Comparing Historical Volatility with Implied Volatility
Summary
The document explains how to estimate historical volatility for comparison with option implied volatility under the Black–Scholes framework. Assuming geometric Brownian motion, log returns have variance that grows with time, so the answer recommends estimating volatility from a time series of log returns and scaling the variance to an annual measure. The stated estimator subtracts the sample mean and uses a degrees-of-freedom adjustment.
The question contrasts that approach with the absolute price change over the option’s life and the square root of summed squared returns. The replies distinguish a terminal percentage move, which may be useful for comparing with an unhedged option position’s breakeven, from realized variance accumulated along the path. The material assumes the classical model and a particular trading-day convention for annualization; it does not discuss alternative volatility estimators, changing volatility, or how best to compare forecasts with realized outcomes in practice.
Key ideas
- Under geometric Brownian motion, log returns have variance proportional to elapsed time.
- Historical volatility for the Black–Scholes framework is estimated from log returns.
- Annualizing the estimate requires a time scaling convention, which the answer sets at 260 trading days.
- An absolute multi-day price change measures terminal movement, not pathwise realized variance.
- Summed squared returns capture variation along the path, even if price later returns to its starting level.
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Full text
# Empirical equivalent for implied vol
# 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 ?
- $|(S_t - S_{t-k})/S_{t-k}|$
- $\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 by Jan Stuller (score 3, accepted)
https://quant.stackexchange.com/a/58506
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 by user42108 (score 0)
https://quant.stackexchange.com/a/58522
"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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.