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Suppose a binormal population $X_1, X_2 \sim \mathcal{N}(0,\Sigma)$ where $\Sigma$ is obtained from $\sigma^2$ the variance of each item, assumed equal, and $\rho$, the correlation between pairs. We sample $n$ pairs from that population.

We can compute two sorts of variances from these data. (1) The variance of the difference $V_D = var(X_1 - X_2)$ which is known to follow a $\chi^2$ distribution with df equals to $n-1$. More precisely,

$$ V_D \sim \sigma_D^2 \frac{\chi^2_{n-1}}{n-1}$$

where $\sigma_D$, the variance of the difference is related to the population variance and the population correlation by $\sigma_D^2 = 2\sigma^2 (1-\rho)$.

(2) The pooled variance, that is in the present case, the mean of the two sample variances, $var(X_1)$ and $var(X_2)$. From Ben here, we have this excellent approximation:

$$V_p \sim \sigma^2 \frac{\chi^2_\nu}{\nu}$$

where $\nu = 2(n-1)/(1+\rho^2)$.

My question: what is the distribution of $Z$, the ratio of these two variates, each following correlated chi-square distributions,

$$Z= \frac{V_D}{V_p}$$

Can we also know the correlation between $V_D$ and $V_p$ and how this correlation depends on the population $\rho$? (as $\rho$ tends to -1, the $v_d$ vs. $v_p$ sample correlation tends to +1).

Edit

Note that an approximate solution involving a chi-square distribution may be more useful than the exact solution, even though the exact solution is probably possible to find.

A simple simulation showing the $v_D/v_p$ ratio from 100,000 samples from a population with $\rho=0.5$, $n=10$ (and $\sigma$=1) suggests that it is similar-looking to a $\chi^2$ distribution (see below). The mean here is expected to be $2 (1-\rho) = 1$. The sample correlation between the $v_D$s and the $v_p$s is found to be $ 0.316 \approx 1 / \sqrt{10}$.

enter image description here

Note

In case it may help, $Z$ can be rewritten as

$$Z = 2\left(1-2\frac{cov(X_1,X_2)}{var(X_1)+var(X_2)}\right)$$

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If you assume that $X_1$ and $X_2$ have the same variance $\sigma^2$, namely that $\Sigma = \sigma^2\begin{pmatrix} \;1 & \rho\\ \rho & \;1 \end{pmatrix}$, then you can see by diagonalizing $\Sigma$ that $x_1 := X_1+X_2$ and $x_2 :=X_1-X_2$ are independent with variances $2\sigma^2(1 \pm \rho)$ respectively.

Furthermore from the properties of the covariance of the sum we have: $$\text{svar}(X_1) + \text{svar}(X_2) = \frac{1}{2}(\text{svar}(x_1) + \text{svar}(x_2))$$

(I use $\text{svar}$ to distinguish sample variance from variance)

Therefore you can express $V_D$ and $V_p$ in term of the two independent variables $s_1^2 = \text{svar}(x_1)$, $s_2^2 =\text{svar}(x_2)$ such that

$$ V_D = s_2^2 , \quad V_p = \frac{1}{4}(s_1^2 + s_2^2) $$

Where by the well known distribution of the sample variance we know that

$$ \frac{n-1}{2\sigma^2(1+\rho)}s_1^2 \sim \chi^2_{n-1}, \quad \frac{n-1}{2\sigma^2(1-\rho)}s_2^2 \sim \chi^2_{n-1}$$

or equivalently $s_{1,2}^2 \sim \text{Gamma}((n-1)/2,4\sigma^2(1 \pm \rho)/(n-1))$.

Your desired ratio is $$ Z = 4\frac{s_2^2}{s_1^2 + s_2^2} $$

Notice that this implies that $0 \le Z \le 4$, So a $\chi^2$ approximation will probably not work very well (note also that when $\rho \to 1$ $s_2^2 \to 0$ so $Z$ becomes concentrated at 0, and likewise when $\rho \to -1$ $s_1^2 \to 0$ so $Z$ becomes concentrated at 4).

In fact the distribution of such a ratio of gamma variables is known in closed from and is given in this case by : (See e.g. here)

$$Z/4 \sim f(z)$$

$$f(z) = \frac{z^{k-1}(1-z)^{k-1}}{r^k B(k,k)} \left(1 + \frac{1-r}{r}z \right)^{-2k}, \quad 0 < z < 1.$$

Where $k = (n-1)/2$ , $r = \frac{1-\rho}{1+\rho}$ and $B(\cdotp ,\cdotp)$ is the Beta function.

Furthermore you can easily calculate the correlation between $V_D$ and $V_p$ : $$\text{Cov}(V_D,V_p) = \text{Cov}( s_2^2 ,\frac{1}{4}(s_1^2 + s_2^2)) = \frac{1}{4} \text{var}(s_2^2) = \frac{2\sigma^4(1-\rho)^2}{n-1}$$

$$\text{var}(V_p) = \frac{1}{16} (\text{var}(s_1^2) + \text{var}(s_2^2))= \frac{\sigma^4(1+\rho^2)}{n-1}$$

$$\rho_{V_D,V_p} = \frac{\text{Cov}(V_D,V_p)}{\sqrt{\text{var}(V_D)\text{var}(V_p)}} = \frac{1-\rho}{\sqrt{2(1+\rho^2)}}$$

(Note that when $\rho=1/2$ you get exactly $1/2\sqrt{2*5/4)} = 1/\sqrt{10}$ )

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  • $\begingroup$ Încredible! the fit is perfect!!! $f(z)$ superimposes perfectly on the histograms from simulations (why am I doubting?). Many, many thanks! $\endgroup$ Feb 15, 2022 at 15:54
  • $\begingroup$ happy to help :-) $\endgroup$
    – J. Delaney
    Feb 15, 2022 at 16:04
  • $\begingroup$ Digging a little bit, I found that this $f$distribution is a Generalized Beta distribution [en.wikipedia.org/wiki/Generalized_beta_distribution] with parameters $a=1$, $b=r$, and $c=1-r$. $\endgroup$ Feb 15, 2022 at 19:26
  • $\begingroup$ Yes interesting. more specifically it is in the Beta family sub-class (B) of this generalized Beta distribution $\endgroup$
    – J. Delaney
    Feb 16, 2022 at 9:58

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