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:

  1. 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 $\bar x$.
  2. Use the fact that $\frac{x_{n+1} - \mu}{\sigma} \sim \mathcal N(0,1)$.
  3. Use the fact that $\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1}$
  4. Combine these facts to get a distribution of $x_{n+1}$.

An alternative approach I thought of goes as follows:

  1. $x_{n+1}\sim N(\mu, \sigma^2)$
  2. $\bar x \sim N(\mu, \frac{\sigma^2}{n})$
  3. $(x_{n+1} - \bar x) \sim N(0, \sigma^2 + \frac{\sigma^2}{n})$
  4. Find the distribution of $\sigma^2 \mid s^2$ using the fact that $\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1}$
  5. Integrate $N(0, \sigma^2 + \frac{\sigma^2}{n})$ over all possible $\sigma^2$ weighted by the distribution of $\sigma^2 \mid s^2$
  • 3
    $\begingroup$ Somewhere a dead monk named Bayes is turning in his grave at these hoops the Frequentists have made you jump through. $\endgroup$
    – bdeonovic
    Oct 28, 2021 at 18:55
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    $\begingroup$ The point about Bayes is apt because the question as posed does not have a solution: the conditional distribution depends on the unknown parameters. The classical approach is to base testing or prediction on the estimated joint distribution of $(\bar x, x_{n+1}),$ while the Bayesian approach updates a prior distribution on the parameters--and gives the kind of answer you are looking for. $\endgroup$
    – whuber
    Oct 28, 2021 at 19:07
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    $\begingroup$ oh no its not your fault @Danny, I'm just lamenting the state of statistical education $\endgroup$
    – bdeonovic
    Oct 28, 2021 at 19:13
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    $\begingroup$ You don't have "samples" $x_1,\ldots,x_n$; rather you have a sample $x_1,\ldots,x_n.$ And $x_{n+1}$ is not one additional sample; it is one additional observation. $\endgroup$ Oct 28, 2021 at 20:24
  • 2
    $\begingroup$ The word sample has a particular meaning in statistics. You claimed to be a lowly physicist but then attempt to correct a statistician on the meaning of the term in this statistical setting.... That's a risky strategy. en.wikipedia.org/wiki/Sample_(statistics) $\endgroup$
    – Glen_b
    Oct 29, 2021 at 1:37

3 Answers 3


The distribution of $x_{n+1}$ conditional on $\overline x=(x_1 +\cdots +x_n)/n$ and $s^2 = \big((x_1-\overline x)^2+\cdots+(x_n-\overline x)^2\big)/(n-1)$ is $\operatorname N(\mu,\sigma^2),$ since the observations $x_1,\ldots,x_n,x_{n+1}$ are independent. Thus I suspect that what you actually want is the distribution of $(x_{n+1}-\overline x)/s,$ upon which you can base a prediction interval whose endpoints are $\overline x\pm c\cdot s_n,$ where $c$ is chosen so as to get a desired probability that $x_{n+1}$ is in the interval.

Notice that $x_{n+1}-\overline x\sim\operatorname N\left(0,\sigma^2\left( 1 + \frac 1 n \right)\right),$ and since $s^2$ is independent of $\overline x$ and of $x_{n+1},$ and $(n-1)s^2/\sigma^2\sim\chi^2_{n-1},$ you have $$ \frac{(x_{n+1}-\overline x)/\sqrt{1+\frac 1 n}}{s/\sqrt n} \sim t_{n-1}. $$ Thus if you choose $c$ so that $\Pr(-c<t_{n-1}<+c)= 1-\alpha,$ then you have $$ \Pr\left(x_{n+1} \text{ is between } \overline x \pm c\cdot\frac s {\sqrt n}\cdot\sqrt{1+\tfrac 1 n} \right) = 1-\alpha. $$ This is a prediction interval for $x_{n+1}.$

  • $\begingroup$ I think precisely, I want the distribution of $x_{n+1} - \overline x$ in terms of $s$ rather than $\sigma$. Maybe I'm thinking about it incorrectly, but either way, your answer gets me there, along with @Mang's. $\endgroup$ Oct 28, 2021 at 23:30

As @whuber points out, this question is much more naturally answered in a Bayesian context. In that case the distribution you are interested in is known as the posterior predictive distribution.

$$ p(x_{n+1}|x_1,\ldots,n) = \int_\theta p(x_{n+1}|\theta,x_1,\ldots,x_n)p(\theta|x_1,\ldots,x_n)d\theta $$

where $\theta = (\mu,\sigma^2)$. This is about as close as you can get in the general case. If you are willing to accept a particular prior distribution for $\mu$ and $\sigma^2$ (Normal for $\mu$ and Inverse-Gamma for $\sigma^2$) you can get a closed form distribution (it is a $t$ distribution) which you can look up here


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 $X_{n+1}$.

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.


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