Why Realized Volatility from Brownian Paths Varies Around the Model Volatility

# What is the annualized realized volatility of simulated Brownian motion paths?


# What is the annualized realized volatility of simulated Brownian motion paths?












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?

## Answer by Magic is in the chain (score 3)

https://quant.stackexchange.com/a/47175

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:

$Y_0=B_0=0$

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:

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

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

$\vdots$

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

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%"