Using a Single Return to Estimate Short-Interval Volatility

# Why is $dS/S$ an estimate of realized volatility?


# Why is $dS/S$ an estimate of realized volatility?












For one period, $dS/S$ is an estimate of realized volatility, which we can annualize by dividing with $\sqrt{\Delta t}$.

But.... why? How is $dS/S$ an estimate of volatility? Volatility is, to me, how big the random fluctuations are. But if I have a stock price that goes 100, 110, 120, 130, 140, 150, 160, every day, then that's not a random fluctuation, and the volatility is 0 .... yet, using above formula we get that the volatility is 0.1 for the first day.

## Answer by Aksakal almost surely binary (score 2)

https://quant.stackexchange.com/a/33973

The reason is that the drift component is very small compared to volatility on daily returns. Here's the details.

First, the obvious part. If you assume that the stock prices are lognormal process, then you have: $$r_t=\Delta S/S_t=\mu \Delta t+\sigma_{\Delta t}\xi_t$$ where $\xi_t\sim\mathcal{N}(0,1)$ - the standard normal random variable, and $r_t$ - periodic return.

The equation for the variance is: $$Var[r_t]=Var[\mu\Delta t]+\sigma_{\Delta t}^2Var[\xi_t]=\sigma_{\Delta t}^2$$

We got this far to see that the variance of the returns is what we're looking for. Now, we need an estimator. Let's look at the well known equation for the variance: $$Var[x]=E[x^2]-E[x]^2$$ If we have only one observation then the estimator for first term looks like this when we start with a definition: $$\bar E[r_t^2]=\frac{1}{1}\sum_{t=1}^1r_t^2=r_1^2$$

Now the second term happens to be very small on daily returns and is often ignored $E[r_1]=E[\mu\Delta t]+E[\sigma_{\Delta t}\xi_t]=\mu\Delta t$, so we're going to drop it.

Hence, the simple estimator of the volatility $$\hat\sigma_{\Delta t}=\sqrt{r_1^2}=r_1=\frac{\Delta S_1}{S_1}$$

This only works because the daily volatility is much bigger than the daily drift, e.g. for SPX your daily vol is on the order of $\sigma_{daily}\approx 0.2/\sqrt{250}\approx 0.01>>\mu_{daily}\approx 0.12/250\approx 0.0005$. That's why this estimator is quite good despite certainly being biased. If you use longer periods then this estimator will not work as well. For instance for annual SPX return you get $\sigma\approx 0.2\sim\mu\approx 0.12$, i.e. quite similar magnitudes, so the second term in the variance equation $E[r]^2$ can't simply be dropped.

Here, I'm assuming you have no problems with time scaling $\sqrt{\Delta t}$ and only are interested in why we can use just one return observation as the estimator of its volatility.

## Answer by NSZ (score 0)

https://quant.stackexchange.com/a/33383

The basic assumption behind this estimate is that the stock price has to be lognormally distributed: $$dS_t/S_t=\mu dt+ \sigma dW_t$$ Therefore if you estimate the standard deviation of the time series you would get an estimate of $\sigma \sqrt{\Delta t}$, then to get the annualized volatility just divide by $\sqrt{\Delta t}$. The problem of your stock is that the distribution is not lognormal. You have a process normally distributed : $$dS_t=\mu dt+ \sigma dW_t$$ with $\sigma=0$ and $\mu=3650$ so that in one day you go up by 10. However you have $$dS_t/S_t=1/S_t(\mu dt+ \sigma dW_t)$$ Therefore you cannot estimate volatility using $\Delta S_t/S_t$ but you should just use $\Delta S_t$ in this case.

## Answer by Chris Degnen (score 0)

https://quant.stackexchange.com/a/33974

Contextually related info, from Computational Financial Mathematics using Mathematica, page 54.

`mean value = p e^(a t)` depends on `σ^2 = dS^2/S^2 dt`