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Why Black–Scholes Produces Lognormal Stock Prices

Article Quant Q&A · Author: confused

Summary

The explanation clarifies that the Black–Scholes stock process is written for the ordinary price level, with drift and volatility proportional to the current price. The logarithm enters when Itô’s lemma is applied to that process: the log price then has a constant drift adjusted downward by half the variance and a diffusion term proportional to Brownian motion.

Because the log price follows a normal distribution under this model, the price itself has a lognormal distribution. The response presents lognormality as a feature that supports a closed-form option-pricing solution, rather than as a separate transformation inserted into the hedging argument. It also notes that some derivations can work under broader assumptions, so the explanation is specific to the standard lognormal setup and does not establish that every option-pricing model requires lognormal prices.

Key ideas

  • The Black–Scholes stochastic differential equation models the ordinary stock price, not its logarithm.
  • Applying Itô’s lemma to the logarithm gives a Brownian process with a variance-adjusted drift.
  • A normally distributed log price implies a lognormally distributed stock price.
  • The lognormal assumption enables a closed-form solution in the standard Black–Scholes model.
  • Other derivations may use different distributional assumptions.

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Full text
# How does the Black Scholes Model Incorporate Log Prices Into Model?


# How does the Black Scholes Model Incorporate Log Prices Into Model?












I am still not understanding the link between log prices and how that is incorporated into the BS model. I understand why log(S) is assumed because it makes math easier and it prevents ending prices from going negative. However, I don't get where that transformation actually occurs.

When pricing options using BS model, I always see that the underlying evolution process follows:

`dS = μSdt + σSdX`

The S in the above, I am led to believe, is just regular stock prices correct? And not log(S)? This makes sense as dividing S across gives us the simple return. The above is how we assume stock prices change, which is necessary to apply Ito's Lemma.

To derive the option price, we first create a portfolio of a long call and short stocks, which after applying Ito's Lemma to the option price, allows us to find the amount of stocks to sell to make the portfolio riskless. This eventually gives us the price of the option.

Nowhere in any of the BS derivations that I have seen do I see log(S) come into play. I see the derivation, and then, as a side point, the author remarks that BS model assumes ending stock prices follow a lognormal distribution. I don't see where log(S) is actually incorporated into the derivation process.

Does it come into play within the dX term where it's not norm(0,1) but some other distribution? Or is dS actually dlog(S)? Or does the transformation come somewhere else?

Thanks!

## Answer by alf (score 1)

https://quant.stackexchange.com/a/45651

Well, the distribution does not need to be lognormal. There are articles that do not assume or require that: the only thing you actually need is that $dX^2 = dt$ (otherwise your price either goes awry or converges to zero). What lognormal distribution gives you is a closed-form solution.

But if you're back to a lognomal walk, the very name of it comes from the Browniam motion with drift,

$$ dS = \mu dt + \sigma\ dW, $$

where $dW$ is the standard Brownian motion.

It's easy to see that the final distribution of that is normal: we sum up "many" random variables with the same distribution, so we'll end up with a normal distribution with mean $\mu t$ and standard deviation $\sigma t$.

Now the lognormal walk,

$$ dS = \mu S dt + \sigma S\ dW $$

If you take a look at $\log S,$ you'll note that (from Itô's lemma)

$$ \begin{aligned} d\log S &= \frac{d\log S}{dS} dS + \frac{1}{2} \sigma^2 S^2 \frac{d^2\log S}{dS^2} dt \\ &= \frac{1}{S}\left( \mu S dt + \sigma S dW\right) - \frac{1}{2} \sigma^2 dt \\ &= \left( \mu - \frac{\sigma^2}{2}\right) dt + \sigma dW \end{aligned} $$

That is, now $\log S$ follows a Brownian motion, and the distribution of it is normal. Hence the "lognormal."

As far as I'm aware, there's nothing else in the name.

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.