# Prediction interval with non normal data

(In the context of simple or multiple linear regression) Perhaps a very simple question to most of you. I know that if your sample size is large enough, despite your data not following a normal distribution, it would be viable to still compute the confidence interval due to the Central Limit Theorem.

What I heard is that this is not the case for a prediction interval. Would someone be able to explain me in plain English what the reason for this is? And if possible provide some mathematics after? :)

Much appreciated!

• After repeating your experiment many times, you can characterize its results accurately and precisely: that's the CLT (or strong LLN) in a nutshell. You cannot, however, make the results of the next repetition completely certain: it will be just as random as every other repetition of your experiment. For the mathematics, please search our site for "prediction interval." (For instance, I worked out a Normal prediction interval from first principles at stats.stackexchange.com/questions/252046. The same reasoning applies to any distribution family, Normal or not.)
– whuber
Commented Dec 21, 2016 at 15:17
• Thanks for yor response. In a textbook I have, it says: "The prediction limits, unlike the confidence limits for a mean response E(Y) are sensitive to departures from normality of the error terms distribution." Commented Dec 21, 2016 at 16:46
• In this case I think that Bill Huber's answer is relevant to this problem of prediction but it is not to the question by the OP of the referenced post. Please see my new comments about outlier detection in the post. Commented Dec 21, 2016 at 17:32
• Bill Huber's point is that prediction for a new Y has greater variability than the estimation of E(Y). In regression you have an unknown new Y but an observed value or vector of values for X. This also means that a prediction interval for the new Y should be wider than the confidence interval for E(Y). Commented Dec 21, 2016 at 18:21

Recall the differences between prediction and confidence intervals: a confidence interval is an interval for a mean which converges to a fixed value as the sample size goes to infinity. The prediction interval is an interval for an observation which goes to a non-singular interval as the sample size goes to infinity.

Since the confidence interval applies to a mean, the CLT gives us a normal approximate sampling distribution for that sample mean. If you incorrectly apply a normal approximation to a prediction interval, you do not account for the heteroscedasticity, skewness, or the kurtosis of the distribution, or even the scale of the response levels. For instance, if I have binary data, it makes sense to create a normal interval for the sample proportion. But no amount of correction will give me a sensible "prediction" interval for a binary valued response: you would (instead) report the probability $Y=1$ or $0$.

Consider a very simple case -- constructing a confidence interval for a common population mean compared to constructing a prediction interval for a single new observation.

With the assumption of independence and a common distribution with finite variance ($Y_i\sim F$, with $E(Y_i)=\mu,\text{Var}(Y_i)=\sigma^2$), then in the limit as $n\to\infty$, $\frac{\bar{Y}-\mu}{s/\sqrt{n}}$ will converge to a standard normal random variate, and from that we can back out an approximate large-sample confidence interval for $\mu$ based on a sample.

In particular, for a certain skewed distribution, here's a simulated distribution of sample means for $n=50$ (NB there's nothing special about $n=50$, or any other specific sample size):

We see that in this case the sample means appear not very far from normal even at n=50 (still slightly skew though, as the original distribution was pretty skewed). In cases similar to this one, "large sample" methods for a confidence interval may work reasonably well at sample sizes of a few dozen. In other cases it might require substantially larger samples.

However, even if we knew $\mu$ exactly, we won't know from results like the above where $Y_{n+1}$ might be. We can't make use of such arguments as we used for the mean because we are dealing with the distribution of a single observation. We can still construct a pivotal-type quantity like $\frac{Y_{n+1}-\bar{Y}}{s/\sqrt{n}}$, but there's no large sample result to invoke here.

[That's not to suggest nothing whatever can be done to try to construct some form of interval for a future observation, just that we can't invoke large-sample results relating to convergence of standardized means to normality to do it. If you have a distributional model for $F$, you may be able to construct a sensible interval for a future observation. Or you may be able to construct some form of nonparametric interval in some situations]

Somewhat similar issues hold for say a regression model.