TITLE: Expected determinant of random symmetric matrix with different Gaussian distributions of the diagonal and non-diagonal elements
QUESTION [6 upvotes]: Consider a random matrix $A \in \mathbb{R}^{N \times N}$ where the elements are  random gaussian variables. The mean and variance of the elements are different on the diagonal and the off-diagonal:
$\begin{align}
&\left<A_{ij}\right>=p &\text{and } & \text{var}(A_{ij})=\frac{1}{N}p(1-p) &\text{  if  } \,  i=j \\
&\left<A_{ij}\right>=p^2 &\text{and } &\text{var}(A_{ij})=\frac{1}{N} p^2(1-p^2) &\text{ if } \, i\neq j
\end{align}$ 
where $p \in (0,1)$ is some fixed paremeter. The matrix is symmetric $A_{ij}=A_{ji}$.
Now I would like to know 
$\begin{align}
\mathbb{E}[\lim_{N\rightarrow \infty}  |\det  A|],
\end{align}$
i.e. the asymptotic behavior as $N$ becomes large. Since the variances are approaching zero for $N \rightarrow  \infty$ I assume that the expectation of the absolute value of the determinant also approaches zero.
Is there a method to determine this behavior or does someone know how to do it? 

Edit: I attached the following figure showing the result of Carlo Beenakker (lines)
$\mathbb{E} [|\det  A|] \sim N p^{N+1}(1-p)^{N-1}+p^{N}(1-p)^N $
and results from simulations (crosses) for different $p$. They seem to fit very well.

REPLY [4 votes]: There is a similar question here : Expected value of determinant of simple infinite random matrix
I rewrite the heuristique from the random matrix theory: 
$$A=p^2 1 + (p-p^2)I_N+\sqrt{p^2(1-p^2)}\tilde{A}+\frac{1}{\sqrt{N}}D$$
where $1$ is the matrix with entries all equal to 1, $D$ a diagonal matrix with bounded variance. And $A$ is a matrix iid random entries with zero mean and variance $1/N$. Therefore its eigenvalues should be distributed similarly as the semicircul law. Because the first matrix is a rank one perturbation and $\| \frac{1}{\sqrt{N}}D \|$ is small, we could ignore them at first approximation.
$$\frac{1}{N}\log(|\det(A)|)\sim \frac{1}{N}(\log(|\det(\frac{p-p^2}{\sqrt{p^2(1-p^2)}}I_N+\tilde{A}))|)+\log(\sqrt{p^2(1-p^2)}) $$
We now use the semicircul law and we should get
$$\frac{1}{N}(\log(|\det(\Lambda I_N+\tilde{A})|))=\frac{1}{N}\sum_{\lambda \in \sigma (\tilde{A})} \log (|\Lambda+\lambda|) \sim 
\int_{-2}^{2} \log(|\Lambda+\lambda|) \rho(\lambda) d\lambda$$ where $\Lambda=\frac{p-p^2}{\sqrt{p^2(1-p^2)}}$ and $\rho(\lambda ) = \frac{\sqrt{4-\lambda^2}}{2\pi}$ is the semicircul law. 
So finally we should obtain
$$\frac{1}{N}\log(|\det(A)|)\sim \log(\sqrt{p^2(1-p^2)})+\int_{-2}^{2} \log(|\Lambda+\lambda|) \rho(\lambda) d\lambda$$