Another old question on this site (How to simulate stock prices with a Geometric Brownian Motion?) inspired me to ask the following question: if we assume that regular returns could be normally distributed, doesn't that entirely invalidate the idea behind the GBM model?

And vice versa, if we like the GBM model and we assume that stock-prices are log-normally distributed, doesn't that imply that regular returns cannot be normally distributed?


Let's denote $R_i$ as regular returns and let's assume that these are normally distributed:

$$R_i=\frac{S_{i+1}-S_i}{S_i}=\mu \Delta t + \sigma W(t)$$.

Let's denote $r_i$ as log-returns, defined as $r_i = ln \left( \frac{S_{i+1}}{S_i} \right)$. Then:

$$ R_i = e^{r_i} - 1 $$

$$ r_i=ln(R_i+1) $$

If we assume that $R_i$ are normally distributed, then $ln(R_i+1)$ is undefined, because Normal distribution produces negative values and $ln(negative)$ is undefined.

(Edit: as per the comments below, I now realize this is a "stupid" thought since regular returns are trivially bounded below by -1, so the log can never be negative: I initially just focused on the hypothetical idea of regular returns being normally distributed, i.e. unbounded.

However the following point is still valid: if $R_i$ is assumed approximately "normally" distributed but bounded by -1 from below, then $ln(R_1 +1)$ still won't be log-normally distributed, so the claim that "assuming $R_i$ to be normally distributed invalidates the assumptions of the GBM model" still holds).

So by this reasoning, believers in the GBM model would argue: regular returns cannot be normally distributed, because we like the idea of stock prices being log-normal (i.e. we like that the future stock-price distribution conditioned on today's value is log-normal: cannot be negative & doesn't have an upper bound, which reflects the real-world behavior we'd expect from stocks). Therefore, based on the GBM model, regular returns have to be log-normally distributed (shifted by "-1").

Reasoning the other way, I am pretty sure that I have seen some papers (apologies, don't have a link and can't remember the name of the authors) that argue that empirical evidence suggests that regular returns are normally distributed. In fact, just a quick philosophical thought: why shouldn't they be? Human beings use regular returns to look at investments, NOT log returns. It would seem sensible at first thought that these regular returns can be negative as well as positive, with a large probability mass centered on zero (or inflation, if $\mu$= inflation): i.e. a "normal" distribution. So if we entertain the idea of regular returns to be normally distributed, that would seem to invalidate the idea of the GBM model.

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    $\begingroup$ $R_i+1$ is not negative unless the stock goes bankrupt. $\endgroup$ Commented Jun 25, 2020 at 21:22
  • $\begingroup$ @stackoverblown: totally right, I was being "stupid" there, just thinking of Normal distribution with no bound. $\endgroup$ Commented Jun 26, 2020 at 10:00

2 Answers 2


You're right but a GBM doesn't assume that percentage returns are normally distributed. It's about log-returns.

  • If the log-return $r_t=\ln\left(\frac{S_{t+dt}}{S_t}\right)$ is normally distributed (GBM assumption), then $r_t$ can indeed be any arbitrarily large (positive or negative) number with positive probability. This also implies that stock prices are log-normally distributed.
  • Let now $\tilde{R}_t=e^{r_t}=\frac{S_{t+dt}}{S_t}$ be the gross return, which is obviously positive.
  • Let $R_t=\tilde{R}_t-1$ be the percentage return, which is bounded below by $-1$ from the above.

If we assume $\mathrm{d}S_t=\mu S_t\mathrm{d}t+\sigma S_t\mathrm{d}B_t$, we know that $r_t$ is normally distributed. However, $R_t=f(r_t)$ with $f(r)=e^r-1$ is not normally distributed. Just derive the distribution for $R_t$ and compare it to the log-normal density.

