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Annualizing Historical Volatility from Daily Stock Returns

Article Quant Q&A · Author: Appliqué

Summary

The question derives daily log returns under geometric Brownian motion and estimates their sample variance from a month of stock prices. Since the variance of a return over a daily interval is the annual variance scaled by the length of that interval, the sample variance must be divided by the time increment to estimate annual variance. Equivalently, daily volatility is annualized by multiplying by the square root of the number of trading days per year.

The answer confirms that annualization is needed when the Black–Scholes input is annualized volatility, and notes that the assumed trading-day count can vary. The discussion distinguishes variance scaling from volatility scaling: variance uses the time factor directly, while standard deviation uses its square root. It offers no price series or empirical comparison, and does not examine other estimation choices such as lookback length, return conventions, or volatility forecasting methods.

Key ideas

  • Daily log-return variance estimates variance over a single daily interval.
  • Annual variance is obtained by dividing daily variance by the daily time increment.
  • Annual volatility is daily volatility multiplied by the square root of the annual trading-day count.
  • The trading-day convention affects the annualized estimate.

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Full text
# Black-Scholes volatility implied by stock prices only


# Black-Scholes volatility implied by stock prices only












I was solving Problem 2.47 from T.F. Crack's "Heard on the Street". I think that the answer given in the book is not correct and I would be thankful if you tell me, where I am mistaken.

> Question 2.47. You have 30 days of "representative" stock price data. How do you calculate historical volatility $\widehat σ^2$ to use in Black-Scholes?

In the Black-Scholes model stock prices follow a geometric Brownian motion $$ \frac{dS_t}{S_t} = \mu dt + σ dW_t. $$ Solving this SDE I get the following expression for one-day continuously compounded returns: $$ \log\frac{S_{t+∆ t}}{S_t} = (\mu-\frac{σ^2}{2})∆\!t + σ \sqrt{∆ t} Z_t, $$ where $Z_t$ are independent random variables from $N(0,1)$, $∆ t = 1/250 \approx 0.004$ and $\sqrt{∆ t} \approx 0.063$. Using the observations I estimate $(\mu - σ^2/2)∆\!t$ and $σ^2 ∆\!t$ by the sample mean and variance: $$ M := \frac{1}{29}\sum_{i=1}^{29}X_i, \\ V := \frac{1}{28} \sum_{i=1}^{29}(X_i - M)^2, $$ where $X_i$ are the observed continuously compounded returns $\log(S_{t+∆ t}/S_t)$ for the period of $30$ days. Then I use $V/{∆t}$ as an estimator of $σ^2$. However, solution given in the book does not use the quantity $∆ t$ at all, and the quantity $V$ is used as an estimator of historical volatility. Are they mistaken?

## Answer by Magic is in the chain (score 1)

https://quant.stackexchange.com/a/45236

Correct you will need to annualise the daily volatility, which is done by dividing the volatility calculated using daily returns by the square root of one over number of days in a year, dt in your example. Which is the same thing as multiplication by the square root of the number of days in a year. Usually assumed to be 252 but vary.

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.