Why does this determinant have a continuous density at zero? This question is a simplification of my previous question. I think this is easy, but I don't have a strong enough background in probability.
Let $A$ be a random $n\times n$ real matrix that satisfies the following rules:


*

*All entries $A_{ij}$ are standard Gaussians, with mean $0$ and variance $1$. Furthermore, they are jointly normally distributed.

*The matrix is symmetric, that is $A_{ij} = A_{ji}$.

*Any two entries on the diagonal, $A_{ii}$ and $A_{jj}$ (where $i \ne j$), are correlated, and their covariance is positive and constant (that is, the same for any pair $i, j$).

*Any two entries in the matrix that weren't covered by the previous two rules are uncorrelated.


Since they are jointly normal, the above rules describe the distribution of $A$ completely (am I correct?).
My question concerns the distribution of the determinant, $\det A$: I need to prove it has a bounded density near $0$. That is, for all small enough $\epsilon>0$, $$\mathbb{P}\left\{|\det A| \le \epsilon\right\} \le M\epsilon$$ for some constant $M$.
(For example, if $n=2$ and $A = \left(\begin{smallmatrix}X&X\\X&X\end{smallmatrix}\right)$ where $X$ is a standard Gaussian then this would be false. But this example violates the 4th rule.)
Any help will be appreciated!
 A: Ok, spectral techniques work, even if they produce a huge constant $M$.
Consider the matrix $B$ in which $b_{ii}=1, b_{ij}=\operatorname{sign}(a_{ii})\cdot\frac{a_{ij}}{a_{ii}}$. The off-diagonal entries are Cauchy distributed, hence, by the Gershgorin theorem:
$$\mathbb{P}\left[|\det B|\geq \left(1-\frac{n-1}{\lambda n}\right)^n\right]\geq \left(\frac{1}{\pi\lambda n}\right)^{(n^2-n)}$$
since the $RHS$ is a lower bound for the probability that all the off-diagonal entries have an absolute value below $\frac{1}{\lambda n}$. By choosing $\lambda$ roughly the size of $n$, we have that:
$$ \mathbb{P}\left[|\det B|\geq \frac{1}{e}\right]\geq\left(\frac{C}{n^2}\right)^{n^2}.$$
Now we just have to deal with the distribution of $D=\prod_{i=1}^{n}a_{ii}$. It happens that the product of two indipendent normal distribution has a density function that bursts in zero like $-\log|u|$, while the density of $|X|^m$, with $X\sim N(0,1)$, bursts in zero like $|u|^{1/m-1}$. Despite ugliness, this two facts gives, by interpolation, that the density of $D$ belongs to $L^1$, so the trivial inequality:
$$\mathbb{P}[|\det A|\geq \varepsilon]\geq \mathbb{P}[|D|\geq e\varepsilon]\cdot\mathbb{P}[|\det B|\geq 1/e]$$
proves your claim with a gargantuan constant $M$, depending only on $n$.
