• Posted by Konstantin 28.01.2009 5 Comments

    Imagine that you have just derived a novel IQ test. You have established that for a given person the test produces a normally-distributed unbiased estimate of her IQ with variance 102. That is, if, for example, a person has true IQ=120, the test will result in a value from a N(120,102) distribution. Also, from your previous experiments you know that among all the people, IQ has a N(110,152) distribution.

    One a sunny Monday morning you went out on a street and requested the first bypasser (whose name turned out to be John) to take your test. The resulting score was t=125. The question is: what can you conclude now about the true IQ of that person (assuming, of course, that there is such a thing as a "true IQ"). There are at least two reasonable approaches to this problem.

    1. You could apply the method of maximum likelihood. Here's John, standing beside you, and you know his true IQ must be some real number a. The test produced an unbiased estimate of a equal to 125. The likelihood of the data (i.e. the probability of obtaining a test score of 125 for a given a) is therefore:

          \[P[T=125|A=a]=\frac{1}{\sqrt{2\pi 10^2}}\exp\left(-\frac{1}{2}\frac{(125-a)^2}{10^2}\right)\]

      The maximum likelihood method suggests picking the value of a that maximizes the above expression. Finding the maximum is rather easy here and it turns out to be at a=125, which is pretty natural. You thus conclude that the best what you can say about John's true IQ is that it is approximately 125.

    2. An alternative way of thinking is to use the method of maximum a-posteriori probability, where instead of maximizing likelihood P[T=125|A=a], you maximize the a-posteriori probability P[A=a|T=125]. The corresponding expression is:

       \begin{multiline} P[A=a|T=125] \sim P[T=125|A=a]\cdot P[A=a] = \\ = \frac{1}{\sqrt{2\pi 10^2}}\exp\left(-\frac{1}{2}\frac{(125-a)^2}{10^2}\right)\cdot \frac{1}{\sqrt{2\pi 15^2}}\exp\left(-\frac{1}{2}\frac{(110-a)^2}{15^2}\right) \end{multiline}

      Finding the required maximum is easy again, and the solution turns out to be a=120.38. Therefore, by this logic, John's IQ should be considered to be somewhat lower than what the test indicates.

    Which of the two approaches is better? It might seem utterly unbelievable, but the estimate provided by the second method is, in fact, closer to the truth. The straightforward "125", proposed to by the first method is biased, in the sense that on average this estimate is slightly exaggerated. Think how especially unintuitive this result is from the point of view of John himself. Clearly, his own "true IQ" is something fixed. Why on Earth should he consider "other people" and take into account the overall IQ distribution just to interpret his own result obtained from an unbiased test?

    To finally confuse things, let us say that John got unhappy with the result and returned to you to perform a second test. Although it is probably impossible to perform any real IQ test twice and get independent results, let us imagine that your test can indeed be repeated. The second test, again, resulted in a score of 125. What IQ estimate would you suggest now? On one hand, John himself came to you and this time you could regard his IQ as a "real" constant, right? But on the other hand, John is just a person randomly picked from the street, who happened to take your test twice. Go figure.

    PS: Some additional remarks are appropriate here:

    • Although I'm not a fan of The Great Frequentist-Bayesian War, I cannot but note that the answer is probably easier if John is a Bayesian at heart, because in this case it is natural for him to regard "unknown constants" as probability distributions and consider prior information in making inferences.
    • If it is hard for you to accept the logic in the presented situation (as it is for me), some reflection on the similar, but less complicated false positive paradox might help to relieve your mind.
    • In general, the correct way to obtain the true unbiased estimate is to compute the mean over the posterior distribution:

          \[E[a|T=125] = \int a \mathrm{dP}[a|T=125]\]

      In our case, however, the posterior is symmetric and therefore the mean coincides with the maximum. Computing the mean by direct integration would be much more complicated.

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