Bayesian daily-return estimate with a Normal prior

Question

You estimate the true average daily return $\mu$ of a strategy, in basis points. Your prior is Normal with mean 4 and variance 2. You observe 8 trading days whose returns average 10 bps, where the daily return noise has known variance 8. Both the prior and the daily noise are Normal.

Using Bayesian updating, what is the posterior mean estimate of the average daily return $\mu$ ?

Accepted Answer

Intuition first. Normal prior plus Normal data gives a Normal answer (conjugate). The updated mean is a precision-weighted tug-of-war, and precision means "one over variance." With n days, the data's precision is n/\sigma^2 because averaging shrinks noise. The rule. With prior mean \mu0, prior variance 	au0^2, and n days averaging \bar x with per-day variance \sigma^2: 	ext{posterior mean} = \frac{\dfrac{\mu0}{	au0^2} + \dfrac{n\,\bar x}{\sigma^2}}{\dfrac{1}{	au0^2} + \dfrac{n}{\sigma^2}}. Plug in. Prior precision = 1/2 = 0.5; data precision = n/\sigma^2 = 8/8 = 1.0. Total precision = 0.5…

Bayesian daily-return estimate with a Normal prior

Your answer

More Statistics questions