I receive lots of error reports for my manuals. Often, these are typos or minor grammatical errors. Sometimes, the arithmetic in a solution is incorrect. But the most interesting errors are mathematical errors. For example, I used to refer in one of my manuals to an unbiased sample standard deviation, until a student pointed out to me that no such thing exists. One can calculate an unbiased sample variance by dividing the sum of square deviations from sample mean by n − 1 rather than by n, but once one applies any nonlinear operation to an unbiased estimator, it is no longer unbiased. So I had to change my wording to “square root of unbiased sample variance”.
Here is a recent example of something that started out as a typo, then became a mathematical erratum, and ended up as just a typo. In the first two printings of the Exam C manual 16th edition, a question from CAS 4B Spring 1998 was listed as exercise 33.36. The question was:
You are given the following: The random variable X has the density function
A random sample of size n is taken of the random variable X.
Determine the limit of â as the sample mean goes to infinity, where â is the maximum likelihood estimator of α.
A. 0 B. 1/2 C. 1 D.2 E. ∞
Exam C students know that the maximum likelihood estimator of
In the manual, the estimator was written n/Πxi. A student wrote me regarding this typo. However, as I reviewed the question, a more frightening thought came to my mind. The sum of ln(1 + xi) is the logarithm of the product of 1 + xi. So it is like a geometric mean. We all know that the geometric mean is less than or equal to the arithmetic mean. So just because the sample mean goes to infinity, why must this denominator go to infinity? It seems like it need not go to infinity, since the geometric mean is less than the (arithmetic) sample mean! And then, the answer is not necessarily A! To clinch the fact that the question was defective, I made up a counterexample, a case in which the sample mean goes to infinity but n/Πln(1+xi) does not go to 0. Namely, suppose the sample consists of n−1 very tiny numbers, say, and one number equal to . Then the sample mean is n, which goes to infinity as n →∞.
Now, ln(1 + ε) is essentially 0, and ln( ) is essentially lnn2 = 2lnn. Then n/Πln(1 + xi) = n/(2lnn), and we know from calculus that this goes to ∞, not 0, as n →∞.
So this old exam question is defective! I removed it, and it does not appear in the Exam C 16th edition 3rd printing manual.
Ah, but I reacted too quickly. Do you see why the question is not defective?
The question, in its second bullet, says that a random sample of size n is taken. It does not say that the sample size goes to infinity. The sample size is fixed. Only the data are varied. The xi are changed so that the sample mean gets higher and higher. But then at least one xi is greater than or equal to the sample mean, and Πln(1 + xi) is then greater than or equal to ln(1 + x). As x goes to infinity, ln(1 + x) will go to infinity and the maximum likelihood estimator must therefore go to 0.
So the old exam question is a good question after all. It will be restored to the next version of the Exam C manual.