# Trouble proving$\frac{X_{n+1}-\bar{X}}{S\sqrt{1+\frac{1}{n}}}$ is T(n-1)

Suppose $$x_{n+1}$$ is an additional observation, which is distributed as $$N(\mu,\sigma^2)$$, and is independent of $$X_1,X_2,...,X_n$$, also $$X_1,X_2,...,X_n$$ are iid from $$N(\mu,\sigma^2)$$.

I have tried to change the form of $$\frac{X_{n+1}-\bar{X}}{S\sqrt{1+\frac{1}{n}}}$$, in order to get N(0,1) on top and Chi-square at the bottom, however, my effort went nowhere. $$\frac{X_{n+1}-\bar{X}}{S\sqrt{1+\frac{1}{n}}}$$=$$\frac{X_{n+1}-\bar{X}}{\sigma\sqrt\frac{S^2}{\sigma^2}\sqrt{1+\frac{1}{n}}}$$

Got stuck at dealing with $$\frac{X_{n+1}-\bar{X}}{\sigma\sqrt{1+\frac{1}{n}}}$$

Can anyone help to take a look at it?

Note that $$X_{n+1}-\bar{X}\sim N(0, \sigma^2(1+n^{-1}))$$ where the second parameter is variance since a linear combination of independent normals is normally distributed. It follows that $$Z=\frac{X_{n+1}-\bar{X}}{\sigma\sqrt{1+n^{-1}}}\sim N(0, 1).$$ Next if $$S^2$$ denotes the sample variance of $$X_{1}, \dotsc, X_n$$, we have that $$W=\frac{(n-1)S^2}{\sigma^2}\sim \chi ^2_{(n-1)}$$ and further $$Z$$ is independent of $$W$$ whence $$T=\frac{Z}{\sqrt{W/n-1}}=\frac{X_{n+1}-\bar{X}}{S\sqrt{1+\frac{1}{n}}}\sim t_{(n-1)}$$ as desired.