Ratio of correlated sample variances (gamma distributed) for $N$ samples of two correlated random variables $X \sim N\left(0,\sigma_X^2\right)$ and $Y \sim N\left(0, \sigma_Y^2\right)$ with correlation $\rho$, I am analyzing the ratio of the sample variances, $r=\frac{s_Y^2}{s_X^2}$. Currently, I am foccusing on $E\left[r\right] = E\left[\frac{s_Y^2}{s_X^2}\right]$.
I have found that $s_X^2 \sim \Gamma\left(\frac{N-1}{2}, \frac{2 \sigma_X^2}{N-1}\right)$, $s_Y^2 \sim \Gamma\left(\frac{N-1}{2}, \frac{2 \sigma_Y^2}{N-1}\right)$ since $\frac{(N-1)S_X^2}{\sigma_X^2}\sim \chi_{N-1}^2$ (and likewise for $s_Y^2$).  
The ratio of the sample variances is therefore the ratio of two dependent gamma distributions. I have found some papers on this topic, for instance [1, 2]. These papers do however always assume the gamma distributions having different shape and same scale parameters (scale 1 to be precise) - just the opposite of my problem.  
I have tried using $\frac{N-1}{2 \sigma_X^2} s_X^2  \sim \Gamma \left(\frac{N-1}{2}, 1\right)$, $\frac{N-1}{2 \sigma_Y^2} s_Y^2 \sim \Gamma \left(\frac{N-1}{2}, 1\right)$ - the ratio of these should be beta prime distributed.
Using the formula for the expected value of a beta prime distribution this leads to
$E\left[\frac{s_Y^2}{s_X^2}\right]
=E\left[\frac{\frac{N-1}{2 \sigma_X^2}}{\frac{N-1}{2 \sigma_Y^2}} \frac{\frac{N-1}{2 \sigma_Y^2}}{\frac{N-1}{2 \sigma_X^2}} \frac{s_Y^2}{s_X^2}\right] 
=\frac{\sigma_Y^2}{\sigma_X^2} E\left[\frac{\frac{N-1}{2 \sigma_Y^2}}{\frac{N-1}{2 \sigma_X^2}} \frac{s_Y^2}{s_X^2}\right]
=\frac{\sigma_Y^2}{\sigma_X^2} \frac{\alpha}{\beta - 1}$ 
with $\alpha = \beta = \frac{N-1}{2}$,
which doesn't seem to be the correct result (shouldn't the ratio in some way depend on the correlation?) and which differs from the results of numerical experiments.
How can the expected value of the ratio be calculated? At which point am I mistaken?
I would very much appreciate any help - thank you in advance.
Sources:
[1] Lee et al.: Distribution of a ratio of correlated gamma random variables. SIAM Journal on Applied Mathematics 36.2 (1979): 304-320.
[2] Tubbs: Moments for a ratio of correlated gamma variates. Communications in Statistics-Theory and Methods 15.1 (1986): 251-259.
 A: If we assume that the underlying normals are jointly normal, then the result is very simple.
This answer uses results from A. H. Joarder (2007/2009), Moments of the product and ratio of two correlated chi-square variables (open access).  
We have
$$r\equiv \frac{s_y^2}{s_x^2} = \frac {\sigma^2_y}{\sigma^2_x}\frac{(n-1)s_y^2/\sigma^2_y}{(n-1)s_x^2/\sigma^2_x}= \frac {\sigma^2_y}{\sigma^2_x} \frac {U_y}{U_x},\;\; U_y \sim \mathcal \chi^2_{(n-1)},\;\; U_x \sim \mathcal \chi^2_{(n-1)}$$
So 
$$E(r) = \frac {\sigma^2_y}{\sigma^2_x} E\left(\frac {U_y}{U_x}\right)$$
In other words, we need the expected value (first raw moment) of the ratio of two correlated chi-squares with equal degrees of freedom.
Note that standardizing the normals does not affect their correlation coefficient, denote it simply $\rho \in (-1,1)$. Then according to the above paper (which references another paper), $$\operatorname{Corr}(U_y, U_x) = \rho^2$$
and by Corollary 3.8 p. 590, if $n-1 >2$ (i.e we need a sample size at least equal to $4$) we have
$$E\left(\frac {U_y}{U_x}\right) = \frac {n-1 -2\rho^2}{n-1-2} = \frac {n-1 -2\rho^2}{n-3}$$
So
$$E(r) = \frac {\sigma^2_y}{\sigma^2_x}\frac {n-1 -2\rho^2}{n-3}$$
The paper contains also expressions for the next three raw moments, increasing slightly the requirement on the size of the sample. Results appear to depend on the two chi-squares having the same degrees of freedom.
