Bayesian test-ability estimate with a Normal prior

Question

You estimate a student's true ability level $\mu$ on a subject, measured on a 0 to 100 scale. Your prior is Normal with mean 70 and variance 25 (standard deviation 5). The student takes one exam and scores 90, but exam scores are noisy with variance 75. Both the prior and the exam noise are Normal.

Using Bayesian updating, what is the posterior mean estimate of the true ability $\mu$ ?

Accepted Answer

Intuition first. A Normal prior meeting Normal data gives a Normal answer (conjugate). The updated mean is a tug-of-war between prior and measurement, won by whichever is more precise. Precision means "one over variance": smaller variance equals more pull. The rule. With prior mean \mu0 and variance 	au0^2, and measurement x with variance \sigma^2: 	ext{posterior mean} = \frac{\dfrac{\mu0}{	au0^2} + \dfrac{x}{\sigma^2}}{\dfrac{1}{	au0^2} + \dfrac{1}{\sigma^2}}, a precision-weighted average of the two values. Plug in. Prior precision = 1/25 = 0.04; exam precision = 1/75 \approx 0.01333. To…

Bayesian test-ability estimate with a Normal prior

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