# Empirical equivalent for implied vol

Implied volatility is supposed to show volatility of the underlying over next k days where k - maturity of the option. Say our stock price is $$S_t$$ and percentage return is $$r_t$$. Then which empirical estimate below should be used to compare with implied vol ?

1. $$|(S_t - S_{t-k})/S_{t-k}|$$
2. $$\sqrt{\sum_{t=2}^{t=k}r_t^2}$$

I believe 1st shows k days volatility, since it will be equal to 0 if spot came back to the same value . However, what does 2nd (total variance) actually represent in this case ?

• implied volatilty depends also on the strike of the option so you can't have a good estimation using your formulas. To get a good estimation of future volatilty you have to look at volatilty swaps quotation that is computed using the 2nd formula! – Valometrics.com Oct 6 '20 at 17:09
• Quadratic variation might be a good proxy. – rubikscube09 Oct 8 '20 at 18:17

For option pricing in the classical Black-Scholes model, you assume the underlying stock follows Geometric Brownian Motion:

$$S_t = S_0 + \int_{h=0}^{h=t} S_h \mu dh + \int_{h=0}^{h=t} S_h \sigma dW_h = S_0 \exp \left( \mu t + 0.5 \sigma^2 t + \sigma W(t) \right)$$

Take the log of the solution above and you get:

$$\ln\left( \frac{S_t}{S_0} \right) = \mu t + 0.5 \sigma^2 t + \sigma W(t)$$

From the above, you see that the log return $$\ln \left( \frac{S_t}{S_0} \right)$$ is normally distributed with mean $$(\mu t + 0.5 \sigma^2 t)$$ and variance $$\sigma^2t$$. Therefore, if you'd like to use historical data to "calibrate" your volatility $$\sigma$$ for the B-S model, you'd need to compute the standard deviation of the log returns, not simple returns. For a historical time series of "n" days, the formula for your volatility estimator $$\hat{\sigma}$$ would be:

$$\hat{\mu} = \frac{1}{n} \sum_{i=1}^{i=n} \ln \left( \frac{S_{t_i}}{S_{t_{i-1}}} \right)$$

$$\hat{\sigma}^2= \frac{1}{n-1} \sum_{i=1}^{i=n} \left( \ln \left( \frac{S_{t_i}}{S_{t_{i-1}}} \right) - \hat{\mu} \right)^2 * 260$$

Above, we multiply by 260 because we assume 260 trading days pear year and we scale the variance of the log-returns to annualize (because the assumed unit of time in the Black-Scholes world is 1-year).

• Thank you Jan. Although I agree with your derivations, I would like to emphasize that I was asking specifically for the case of k days change, while your solution works for k=1 only and variance is effectively variance of daily (not k days) returns. Annualization does convert it to yearly volatility, but it is still based on daily returns, not k period returns. Could you please generalize to k>1 so I can accept your answer ? – Kreol Oct 6 '20 at 19:49
• Hmmm... perhaps you should clarify your question... what are you trying to do? Why interested in the k-day change? – noob2 Oct 7 '20 at 7:07
• In the Black-Scholes world, in my experience, if you're interested in Volatility over (say) 3 months, you compute a sample of daily log-returns, and then scale it by 3/12. In other words, you always end up scaling the vol computed on daily time-series. Reason for that is: (i) the model explicitly assumes that vol is proportional to time (ii) if you wanted to compute (say) quarterly vol based on non-overlapping returns, you'd often run into a problem of not long-enough historical data available. Does that answer your question? – Jan Stuller Oct 7 '20 at 7:09

"I believe 1st shows k days volatility, since it will be equal to 0 if spot came back to the same value . However, what does 2nd (total variance) actually represent in this case ?"

Your first formula is simply absolute % change over k-days. This is sometimes used to compare against the breakeven on an options position (e.g. a straddle) if you aren't going to delta hedge the option. Does this help? Not sure I fully understand the question.

In response to Jan's answer, it's common practice just to drop the mean and look at the square of returns.