Given historical asset prices at consistent time intervals, one can estimate annual volatility as:

SampleStDev(log(Si/Si-1)) / sqrt(interval)

What's the correct way to do this when the time intervals are inconsistent (e.g. S2 was observed 1 day after S1, but S3 was observed 5 days later)? Should longer time intervals get greater weight in the estimate?


2 Answers 2


To make @nbbo2's answer more precise, let's assume that we observe various sums $z_{k,h}$ of independent and identically distributed random variables $x_i$ (i.e. returns),

$$ z_{k}\equiv \sum_{i=k-h_k+1}^k x_i $$ where $h_k$ is the horizon of aggregation. For simplicity, let's aggregate over one, two, three, ... individual daily returns, $x_i$. On a "normal" day, we'd have $h=1$, and on a Monday, we usually have $h=3$.

Assuming normally distributed individual return contributions, and "smallest time fraction" $\Delta$, e.g. $\Delta = 1/255$, we have

$$ x_i\sim N(\mu\Delta,\sigma^2\Delta)\Rightarrow z_k\sim N(\mu\Delta_k,\sigma^2\Delta_k) $$ where $\Delta_k=\sum_{i=k-h_k+1}^k\Delta=\Delta \times h_k$.

We can now find the maximum-likelihood-estimators. Given $n$ observations $z_1,\ldots,z_n$ and knowledge of all the individual observation windows $\Delta_1,\ldots,\Delta_n$ the log likelihood is

$$ l(z)=-\frac{n}{2}\ln(2\pi)-\frac{n}{2}\ln(\sigma^2)-\frac{n}{2}\sum_k^n\ln(\Delta_k)-\frac{1}{2}\sum_k^n\frac{(z_k-\mu\Delta_k)^2}{\sigma^2\Delta_k} $$

The maximum likelihood estimators are found as:

$$ \hat{\mu}=\frac{\sum_k z_k}{\sum_k \Delta_k} $$ and $$ \hat{\sigma^2}=\sum_k\frac{\left(z_k-\hat{\mu}\Delta_k\right)^2}{n\Delta_k} $$

  • $\begingroup$ In the mean time, I had decided to just take a weighted variance Σw(x-u)^2/ΣΔk with weight equal to 1/SE^2 = Δk. It's interesting to note that this turns out equivalent to the MLE estimator. $\endgroup$ Commented May 13, 2022 at 1:56
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    $\begingroup$ When trying to apply your formula, I realized I had assumed you meant zk instead of xk in your variance formula. If you did, can you edit it, and if not, can you restate your formula in terms of z's since x's are unobservable? $\endgroup$ Commented May 13, 2022 at 3:33
  • $\begingroup$ Thanks for spotting that. $\endgroup$ Commented May 13, 2022 at 4:32

Assume GBM. Suppose there are $n+1$ observations, indexed by $j=0,\cdots,n$. The stock price is $S_j$ at time $T_j$ where time is measured in yearly units.

The annualized variance (assuming zero mean) is

$V=\frac{1}{T_n-T_0}\sum_{j=1}^n \ln(S_j/S_{j-1})^2/(T_j-T_{j-1})$

this is true whether the intervals are equal (i.e. $T_j-T_{j-1}=\Delta T, \forall j$) or unequal but fixed in length (as long as they are nonrandom).

When the intervals are equal it simplifies to the more familiar form

$V=\frac{1}{\Delta T} Var(\ln(S_j/S_{j-1})) $ since $T_n-T_0=n\cdot\Delta T$.


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