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.

  • 1
    $\begingroup$ 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 $\mathbf{X}$. $\endgroup$
    – Dayne
    Oct 14, 2019 at 9:59
  • 1
    $\begingroup$ $a'X$ and $\sqrt{X'X}$ are not independent. $\hat \sigma^2$ is independent on $\sqrt{X'X}$, but total sum of squares isn't. $\endgroup$
    – carlo
    Oct 14, 2019 at 11:09
  • $\begingroup$ @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. $\endgroup$
    – EzioBosso
    Oct 14, 2019 at 11:31

1 Answer 1


Continuing from my comment above:

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

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

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


\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.


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