If I want to price an option with the B-S model, why do I have to use the standard deviation of the log-returns of the underlying for the sigma parameter and not just the standard deviation of the absolute price level of the underlying?


2 Answers 2


Black Scholes assumes the price series is lognormally distributed, and so $ln(S_t) = X$, where $X$ is normally distributed

Notice that to price the option, you want to get the probability of the underlying price reaching the strike, so basically you want the probability of a jump from the current level to the strike level, ie:

$$X_i = ln(\frac{K}{S})$$

This variable is normally distributed and to convert it to standard normal we'll subtract the mean and divide by the standard deviation,

$$ \frac{X_i - \bar{X}}{std(X)} $$

which in the BS framework would be $\bar{X} = (r - \frac{1}{2}\sigma^2)t$ and $std(X) = \sigma \sqrt{t}$

Replacing this above gives you:

$$d = \frac{ln(K/S) - (r-\frac{1}{2} \sigma^2)t}{\sigma \sqrt{t}}$$

This would allow you to determine the probability of underlying finishing below the strike by using the standard normal distribution, ie, $P[S_t < K] = N(d)$

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Because the normal distribution is simetric, $P[S_t \geq K] = 1 - P[S_t < K]$, so you just have to change it to $1 - N(d)$ or $N(-d)$, which is basically the N(d2) in the BS option pricing formula

All this to say, that you are modelling the price returns so you should use the volatility of the returns, and not the absolute price change


note that for options on interest rates , often people use the Normal model rather than black-scholes , and for the Normal model, the standard deviation of the rate (rather than stdev of log returns of it), is the input volatility.


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