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

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.

• It is a good idea to define $\bar X$ and $S^2$. – StubbornAtom Oct 2 '18 at 19:00

## 2 Answers

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.

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.

• (+1) A more direct route. – StubbornAtom Oct 2 '18 at 20:00