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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.

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    $\begingroup$ Consider rewriting the whole thing in terms of $D_i = X_i-Y_i$ (or written in terms of things related to $D$); you might find things simplify enough to see what's going on. $\endgroup$ – Glen_b -Reinstate Monica Nov 19 '19 at 0:17
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    $\begingroup$ I suggest you literally work with D, getting rid of everything in X and Y. Be careful about terms like $\text{Var}(U-\bar{U})$. Make sure you justify your steps! $\endgroup$ – Glen_b -Reinstate Monica Nov 19 '19 at 5:06
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    $\begingroup$ As @Glen_b-ReinstateMonica suggested (+1), working directly with $D_i=X_i-Y_i$ restates the problem as a single Normal sample problem. And hence to the definition of Student's $t$ distribution. $\endgroup$ – Xi'an Nov 28 '19 at 19:35
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    $\begingroup$ Ideally effort put into a post helps someone else other than yourself (typically posts here help many other people), so better than deletion would be to post an answer to the question. However, keep in mind the aim for self-study questions is to give guidance and hints. As the original poster you can be more explicit in an answer than I would, but I'd lean toward answering briefly, but making sure to include the most critical insights. If you use a hint that was given to you by someone else (i.e. Glen_b or Xi'an) simply credit them (mention their name) when you use it. $\endgroup$ – Glen_b -Reinstate Monica Nov 28 '19 at 22:59
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    $\begingroup$ The variance of $X_i-Y_i$ is incorrect; it should be $2\sigma^2\color{blue}{(1-\rho)}$ with the covariance. Final answer is thus $\sqrt{n-1}\,T\sim t_{n-1}$ assuming all parameters are unknown. $\endgroup$ – StubbornAtom Jan 12 at 20:14
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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.

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  • $\begingroup$ This is a great answer - thank you! $\endgroup$ – Edison Feb 13 at 6:08

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