Say I have an asset return time series:
Jan2020: -5%
Feb2020: +5%
Mar2020: -5%
Apr2020: +5%
May2020: -5%
Jun2020: +5%
Q3 2020: +20%
Oct2020: +5
Nov2020: -5
Dec2020: +5

Note that 3 months of data is an aggregate i.e., we don't have July, Aug, Sept. Is there a simple unbiased estimator for the variance of monthly returns of such a series?


This can be done through Bayesian inference, on a maximum likelihood estimation basis. Assuming a (log)normal distribution, you could calculate the likelihood that any prior (however stupid) estimate of mu and sigma (and thus variance). Since the normal distribution is its own conjugate prior, one can always come up with an iterative better guess for the parameters. But it's iterative; and the update formulae ain't pretty.


A simpler way - that is unbiased - would be fall back on the observation that the variance calculation can be decomposed into a sum of the squares versus mean formulation. Both of which are easy to meausre.

Your mean is easy to compute, give the aggregation problem. Just third/log-third the quarterly one. Then calculate the sum of the squares of your returns - irrespective of timeframe. This just assumes temporall return independence, which is implicit measuring monthly returns in the first place. If this, then quarterly returns should be root-3 more volatile than monthly returns; and monthly variances 3x. The sum-of-squares follows.


Given this average and sum of squares, the variance is as below:

enter image description here


If variance is stationary, specifically sample variance in Q3 was the same as in other periods:

  1. Determine sample mean including missing observations as product^1/12: 1.9%
  2. Calculate unbiased variance for known observations: -11.7%, or stdev 5.3%

This is the same as ignoring Q3 altogether. I'm curious for my own education, how is this straight-forward estimate suboptimal to the maximum likelihood calculation?

  • $\begingroup$ Because my example is oversimplified. The actual situation I'm in is a data set where every observation has a different term. For example, period 1 is 3 days, period 2 is 8 days, period 3 is 4 days, etc. $\endgroup$ – pandichef Jan 23 at 18:59
  • $\begingroup$ Step 1 remains the same. Pick the most frequent period, calculate variance, and then multiply by √n to get variance for the base period (e.g. 1 day in the latest example). Or you can perform this calculation for each distinct period and calculate weighted 𝜎 at the end. $\endgroup$ – Sergei Rodionov Jan 24 at 5:41

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