How can I find the distribution of the quantity $T(X,Y)$ given a sample from a bivariate normal distribution?

Question

Let $(X_1,Y_1),\cdots,(X_n,Y_n)$ be a sample from a bivariate normal distribution with parameters $E(X_i)=\mu_1, E(Y_i)=\mu_2, Var(X_i)=Var(Y_i)=\sigma^2,$ and $Cov(X_i,Y_i)=\rho\sigma^2, i=1,\cdots,n$. Find the distribution of $T(X,Y)=\sqrt{n}\frac{ (\bar{X}-\mu_1)-(\bar{Y}-\mu_2) }{ \sqrt{ \Sigma^n_{i=1}(X_i-Y_i-\bar{X}+\bar{Y})^2 } }$.

My work so far:

I think $T(X,Y)$ resembles a t distribution, so I started solving this by trying to make the numerator a standard normal distribution.

$X \sim N(\mu_1,\sigma^2)$ and $Y \sim N(\mu_2,\sigma^2)$

$\bar{X} \sim N(\mu_1,\sigma^2/n)$ and $\bar{Y} \sim N(\mu_2,\sigma^2/n)$

$\frac{\bar{X}-\mu_1}{\sigma/\sqrt{n}} \sim N(0,1)$ and $\frac{\bar{Y}-\mu_2}{\sigma/\sqrt{n}} \sim N(0,1)$

$\frac{(\bar{X}-\mu_1)-(\bar{Y}-\mu_2)}{\sigma \sqrt{2/n}} \sim N(0,1)$

So that worked well so far, but I am having troubles with the denominator. For a t distribution, the denominator should be $\sqrt{V/p}$, where $V$ is a Chi-squared distribution with $p$ degrees of freedom. Since a sum of Chi-squared distributions is another Chi-squared distribution with the degree of freedom being the sum of the degrees of freedom of the added distributions, then $(X_i-Y_i-\bar{X}+\bar{Y})^2$ should be a Chi-squared distribution. However, I am not getting that it is one. Am I approaching this problem incorrectly?

Update:

I am getting that $(X_i-Y_i)+(\bar{Y}-\bar{X}) \sim N(0,2\sigma^2(1+n)/n)$, so

$\frac{ \frac{\sqrt{n}}{2\sigma}(\bar{X}-\mu_1)-(\bar{Y}-\mu_2) }{ \sqrt{\frac{n}{2\sigma^2(1+n)}\Sigma^n_{i=1}(X_i-Y_i-\bar{X}+\bar{Y})^2} }=\frac{ \sqrt{n}(\bar{X}-\mu_1)-(\bar{Y}-\mu_2) }{ \sqrt{\Sigma^n_{i=1}(X_i-Y_i-\bar{X}+\bar{Y})^2} } \cdot \frac{\sqrt{2(1+n)}}{2} \sim t_n$.

This closely resembles $T(X,Y)$, but there is the additional $\frac{\sqrt{2(1+n)}}{2}$ term. Did I do something wrong? If not, what is the next step?

A Better Update Thanks to Xi'an and Glen_b

Let $D_i=X_i - Y_i$, so $D_i \sim N(\mu_1-\mu_2, 2\sigma^2(1-\rho))$ and $\bar{D} \sim N(\mu_1-\mu_2, 2\sigma^2/n)$.

We can now write $\sqrt{n-1}T(X,Y)=\frac{(\sqrt{n2\sigma^2(1-\rho)}/\sqrt{2\sigma^2(1-\rho)})(\bar{D}-(\mu_1-\mu_2))}{\sqrt{\frac{1}{n-1}\Sigma^n_{i=1}(D_i-\bar{D})^2}}$. From here, I do not get the same answer as Stubborn_Atom.

Answer 1

Consider $D_i = X_i - Y_i$. Then

\begin{align*} D_i &\sim \mathcal{N}\left(\mu_1-\mu_2, 2\sigma^2(1-\rho)\right)\\ \implies \bar{D} &\sim \mathcal{N}\left(\mu_1-\mu_2, \frac{2\sigma^2}{n}(1-\rho)\right)\\ \end{align*}

Let $\mu_0 = \mu_1-\mu_2$, $\sigma_0^2 = 2\sigma^2(1-\rho)$, and $s=\sqrt{ \frac{\Sigma^n_{i=1}(D_i-\bar{D})^2}{n-1} }$.

\begin{align*} D_i &\sim \mathcal{N}\left(\mu_0, \sigma_0^2\right)\\ \implies \bar{D} &\sim \mathcal{N}\left(\mu_0, \sigma_0^2/n\right)\\ \end{align*}

Now, \begin{align*} T(X,Y) &=\sqrt{n}\frac{ (\bar{X}-\mu_1)-(\bar{Y}-\mu_2) }{ \sqrt{ \Sigma^n_{i=1}(X_i-Y_i-\bar{X}+\bar{Y})^2 } }\\ &= \sqrt{n} \frac{ \bar{D}-\mu_0 }{\sqrt{n-1} \sqrt{ \frac{\Sigma^n_{i=1}(D_i-\bar{D})^2}{n-1} } }\\ \sqrt{n-1} T &= \frac{ \bar{D}-\mu_0 }{ s/\sqrt{n}}\\ &\sim t_{n-1} \end{align*}

which is the definition of a $t$ distribution.