# Showing t-distribution from multivariate standard normals

I came across a paper that assumes the following has a t-distribution:

Let $$W = \frac{\mathbf{a}'\mathbf{X}}{\sqrt{\mathbf{X}'\mathbf{X}}}$$ and $$\mathbf{a}' \in \mathbb{R}^n$$ with $$\mathbf{a}'\mathbf{a} = 1$$. If $$\mathbf{X} \sim N_n(\mathbf{0}, \mathbf{I}_n)$$ then

$$\frac{\sqrt{(n-1)}W}{\sqrt{\left(1 - W^2\right)}},$$

follows a t-distribution with $$n-1$$ degrees of freedom.

I don't quite see how this is true though. I'm assuming that to start you'd need to show that $$\mathbf{a}'\mathbf{X}$$ and $$\sqrt{\mathbf{X}'\mathbf{X}}$$ are independent, but I'm not quite sure how to proceed in the multivariate case.

• Hi. Welcome to CV. Showing that $\mathbf{a}'\mathbf{X}$ and $\sqrt{\mathbf{X}'\mathbf{X}}$ are independent would prove what $W$ follows t-distribution. And if so were the case, the statistic that you have shown would not follow t-distribution. So probably you need to first write your statistic in terms of $\mathbf{X}$. Oct 14, 2019 at 9:59
• $a'X$ and $\sqrt{X'X}$ are not independent. $\hat \sigma^2$ is independent on $\sqrt{X'X}$, but total sum of squares isn't. Oct 14, 2019 at 11:09
• @carlo You're right. On second thought, I think the approach is to rather condition the statistic on $\mathbf{a}'$, i.e. the conditional distribution of the statistic given $\mathbf{a}$ is the $t_{n-1}$ distribution. You can do this since $\mathbf{a}'$ and $\mathbf{X}$ are independent. Oct 14, 2019 at 11:31

Continuing from my comment above:

Let $$T=\frac{(\sqrt{n-1})W}{\sqrt{1 - W^2}}$$

Now, $$1-W^2 = 1-\frac{X'aa'X}{X'X}$$

$$\implies (1-W^2)X'X=X'X-X'AX=X'(I-A)X\,,\qquad A\equiv aa'$$

Therefore,

\begin{align} T&=\frac{(\sqrt{n-1})a'X}{\sqrt{X'X}}\frac{\sqrt{X'X}}{\sqrt{X'(I-A)X}} \\&=\frac{(\sqrt{n-1})a'X}{\sqrt{X'(I-A)X}} \end{align}

See that $$A$$ is a symmetric idempotent matrix with $$\operatorname{tr}(A)=\operatorname{rank}(A)=1$$. Similarly, $$(I-A)$$ is also symmetric idempotent so $$\operatorname{rank}(I-A)=n-1$$.

Now let $$Z=a'X \quad\text{ and }\quad V=X'(I-A)X$$

We have, $$X'X=X'AX+X'(I-A)X = Z^2+V$$

Since $$\operatorname{rank}(A)+\operatorname{rank}(I-A)=n$$, from Cochran's theorem:

$$X'AX$$ and $$X'(I-A)X$$ are independent.

Therefore $$Z$$ and $$V$$ are independent and $$Z^2\sim \chi^2_{(1)}$$ and $$V\sim \chi^2_{(n-1)}$$.

Finally, $$T=\frac{Z}{\sqrt{V/(n-1)}}\,,$$ which by definition now follows t-distribution with $$(n-1)$$ degrees of freedom.