0
$\begingroup$

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 ?

$\endgroup$
2
  • $\begingroup$ 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! $\endgroup$ Oct 6, 2020 at 17:09
  • $\begingroup$ Quadratic variation might be a good proxy. $\endgroup$ Oct 8, 2020 at 18:17

2 Answers 2

3
$\begingroup$

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).

$\endgroup$
3
  • $\begingroup$ 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 ? $\endgroup$
    – Kreol
    Oct 6, 2020 at 19:49
  • $\begingroup$ Hmmm... perhaps you should clarify your question... what are you trying to do? Why interested in the k-day change? $\endgroup$
    – nbbo2
    Oct 7, 2020 at 7:07
  • 1
    $\begingroup$ 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? $\endgroup$ Oct 7, 2020 at 7:09
0
$\begingroup$

"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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.