I saw this following question in an exam. Take a Brownian motion simulation with drift 5% and annualized volatility of 20% for a period of 1 year. Then the annualized realized volatility of the sample path is

A. always <20%

B. always = 20%

C. = 5%

D. approximately 20%, but random

More generally, how to find the relationship between the annualized realized volatility of the simulated path and the volatility parameter in Brownian motion that generated the path?

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    $\begingroup$ It is randomly distributed according to a Chi Square distribution centered around 20%. $\endgroup$ – Alex C Aug 17 at 17:47
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    $\begingroup$ It’s a fundamental theorem in statistics that sample variance of gaussian RVs is a chi-square. All explained here stats.stackexchange.com/a/121676. $\endgroup$ – Ivan Aug 17 at 21:44
  • $\begingroup$ It's sampling statistics. If the population vol is 20%, then the vol of a random sample within that distribution will be close to but slightly different that of the population. The Chi^2 distribution is just the distribution of those sample vols if the samples are random and the underlying returns are normally distributed. $\endgroup$ – demully Aug 17 at 22:29

Let's try a simple approach, ignoring the difference between sample and population variance, and assuming the process is just the standard brownian - with no drift and sigma term. Generalisation should be easy.

We define a process Y as equal to standard brownian, but we are assuming finite sampling with difference between two observations equal to $\Delta t$. So our process starts at zero:


And the increments are normally distributed:

$ \left. Y_k \right|Y_{k-1}=N\left[Y_{k-1},\Delta t\right]$

We can write the process values at observation points $\left(t_1,t_2,\dots,t_n\right)$ using the standard normals, $Z_k$ as follows:

$Y_1=\sqrt{\Delta t}\,Z_1$

$Y_2=\sqrt{\Delta t}\left(Z_1+Z_2\right)$


$Y_n=\sqrt{\Delta t}\left(Z_1+Z_2+\dots+Z_n\right)$

Now we will calculate the sample realised variance (notice i am not paying attention to n and n-1 as per simplifying assumption!) of the process as follows:

$\sigma^2=\frac{1}{n} \sum_{k=1}^n{ \left(Y_k- E \left[ \left. Y_k \right|{Y_{k-1}}\right]\right)^2}$

Which in terms of Z's is:

$\sigma^2=\frac{1}{n} \sum_{k=1}^n{ \left(\sqrt{\Delta t} Z_k\right)^2}$

$=\frac{\Delta t}{n} \sum_{k=1}^n{ Z_k^2}$

And you know the sum of the squares of the n Normal is Chi-Square with n degree of freedom, hence the discussion in the comments. The mean is equal to DF, so if you have $n=\frac{1}{\Delta t}$ observations per year, then the average variance will be equal to $\Delta t$:

$E \left[ \sigma^2\right]=\frac{\Delta t}{n} n=\Delta t$

And your annualised variance will be simply $n \Delta t=1$. So the answer is D, as Alex put it "It is randomly distributed according to a Chi Square distribution centered around 20%"

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    $\begingroup$ Wonderful comments, wonderful answers $\endgroup$ – Fr1 Aug 19 at 2:31

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