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It is well-known that the solution to the stochastic SDE

$$ dS = S_0(\mu dt + \sigma dWt) $$

is

$$ S_t=S_0 e^{(\mu-\frac{\sigma^2}{2})t+W_t} $$

Were $\sigma=0$, this is simply the formula for continuous compounding. It stands to reason that the term $-\frac{\sigma^2}{2}t$ exists because the other way to write the solution is $$ S_t=S_0 e^{(\mu-\frac{\sigma^2}{2})t}e^{W_t} $$

or, even better,

$$ S_t=S_0 e^{\mu t}e^{-\frac{\sigma^2}{2}t}e^{W_t} $$

i.e., what re are really doing is (1) scaling by the determinstic amount; (2) scaling by the variable amount (the last term); and then (3) eliminating the bias introduced by the prior step from the last term by multiplying by the second term.

If so, it seems unfair to claim that 'returns are normally distributed' in the standard explanation. That would imply that in price-space, the price (an RV) would be multiplied by a normally distributed increment. But, in reality, what is really being said is that log P is growing by a constant term plus an (additive) normally distributed Brownina motion, i.e.,,

$$\frac{dS}{S}=\mu t + N(\frac{\mu}{t}, \frac{\sigma^2}{t})$$

in other words, in price space returns are not normally distributed, but themselves long-normally distributed (which would have the desireable property that product lof L-N RVs are themselves L-N, so it is an attractor).

But. other than convenience in stochastic calculus, what is the justification for using log-normality? Is there an empirical argument that returns are log-normally distributed?

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closed as unclear what you're asking by Raskolnikov, LocalVolatility, David Addison, amdopt, Helin Apr 26 '18 at 8:28

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Welcome to QSE! Maybe I miss something but where is the question in your statement? :) $\endgroup$ – Ric Apr 25 '18 at 5:31
  • $\begingroup$ The question is this: "are returns normally distributed, so that we can approximate them in 'if we divide up the interval of time in the range [0,1] into n smaller intervals in price space', then let $P_{t+1}P_{t}*(1+r/n)$, and take the limit as n goes to infinity. does that give a standard GBM? Or is it that 'the returns' are not normally distributed, and what er really are doing is multiplying in each interval the prior price by a log-normal multiplier which has the unfortunate consequence of adding 'too much' return that needs to be removed? $\endgroup$ – eSurfsnake Apr 25 '18 at 15:56
  • $\begingroup$ @eSurfsnake It's unclear what you're asking. Would you mind you putting your comment into the question itself so it's clearer? Also, the title of your question does not appear to be related to the question either. $\endgroup$ – David Addison Apr 25 '18 at 18:04
  • $\begingroup$ @eSurfsnake right: please edit your question ... Besides that I think David Addison gives the right answer: it is about the nonlinear nature of exponential and logarithm and the linear nature of Brownian motion. $\endgroup$ – Ric Apr 26 '18 at 7:40
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    $\begingroup$ @eSurfsnake "Ito, Stratonovich and Friends" by Holger K. von Jouanne-Diedrich who is very active here also gives some nice insights to the topic. You find it e.g. here. $\endgroup$ – Ric Apr 26 '18 at 7:43
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Ito's lemma results in a convexity adjustment since when you take a the exponential of a linear function with Gaussian noise that function becomes convex. GBM is therefore simply a solution to a differential equation in which the logarithmic returns have a normal distribution. Intuitively, the resulting "convexity" adjustment can be seen as consequence of Jensen's inequality.

I.e., write the instantaneous change of $X$ as being normally distributed:

$dX =X \,\mu \,d\tau + X\,\sigma dB_\tau$,

$\frac{dX}{X} =\mu \,d\tau + \sigma dB_\tau$,

$\int\frac{dX}{X} d\tau =\mu \,\tau + \sigma B_\tau$.

$\frac{dX}{X}$ looks related to $1 \over X $ (i.e., the derivative of $\log[X]$), so if the normal rules of calculus applied we would simply write the solution as:

$X_t = X_0 e^{ \mu \, \tau + \sigma B_\tau}$

However, the normal rules of calculus DO NOT apply since $X$ is a random variable, which intuitively means that the higher order derivatives of $f(X,\tau)$ actually do something. This problem thereofore requires Ito's lemma (or other stochastic differentiation methods) in order to describe what those higher level terms do. In the case of GBM, the term $-\sigma^2/2$ is a consequence of the second order terms of a Taylor series approximation of $X_{\tau}$.

How exactly you want to conceptualize this result is a matter of personal preference. While there are a few errors in your derivation of the distribution, the gist is that — as you allude — you can think of the solution as a lognormal distribution with a correction term or you could think of the distribution itself as being "off-normal”.

I.e.:

$\mathbb{E}[X_t] = X_0 e ^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t} \equiv X_0 e ^{\mu t+ Y_t} \equiv X_0 e ^{Z_t} \equiv \mathcal{P}_t $

where $W_t \sim \mathcal{N}[0,\sqrt{\tau}]$,

$Y_t \sim \mathcal{N}[-\frac{\sigma^2}{2}\tau,\sigma \sqrt{\tau}]$,

$Z_t \sim \mathcal{N}[(\mu-\frac{\sigma^2}{2})\tau,\sigma \sqrt{\tau}]$, and

$\mathcal{P}_t \sim \mathcal{logN}[(\mu-\frac{\sigma^2}{2})\tau + \log[X_0],\sigma \sqrt{\tau}] $

Analytically, these methods are equivalent.

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