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?


  • 2
    $\begingroup$ The fact that S will have a lognormal distribution is a non-obvious consequence of the equation $dS = \mu S dt + \sigma S dX$. It is not assumed anywhere, but can be proven. A lognormal distribution is defined as a distribution such that if you take the log, it looks normal (that is the place where the logarithmic transformation comes about, but you are right that this transformation is not used in the Black Scholes derivation. Of course Black and Scholes knew that if you choose this model for S, it will have some desirable properties including the fact that S stays positive, etc.) $\endgroup$
    – Alex C
    Commented May 15, 2019 at 14:45
  • $\begingroup$ Also dX is not really norm(0,1) it is a tiny tiny (infinitesimal) increment which is proportional both to a norm(0,1) variable (usually called $\epsilon$) and to $\sqrt{dt}$. It is quite an unusual thing, takes some getting used to. $\endgroup$
    – Alex C
    Commented May 15, 2019 at 14:53
  • 1
    $\begingroup$ Would you like to see the proof that $dS=\mu S dt + \sigma S dX$ leads to a lognormal S? The proof that Black and Scholes omitted in their paper ;) $\endgroup$
    – Alex C
    Commented May 15, 2019 at 15:08
  • $\begingroup$ For a general understanding of some of the above mentioned points and to gain some intuition my paper "Ito, Stratonovich and Friends" might be helpful: papers.ssrn.com/sol3/papers.cfm?abstract_id=2956257 $\endgroup$
    – vonjd
    Commented May 15, 2019 at 16:15
  • 1
    $\begingroup$ To understand $\frac{dS}{S} = \mu dt + \sigma dW$ perhaps first examine the diff-eq $\frac{1}{y} \frac{dy}{dt} = r$. It is equivalent to $\frac{d \log y}{dt} = r$ hence $\log y = rt$ and $y = e^{rt}$. Speaking loosely, $\frac{dy}{y}$ is a percent change and a continuously compounding percent change is a log difference. Continuously compounding percent changes leads to exponential growth, and exponential growth in $y$ is linear growth in $\log y$. $\endgroup$ Commented May 17, 2019 at 1:15

1 Answer 1


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.

  • $\begingroup$ Thanks! Are simple returns also lognormally distributed then or just the underlying prices itself? $\endgroup$ Commented May 28, 2019 at 0:41
  • $\begingroup$ @user6472523: according to BS returns are normally distributed (in reality they are not!) $\endgroup$
    – vonjd
    Commented Jun 3, 2019 at 10:01

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