Normality or Log-Normality of Regular Returns

Question

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?

Specifically:

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.

Answer 1

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...

Answer 2

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.