $\det(A)$. Now with characteristic function and the law of total expectation I come to the solution, that the distribution function has density $\exp(-|x|)/2$. This was not so difficult.

Now I know that $\det(A)$ is the area of a parallelogram with vertices \begin{align} (0,0), (X_{1},X_{2}), (X_{3},X_{4}), (X_{1} + X_{2},X_{3} + X_{4}). \end{align}

Is there a way to obtain the solution using this geometric way?

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  • I believe the fourth point of the parallelogram should be $X_1+X_3,$ $X_2+X_4$. – RideTheWavelet May 17 '18 at 15:46
  • have you a solution to this problem? I'have solved the problem with characteristic function but it would be interesting to obtain a geometric solution – wayne May 17 '18 at 15:52
  • Sorry about that. I believe that the idea there will work, but the distributions of the two factors in the product aren't just half-normal. The first one is Chi with 2 df's, and it would take some work to determine the distribution of $X_{4}$ (it might just be half-normal, but I don't know). – RideTheWavelet May 17 '18 at 16:37
  • with characteristic function it is very easy, but trying it with annother approach seems to me very difficult – wayne May 17 '18 at 16:38
  • I think is to difficult to obtain a solution this way – wayne May 18 '18 at 18:01

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