How do I show that this is a T distribution?

Question

$Y_1,...,Y_n$ is a random sample from a $N(\theta,\theta)$

How do I show that $$\frac{\sqrt{n}(\bar{Y}-\theta)}{s}$$ is a t-distribution with n-1 degrees of freedom?

By the way, $$s^2=\frac{\sum_{i=1}^n(Y_i-\bar{Y})^2}{n-1}$$

Answer 1

Steps:

$1.$ Because the random sample has $X_i \sim \mathsf{N}(\mu, \theta),$ where $\theta$ is the variance, we have $\bar X \sim \mathsf{N}(\mu,\, \theta/n).$ Then $U = \frac{\bar X - \theta}{\sqrt{\theta/n}} \sim \mathsf{N}(0,1).$

$2.$ Also, there is a theorem that $V = (n-1)S^2/\theta \sim \mathsf{Chisq}(n-1)$ and a related theorem that $\bar X$ and $S^2$ are independent, so that $U$ and $V$ are independent.

$3.$ Because $V/(n-1) = S^2/\theta$ is a chi-squared random variable divided by its degrees of freedom, we have $$T = \frac{U}{\sqrt{V/(n-1)}} = \frac{(\bar X - \theta)/\sqrt{\theta/n}}{S/\sqrt{\theta}} = \frac{\sqrt{n}(\bar X - \theta)}{S}.$$

The two theorems (in $2$), about the chi-squared distribution related to $S^2$ and the independence of $\bar X$ and $S^2,$ are needed for your proof. Proofs are via moment generating functions or matrix transformations. I assume you have covered these theorems.

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Note: Some students find "counterintuitive" either the 'loss' of a degree of freedom in chi-squared (from $n$ to $n-1$) or the stochastic independence of the quantities $\bar X$ and $S^2 = \frac{1}{n-1}\sum_i (X_i = \bar X)^2,$ which are clearly not functionally independent. Here is an illustration based on $m = 50,000$ samples of size $n=5$ from a normal population with mean 100 and SD $\sqrt{\theta} = 15.$

In the figure below, the first plot is a histogram of the $m$ values of $V$ showing a good fit to $Chisq(4)$ (green curve) and a bad fit to $Chisq(5)$ (red); the second is a scatterplot of $m$ pairs $(\bar X, S)$ showing lack of association; the third is a histogram of $m$ values of $T,$ derived from the $\bar X$'s and $S$'s (four extreme outliers off scale), along with the PDF or $\mathsf{T}(4).$