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Except for the fact that returns can be negative while prices must be positive, is there any other reason behind modelling stock prices as a log normal distribution but modelling stock returns as a normal distribution?

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  • $\begingroup$ Neither claim in the title is actually true. You should define returns in your question (I've seen two slightly different definitions, and it matters for how an answer should be phrased in this case) $\endgroup$
    – Glen_b
    Commented Nov 27, 2014 at 20:25
  • $\begingroup$ Daily returns...as in...the difference between the day open and the day close expressed as a % of the open. $\endgroup$
    – Victor
    Commented Nov 27, 2014 at 20:38
  • $\begingroup$ You may find updated information about stock returns distributions on 3-minute video published here: indiegogo.com/projects/fat-tails-mathematics/x/17297122# $\endgroup$ Commented Sep 19, 2017 at 9:41
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    $\begingroup$ The existence of a highly upvoted & accepted answer implies this Q is sufficiently clear to be answered. I'm voting to leave open. $\endgroup$ Commented Sep 19, 2017 at 15:15

1 Answer 1

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Some points to start with:

i) these distributional conventions are at best approximations. They can be convenient models, but we shouldn't confuse that with the actual distribution of stock prices or returns.

ii) stock prices are typically increasing (but in any case, have changing mean; the mean isn't stable). So when we're talking about the distribution of stock prices, we're usually not referring to their marginal distribution, but a conditional distribution. So we often tend to mean something more like $y_t$ is approximately lognormal with mean changing with $t$ (specifically, conditionally lognormal, conditional on some previous value and elapsed time). The variance, too, can change, in which case both mean and variance condition on some previous value and time. So for example, by "stock prices are approximately lognormal" we might mean $y_t/y_{t-1} \, \dot{\sim}\, \log N(\mu_\text{daily},\sigma^2_\text{daily})$ or equivalently $y_t \, \dot{\sim}\, \log N(\log(y_{t-1})+\mu_\text{daily},\sigma^2_\text{daily})$

iii) Note that for small $x$, $\log(1+x)\approx x$.

For short period returns, such as daily returns, generally, $\frac{y_{t}-y_{t-1}}{y_{t-1}}$ is quite small, typically on the order of 0.01 - or often less - in absolute value.

When that ratio is small, $\log(y_t)-\log(y_{t-1})=\log(\frac{y_t}{y_{t-1}})\approx \frac{y_t}{y_{t-1}}-1=\frac{y_t-y_{t-1}}{y_{t-1}}$

That is, the return is approximately the change in log stock price (try it with real stock prices and see they're almost identical).

So if

$$y_t \, \dot{\sim} \, \log N(\log(y_{t-1})+\mu_\text{daily},\sigma^2_\text{daily})$$

which implies

$$\log(y_t) \, \dot{\sim} \, N(\log(y_{t-1})+\mu_\text{daily},\sigma^2_\text{daily})$$

then

$$\frac{y_t-y_{t-1}}{y_{t-1}}\approx \log(y_t)-\log(y_{t-1})\,\dot{\sim}\,N(\mu_\text{daily},\sigma^2_\text{daily})$$

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    $\begingroup$ As mentioned, for short period returns you could mistake it for a normal distribution. I have illustrated it in a post on my blog. $\endgroup$ Commented Mar 10, 2016 at 13:15

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