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I'm struggling with what the exact meaning of "stock prices are lognormal" (and its use to show normality of returns). My assumption was that given ${S_t}$ are stock prices and returns are defined as $r_t = \frac{S_t-S_{t-1}}{S_{t-1}}$, then we assume $S_t$ is lognormal, and then:

$$\log(1+r_t)=\log\left(\frac{S_{t-1}+S_t-S_{t-1}}{S_{t-1}}\right)=\log\left(\frac{S_t}{S_{t-1}}\right)=\log(S_t)-\log(S_{t-1}) \tag{1}$$

As this would be the sum of two normal variables, the result is normal, and that allows us to show $\log(1+r_t)$ is normal.

However, I was reading the following link:

In it, the author states (I've replaced his notation with mine for ease of comparison):

If we assume that prices are distributed log normally (which, in practice, may or may not be true for any given price series), then $\log(1+r_i)$ is conveniently normally distributed, because:

$$1+r_i=\frac{S_{t}}{S_{t-1}}=\exp\left(\log\left(\frac{S_t}{S_{t-1}}\right)\right)\tag{2}$$.

From the definitions of lognormal, in order for the inner term of the right-hand side (i.e. $\log\left(\frac{S_t}{S_{t-1}}\right)$) to be normal, we would need $1-r_i$ to be lognormal. But that seems different to me than "prices are lognormal". The following cross validation answer makes a bit more sense of this, namely part ii), where the answerer mentions conditional lognormality, or that the assumption of log normality in prices usually refers to $\frac{S_t}{S_{t-1}}$, and that would satisfy equation 2.

So to summarize, what is the correct way to define the lognormality assumption in prices? My apologies if I'm simply overthinking things. Thank you!

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In reality, neither are stock prices log-normally distributed nor are returns normally distributed. More sophisticated models drop this assumption. For instance, returns are more peaked and have fatter tails than a normal distribution would suggest.

In simple models, such as the Black and Scholes (1973) model, it is however assumed that the stock price satisfies the SDE $\frac{\mathrm{d}S_t}{S_t}=\mu \mathrm{d}t+\sigma \mathrm{d}W_t$ which means that changes in the stock price are proportional to its current price - this is rather reasonable and implies directly that the instantaneous returns $\frac{\mathrm{d}S_t}{S_t}=\mathrm{d}\ln(S_t)$ are normally distributed.

In your quote, $S_t$ is log-normally distributed and so is $\frac{S_t}{S_{t-1}}$ implying that $1+r_t$ is also log-normally distributed. Thus, $\ln(1+r_t)$ is indeed normally distributed as claimed.

Historically, normally distributed prices where first considered (models such as an arithmetic Brownian motion), but Samuelson introduced the geometric Brownian motion to avoid negativity. Black and Scholes built upon this insight. Hence, prices are always positive but returns may be negative.

Note that both, the stock price and its returns are stochastic processes and hence "$S_t$ is log-normally distributed" really means that the random variable $S_t(\omega)$ follows a log-normal distribution for each fixed time point $t>0$.

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  • $\begingroup$ Thanks for the answer! I think this mostly resolves my confusion, with one gap I want to fill in. You mention: "In [my] quote, $S_t$ is log-normally distributed and so is $\frac{S_t}{S_{t-1}}$"; is $\frac{S_t}{S_{t-1}}$ being log-normal a distinct assumption from $S_t$ being log-normal, or does it just follow from $\{S_t\}$ being iid, and the division of two log-normal iid variables is lognormal? $\endgroup$ – deetsb Jan 8 '20 at 15:26
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    $\begingroup$ @deetsb good question. Note that the process $(S_t)$ does not have independent increments but indeed, if you solve the model, $S_t=S_0e^{\left( \mu-0.5\sigma^2\right)t+\sigma W_t}$ and if you compute $\ln\left(\frac{S_t}{S_{t-1}}\right)$, you’ll see that it is normally distributed. So it directly follows from $(S_t)$ being lognormal, just as you said. $\endgroup$ – Kevin Jan 8 '20 at 15:33
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    $\begingroup$ "Just as I said", but definitely not the way I got there. Serves me right for being lazy and trying to hand wave it. Thanks, answer accepted! $\endgroup$ – deetsb Jan 8 '20 at 15:36
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Stock prices cannot be negative which means that they are not normally distributed due to the fact they cannot be negative as result of this stock prices behave similarly to exponential functions. To transform this exponential values back to a normally distributed variable, you need to take the natural logarithm, and therefore can take a lognormal value and distribution.

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  • $\begingroup$ "Meaning stock returns are not normally distributed due to the fact they cannot be negative as result of this stock prices behave similarly to exponential functions" -- You should rewrite this sentence. I have never seen a stock that cannot have a negative return :) $\endgroup$ – amdopt Jan 8 '20 at 13:17
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    $\begingroup$ Yes @amdopt sorry didn't pay attention. $\endgroup$ – Gogo78 Jan 8 '20 at 13:22
  • $\begingroup$ I appreciate the answer, but I'm comfortable with the high level intuition behind the choices and transforms. I'm just a little tripped up in the semantics of the assumption, and found the other answer to be more helpful! $\endgroup$ – deetsb Jan 8 '20 at 15:27

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