Which is the distribution of $({\bar{X}\sqrt{n}}/{S})^2$ where $(X_1,\ldots,X_n)$ is a sample from $N(0,\sigma^2)$?

Question

Let $X_1,X_2,\ldots,X_n$ a random sample of a normal population with mean $0$ and variance $σ ^2$. Consider the random variable define as $T=\frac{\bar{X}\sqrt{n}}{S}$. Which is the distribution of $T^2$?

Standardizing, $Z=\frac{\bar{X}-0}{\sqrt{\sigma^2/n}}=\frac{\bar{X}\sqrt{n}}{\sigma}$, because $\bar{X}\sim N(0,\sigma^2/n)$. Also, $Z^2\sim \chi^2_{(1)}$.

On the other hand, we have that $$\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{(n-1)}$$

How can I get to this distributions from $T^2=\left(\frac{\bar{X}\sqrt{n}}{S}\right)^2$, in order to say that it distributes $F_{(1,n-1)}$?

Note that $\bar{X}$ and $S^2$ are estimators of $\mu$ and $\sigma^2$, respectively.

Answer 1

For $(X_1,X_2,\ldots,X_n)$ i.i.d $N(0,\sigma^2)$ and $$\bar X=\frac{1}{n}\sum_{i=1}^nX_i\quad,\quad S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2$$

, we know the exact distribution of $T$, namely

$$T=\frac{\sqrt{n}\bar X}{S}\sim t_{n-1}$$, a $t$ distribution with $n-1$ degrees of freedom.

This is because $$T=\frac{\sqrt{n}\bar X/\sigma}{\sqrt{\frac{(n-1)S^2/\sigma^2}{n-1}}}$$ is formed as the ratio of a standard normal variable and the square root of a chi-square variable, independent of the normal variable, divided by its degrees of freedom.

Now find the distribution of $T^2$.

If $f_{T}$ is the pdf of $T$, then pdf of $T^2$ would be of the form $$f_{T^2}(y)=\frac{1}{2\sqrt{y}}\left[f_T(\sqrt{y})+f_T(-\sqrt{y})\right]\mathbf1_{y>0}$$

You would find that $T^2\sim F_{(1,n-1)}$, an $F$ distribution with $n-1$ degrees of freedom.

Answer 2

If $U\sim \chi^2_p$ and $V\sim\chi^2_q$ and $U$ and $V$ are independent, then the ratio $$ \frac{U/p}{V/q} $$ has $F_{p,q}$ distribution. This is either the definition of the $F$ distribution, or can be proved as a theorem. Either way, your conclusion follows immediately from the additional fact that $\bar X$ and $S^2$ are independent.