How do I determine the distribution of an additional sample $x_{n+1}$ given sample mean, variance $\bar x, s^2$ of a normally distributed $x$? Suppose I draw $n$ samples $x_1...x_n$ of a random variable $x$, which is normally distributed with an unknown mean $\mu$ and variance $\sigma^2$. From those samples, I compute a sample mean $\bar x$ and a sample variance $s^2$. I wish to compute the distribution of $x_{n+1} \mid \bar x, s^2$, i.e. the distribution of one additional sample $x_{n+1}$, given my measured $\bar x$ and $s^2$.
How should I proceed? This is probably a bad way of proceeding with loads of issues, but here is my thought process so far:

*

*Start with the fact that $\frac{\bar{x} - \mu}{s/\sqrt{n}}$ is $t$-distributed, and then use that to find a likelihood of $\mu$ given $\bar x$.

*Use the fact that $\frac{x_{n+1} - \mu}{\sigma} \sim \mathcal N(0,1)$.

*Use the fact that $\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1}$

*Combine these facts to get a distribution of $x_{n+1}$.

An alternative approach I thought of goes as follows:

*

*$x_{n+1}\sim N(\mu, \sigma^2)$

*$\bar x \sim N(\mu, \frac{\sigma^2}{n})$

*$(x_{n+1} - \bar x) \sim N(0, \sigma^2 + \frac{\sigma^2}{n})$

*Find the distribution of $\sigma^2 \mid s^2$ using the fact that $\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1}$

*Integrate $N(0, \sigma^2 + \frac{\sigma^2}{n})$ over all possible $\sigma^2$ weighted by the distribution of $\sigma^2 \mid s^2$

I believe that the following holds:
$$ T^*:= \sqrt{\frac{n}{n+1}}\frac{(X_{n+1}-\bar{X_n})}{s_n} \sim t_{n-1}$$
So if you define the test statistic $T^*$, you can thus derive confidence intervals, point estimates, and other quantities of interest with respect to $X_{n+1}$.
Proof:
First, note that $X_{n+1}, \bar{X_n}, s_n$ are all mutually independent.
Then, $\bar{X_n} \sim N(\mu, \frac{\sigma^2}{n})$, $X_{n+1} \sim N(\mu, \sigma^2)$, which implies (per your footnote) that $X_{n+1} - \bar{X_n} \sim N(0,\frac{n+1}{n} \sigma^2)$.
Thus, $$Z:= \frac{1}{\sigma} \sqrt{\frac{n}{n+1}} (X_{n+1} - \bar{X_n}) \sim N(0,1)$$
It's well known that $$V:= (n-1) \frac{s^2_n}{\sigma^2} \sim \chi^2_{n-1}$$ where $s^2_n = \frac{1}{n-1}\sum_{i=1}^n (X_i-\bar{X})$.
Since $Z$ and $V$ are independent, by definition of the t distribution, it follows that:
$$T^* = Z \sqrt{\frac{n-1}{V}} \sim t_{n-1}$$

This can be easily verified by simulation as well: Simulate n+1 independent normals using your favorite software, look at the density plot of your test statistic $T^*$ and compare with the $t_{n-1}$ distribution.
