Kalman and Bayes average grades
This post will look at the problem of updating an average grade as a very simple special case of Bayesian statistics and of Kalman filtering.
Suppose you’re keeping up with your average grade in a class, and you know your average after n tests, all weighted equally.
m = (x1 + x2 + x3 + … + xn) / n.
Then you get another test grade back and your new average is
m′ = (x1 + x2 + x3 + … + xn + xn+1) / (n + 1).
You don’t need the individual test grades once you’ve computed the average; you can instead remember the average m and the number of grades n [1]. Then you know the sum of the first n grades is nm and so
m′ = (nm + xn+1) / (n + 1).
You could split that into
m′ = w1 m + w2 xn+1
where w1 = n/(n + 1) and w2 = 1/(n + 1). In other words, the new mean is the weighted average of the previous mean and the new score.
A Bayesian perspective would say that your posterior expected grade m′ is a compromise between your prior expected grade m and the new data xn+1. [2]
You could also rewrite the equation above as
m′ = m + (xn+1 − m)/(n + 1) = m + KΔ
where K = 1/(n + 1) and Δ = xn+1 − m. In Kalman filter terms, K is the gain, the proportionality constant for how the change in your state is proportional to the difference between what you saw and what you expected.
Related posts
- A Bayesian view of Amazon Resellers
- Kalman filters and functional programming
- Kalman filters and bottom-up learning
[1] In statistical terms, the mean is a sufficient statistic.
[2] You could flesh this out by using a normal likelihood and a flat improper prior.
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