# Conditioning within definition explanation

I have a doubt on the meaning of a conditioning within a definition.

In a book I've found the following definition of upper tolerance limit:

$P(P(X<\bar X+kS|\bar X, S)>p)=1-\alpha$

where $X$ is a random variable, $\bar X$ is the sample mean of values taken from the distribution of $X$ and $S$ is their standard deviation. $p$ and $\alpha$ are numbers between 0 and 1.

The question is:

Why do we need the condition on $\bar X$ and $S$, and what does this conditioning mean in this context?

I.e. What's wrong with just writing the definition as

$P(P(X<\bar X+kS)>p)=1-\alpha$

This is just an obfuscating way to express a simple idea: an upper tolerance limit is just an upper confidence limit for a percentile.

The random variable $$X$$ is supposed to have some definite (but unknown) distribution $$F$$. The statistic $$\bar X + k S$$ (derived from a random sample from $$F$$) is intended to estimate the $$p^\text{th}$$ percentile of $$X$$, $$F^{-1}(p)$$. That is, we hope that

$$F(\bar X + k S) = p.$$

Of course that won't be exactly true, because $$\bar X + k S$$ is random. An upper tolerance limit is a procedure intended not to underestimate $$F^{-1}(p)$$. The value of $$\alpha$$ is the chance you can tolerate of the procedure being wrong. In other words, you want it to overestimate its target at least $$1-\alpha$$ of the time. In many cases you can choose $$k$$ to assure this chance is exactly $$1-\alpha$$. Thus,

$${\Pr}_F(F(\bar X + k S) \ge p) = 1-\alpha\tag{1}$$

is the defining criterion for a "$$1-\alpha$$ confidence upper tolerance limit of coverage $$p$$." In English we could read it as

There is a $$1-\alpha$$ chance that the true percentile corresponding to the sample statistic $$\bar X + k S$$ will exceed $$p$$.

If you wanted to make expression $$(1)$$ look more complicated, you could unravel it using the definition of $$F$$; to wit,

$$F(z) = {\Pr}_F(X \le z)\tag{2}$$

for any real number $$z$$. Fixing $$z = \bar X + kS$$ for the moment and plugging it into $$(2)$$ would give

$$F(\bar X + k S) = {\Pr}_F(X \le \bar X + k S).$$

That's a mighty ambiguous expression, though, because $$F$$ determines the distribution of both $$X$$ (thought of as an abstract random variable in $$(2)$$) as well as the distribution of $$\bar X + k S$$ (because that is determined by a random sample from $$F$$). To make it clear we are talking in this context only of $$X$$ as the random variable, with $$\bar X + k S$$ being treated as a constant, we might write

$$F(\bar X + k S) = {\Pr}_F(X \le \bar X + k S\,|\, \bar X, S).$$

Plugging this into $$(1)$$ gives an expression like that in the book. (It differs only in that I have been more careful in distinguishing $$\ge$$ and $$\gt$$, but that is of no matter.)

### References

A standard book is Hahn & Meeker, Statistical Intervals, A Guide to Practitioners (John Wiley & Sons, 1991). Here is its explanation:

The following characterizes a tolerance interval that one can claim contains a proportion $$p$$ of the population with $$100(1-\alpha)\%$$ confidence: "If one calculated such intervals from many independent groups of random samples, $$100(1-\alpha)\%$$ of the intervals would, in the long run, correctly include at least $$100p\%$$ of the population values..."

• Actually $\bar X$ and $S$ are random variables and not constants... So this is basically just a "formal" way to express the concept that we are focusing our attention on the distribution of $X$ rather than on that of $\bar X+kS$ within the $Pr_F$ operator?
– xanz
Commented Jun 11, 2016 at 17:00
• Actually before you take the sample $\bar X$ and $S$ are random variables. All the uncertainty lies in them. The confusing part of how your book defines a UTL is that "$X$" refers to an abstract construct and doesn't even need to appear in the definition $(1)$, which is based solely on $F$.
– whuber
Commented Jun 11, 2016 at 19:46
• Ok, I just had a talk with a professor in stochastic mechanics who told me that the definition of the book (the one with the conditioning) is kind of "weird" and unclear for a matematician and thus not suitable to be published in a paper (exactly I didn't understand what he was complaining about)... Honestly to me this seems right (even if maybe a little bit confusing at first) but I'm no expert. What is your opinion on this?
– xanz
Commented Jun 15, 2016 at 14:37
• The book's definition could be fixed up by first explaining that the author is considering the multivariate distribution of $(\bar X, S, X)$ where $X$ is independent of the $X_i$ from which $\bar X$ and $S$ are derived and has the same distribution as the $X_i$. I happen to think that the reference to $X$ is superfluous (as well as potentially confusing) and would recommend using a more traditional (simpler, clearer) definition similar to that given in my answer. If you would like a reference, please see Hahn & Meeker, Statistical Intervals.
– whuber
Commented Jun 15, 2016 at 14:42