Empirical equivalent for implied vol

Question

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 ?

Answer 1

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).

Answer 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 ?"

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.