Is a 1-sided 90% prediction interval equivalent to a 2-sided 95% prediction interval?

I have located references that indicate that a 1-sided confidence band is equivalent to the 95% 2-sided confidence band on a linear regression. Does this equivalence hold true for prediction intervals over a linear regression model?

Example: Given a set of data with a random sampling of age vs strength values, and using a simple linear regression on the above. If I state: "I want to be 95% sure that future values of strength where age=39 are >= XX", to find XX, can I plot a typical 90% prediction interval, and use it's lower band as the equivalent of 95% 1-sided prediction interval to answer that question? Any references available for such an equivalence?

• It looks like the title and the first paragraph may have (confusingly) switched the percentages: a one-tailed 95% prediction interval can be derived from a symmetric two-tailed 90% prediction interval, but not the other way around. Any reference will document this because the equivalence follows immediately from the definitions.
– whuber
Jan 8 '12 at 4:29

[Edited: This answer has been improved since original posting based on comments.]

You have this backwards. If you have a 95% prediction interval, then you expect that 2.5% of future observations will have values larger than the upper limit, and 2.5% of future observations will have values smaller than the lower limit. If you don't care about values that are smaller than the lower limit, then you can just use the upper limit as the top bound of a 97.5% one-sided prediction interval, as the same percentage will exceed it whether you care about the lower bound or not. On the other hand, if you want to be able (later) to say that 95% of future observations will be below some limit, then you need to get a 90% prediction interval (and subsequently ignore the lower limit).

Note: standard equations for calculating a prediction interval following an OLS regression fit assume that the data (specifically the residuals) follow a normal distribution with mean = 0, and variance = $\sigma^2$. (This means, among other things, that the distribution is symmetrical.) This answer is assuming that as well. If your residuals are not normal, a bootstrapping approach may be appropriate. For example, you could simulate a bootstrapped distribution, sort it, and take the 95th percentile as the upper limit of a 95% prediction interval.

• You assume the two-sided intervals are symmetric (in probability). Note, too, that the question asks specifically about prediction intervals, not confidence intervals. Please note that neither type of interval has the property you ascribe to them in your second sentence. Your definition sounds like a kind of Bayesian tolerance interval.
– whuber
Jan 8 '12 at 4:50
• You're right. Given asymptotic normality, and that the question appears to have OLS regression in mind, I am assuming that the intervals are symmetric. Further, I understand the differing conceptions between Bayesian and frequentist approaches to derive partly from the fact that for parameter estimation, parameters are typically not thought of as random variables by frequentists, but are by Bayesians. However, in some situations (such as predictions of future observations) they are thought of as random variables by both traditions, and so the different interpretations converge. Jan 8 '12 at 5:18
• @whuber, if these assumptions are sufficiently inaccurate that they could lead CV users astray, I can delete this answer. Jan 8 '12 at 15:14
• The main point of my comment is that you use the term "confidence interval" incorrectly in your reply. The characterizations in your second and last sentences are wrong.
– whuber
Jan 8 '12 at 17:30
• @whuber, that's true. Upon rereading the question, I see that it does explicitly ask about a prediction interval. I suppose I just reflexively think 'confidence interval' out of habit. I have edited my answer in keeping with your suggestions. Jan 8 '12 at 18:24

For a regression analysis of a normally distributed variable X it is true because the distribution of X and the distribution of the residuals are both symmetric (X is normal and residuals have a t distribution). Because it is symmetric the 90% two sided prediction interval will have 5% in each tail. A 95% one sided prediction interval will have 5% in one tail so this works out as having the same limit as the two sided 90% prediction interval