# A Prediction Interval for a Single Future Value For a Normal Distribution

Posted by Beetle B. on Sun 16 July 2017

$$\newcommand{\Cov}{\mathrm{Cov}}$$ $$\newcommand{\Corr}{\mathrm{Corr}}$$ $$\newcommand{\Sample}{X_{1},\dots,X_{n}}$$

What if you have $$n$$ observations in a normal distribution and want to predict $$X_{n+1}$$?

The prediction interval is $$\bar{x}\pm t_{\alpha/2,n-1}s\sqrt{1+\frac{1}{n}}$$ for a $$100(1-\alpha)$$

To derive this, see that the expected value of $$X_{n+1}$$ is $$\mu$$. Then $$E(\bar{X}-X_{n+1})=\mu-\mu=0$$ and $$V(\bar{X}-X_{n+1})=V(\bar{X})+V(X_{n+1})=\frac{\sigma^{2}}{n}+\sigma^{2}$$

As $$\bar{X}-X_{n+1}$$ is a linear combination of independent normal distributions, it is itself normal, with $$\sigma^{2}\left(1+\frac{1}{n}\right)$$ as the variance. Replace with $$s\sqrt{1+\frac{1}{n}}$$ and treat it as a t-distribution.

The prediction interval is wider than the confidence interval. This is because the former has two random variables, whereas the latter has one.

Important point: As $$n$$ approaches $$\infty$$, the confidence interval converges to $$\mu$$. But the PI approaches $$\mu\pm z_{\alpha/2}\sigma$$. There is always uncertainty in it.