Consider a matrix of order $n$ whose elements are independently and uniformly chosen from the interval $[a,b]$ wher both $a$ and $b$ are positive .Let $X$ be the r.v. corresponding to the determinant of this matrix.I am interested in knowing the following about $X$
(1).the mi. and the max. of $X$
(2) the pdf of $X$
This is not a complete answer, but a collection of observations (mainly about part (1)), which may be useful for the problem.
Let's denote maximal determinant with $D_n$.
Case $n=1$ is a trivial exceptional case, where determinant is uniformly distributed on $[a,b]$. So we can assume, that $n\geq 2$. In this case the minimal value of determinant is $-D_n$, because we can always permute two rows. Moreover, by the same reason, the distribution of the determinant is symmetric with respect to 0.
Lemma: the answer to 1 has the form $D_n = (b-a)^{n-1} \max_i (c_i a + d_i b)$, where $c_i,d_i$ are some integers, $i=1,\dots,i_{max}(n)$.
Proof: First, note, that the determinant is linear in each of matrix elements. Therefore any local (and hence global) maximum and minimum will be achieved for the extremal values of the elements. That is, we can assume, that each matrix element is either $a$ or $b$. Let $A$ be such a matrix. Do the following:
Note, that $\det A = \pm (b-a)^{n-1} \det B$. By permuting columns we now redefine $B$ to remove $\pm$. $\det B$ has by construction the form $c a + d b$, where $c$ and $d$ are integers, so $\det A = (b-a)^{n-1}(c a + d b)$. Since there was only a finite number of matrices $A$ to start with, the lemma is proven.
By brute-force computation I obtained $$\begin{aligned} D_1 &= (1b+0a),\\ D_2 &= (1b+1a)(b-a),\\ D_3 &= (2b+1a)(b-a)^2,\\ D_4 &= (3b+2a)(b-a)^3,\\ D_5 &= (5b+4a)(b-a)^4. \end{aligned}$$
That is, for these cases there is just one term in the $\max_i$. I don't know, if this will be true for larger values of $n$.
P.S. It seems, that the maximum determinant problem is not solved even for 0-1 matrices, leave alone the probability distribution.