So, the assumptions of a GBM do not lead to percentage returns being normally distributed. Quite the opposite, they are bounded below by $-100\%$ (you can't lose more than you invested). So, $r_t=\ln(R_t+1)$ could only cause a problem if $R_t=-100\%$ but even that can't really happen in a GBM world: this would require the stock price to be zero in the future (bankruptcy). But the range of a log-normally distributed random variable is $(0,\infty)$, it has to be strictly positive. So, if $r_t$ is normal (GBM is true), then $R_t>-1$ and $r_t=\ln(R_t+1)$ is no problem.

I make one final point

  • I wouldn't believe for a second that any kind of return is normally distributed (think of fat tails, asymmetry, heteroscedasticity, etc.) Mandelbrot and Fama already worked on non-normally distributed returns back in the 1960s...
  • $\begingroup$ Isn't log-normal return the same as percentage return approximately? If you prove that the log-normal return is normal, it proves that the percentage return in the short time is normal. $\endgroup$ Commented Jun 26, 2020 at 2:12
  • $\begingroup$ Log returns (not log-normal) returns are different to percentage returns. The latter one is bounded by $-1$ and the former is unbounded. So both variables can (and do) have different distributions. Just compute the distribution of the percentage returns yourself. Also, you don’t prove that log returns are normally distributed but it’s an assumption. An assumption in which nobody really believes... $\endgroup$
    – Kevin
    Commented Jun 26, 2020 at 7:00
  • $\begingroup$ @KeSchn: thanks so much for the great answer. I focused too much on the hypothetical idea of regular returns being "normally distributed" in the mathematical sense that I forgot they're bounded below by -1. So indeed, if the line of reasoning is as you built it: GBM => Log-returns $r_i$ are Normally distributed => $R_i=e^r_i - 1$, then what exactly is the distribution of $R_i$? If a random variable $X$ is normally distributed, then $Y=e^X$ is log-normal, what about $\tilde{Y}:=e^X -1$? "Log-normal shifted"? (I could plot it but I don't have access to my comp right now: will plot tonight). $\endgroup$ Commented Jun 26, 2020 at 9:56
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    $\begingroup$ @JanStuller the percentage returns are not log-normal. Their range is $(-1,\infty)$ but a log-normal random variable must take values in $(0,\infty)$. I don't think that distribution class has a name. Let $X$ normal and $e^X$ log-normal. I remember people calling $e^{X+c}$ shifted log-normal (that variable is indeed log-normal) and I remember people referring to $e^X+c$ as shifted log-normal, that's our case. We can derive a density function for $e^X+c$ though using a density transformation formula. $\endgroup$
    – Kevin
    Commented Jun 26, 2020 at 11:12
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    $\begingroup$ The facts about returns (fat tails, stochastic volatility etc.) are in many textbooks (``stylised facts'') and motivate for example the (G)ARCH literature to better model conditional moments of returns. A normal distribution certainly does not fit real data very well. $\endgroup$
    – Kevin
    Commented Jun 26, 2020 at 11:14

The return $R_i$ as expressed in $$R_{i+1,i}=\frac{S_{i+1}-S_i}{S_i}=\mu \Delta t + \sigma \Delta W(t_{i+1},t_i)$$ is not possible.

To see this, let's get the returns over two small time steps of $\Delta t$ each. Then $$R_{i+2,i+1}=\frac{S_{i+2}-S_{i+1}}{S_{i+1}}= \mu \Delta t + \sigma \Delta W(t_{i+2},t_{i+1})$$ but $$R_{i+2,i}=\frac{S_{i+2}-S_{i}}{S_{i}}= 2 \mu \Delta t + \sigma \Delta W(t_{i+2},t_{i})$$ While the right-hand-side is additive, the left is not because $$R_{i+2,i} \neq (R_{i+2,i+1} + R_{i+1,i})$$.

For the log return $r_{i+1,i}$, $$ r_{i+1,i} = \ln\frac{S_{i+1}}{S_i}$$ however, there is no such problem because by virtue of the logarithmic product rule $$ r_{i+2,i} = ( r_{i+2,i+1} + r_{i+1,i} ) $$ holds.

So the $R_{i+1,i}$ cannot be normally distributed with drift.


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