I was running a bunch of simple simulations in excel the other day in excel. Using the NORM.INV(RAND(),0,1) to simulate daily stock returns I noticed that the more compounded the returns, ie, the more I multiplied the normally distributed variable with themselves in the form (1+ NORM.INV(RAND(),0,1))*(1+ NORM.INV(RAND(),0,1))...(1+ NORM.INV(RAND(),0,1)) the more and more the distribution of the final returns clumped around its mean and the fatter the tails became. Is this the same property the stock market exhibits, stock market returns might be normally distributed dover one unit of time, but the more and more returns compound the distribution changed and becomes leptokurtic??

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    $\begingroup$ 1) Your simulated "returns" are huge (the return will be >100% more than 10% of the time). 2) What did you expect? Compounding is exponential, so long tails aren't surprising (especially with huge returns). $\endgroup$ Commented Mar 12, 2014 at 21:57
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    $\begingroup$ This effect will vanish if you use log returns. Be aware of the difference between the two and when each is appropriate. See also the papers by Meucci on this. $\endgroup$
    – Quartz
    Commented Mar 13, 2014 at 11:21
  • $\begingroup$ what is kurt? did you compute it? $\endgroup$ Commented Mar 13, 2014 at 13:45
  • $\begingroup$ In the two plots what is the standard deviation (sd) of the first random variable (is it $1$?) and what in the second? The axes and the sd should fit together. Or you use histograms. $\endgroup$
    – Richi Wa
    Commented Mar 13, 2014 at 17:27

2 Answers 2


Basically, what you are asking is: What is the distribution of $$ Y = \prod_{i=1}^n X_i $$ where the $X_i$ are i.i.d. and $X_i \sim N(\mu, \sigma^2)$.

In general, $Y$ has a very complicated distribution. Check out the discussion in https://math.stackexchange.com/questions/161757/what-is-the-distribution-of-a-random-variable-that-is-the-product-of-the-two-nor?lq=1



What you can easily calculate are the moments of $Y$ since the $X_i$ are i.i.d.: So $$ \mathbb{E}[Y] = \mathbb{E}\left[\prod{i=1}^n X_i\right] = \prod{i=1}\mu =\mu^n\\ \mathbb{V}[Y] = \mathbb{E}[Y^2] - \mathbb{E}[Y]^2 = \mathbb{E}[\left(\prod{i=1}^n X_i\right)^2] - \mu^{2n} = \prod_{i=1}^n \mathbb{E}[X_i^2] - \mu^{2n} = \left(\sigma^2 + \mu^2\right)^n - \mu^{2n} \\ \mathbb{E}[(Y-\mathbb{E}[Y])^4] = \mathbb{E}[Y^4] - 4\mathbb{E}[Y^3]\mu^n + 6\mathbb{E}[Y^2]\mu^{2n} - 4\mathbb{E}[Y] \mu^{3n} + \mu^{4n} = \prod_{i=1}^n \mathbb{E}[X_i^4] + 6 \left(\sigma^2 + \mu^2\right)^n\mu^{2n} - 3 \mu^{4n} \\ = 3^n \sigma^{4n} + 6 \left(\sigma^2 + \mu^2\right)^n\mu^{2n} - 3 \mu^{4n} $$

The expression for the kurtosis of $Y$ is therefore $$ kurt(Y) = \frac{3^n\sigma^{4n} + 6 \left(\sigma^2 + \mu^2\right)^n\mu^{2n} - 3 \mu^{4n}}{\left(\sigma^2 + \mu^2 \right)^n - \mu^{2n}} $$ which is increasing in n.


As @Joshua Ulrich points out your distribution gets wider. Approximately, what you do is, you simulate

$$ Y = X_1 + \dots + X_n $$ and $X_i$ is standard normal. Of yourse the variance increases with $n$ (and standard deviation with $\sqrt{n}$).

But: greater variance does not mean heavier tails as suprises. If you want to put this in an easy number (besides the more complex tail index) you should look at kurtois, which is given (assuming zero expectation) by $$ kurt(X) = \frac{E[X^4]}{Var[X]^2} $$ thus you normalize for increasing variance. For details look at wikipedia.

E.g. the t-distribution can have heavier tails than normal but with the same variance.

PS: In your compound you do a $-1$ in the end - right?

  • $\begingroup$ Hi Richard, no I simply multiply 1 with (1+norm.inv(rand(),0,1)*......*(1+norm.inv(rand(),0,1). From the graph that shows compounded returns, you can easily tell that distributions doesn't look "bell-shaped" anymore. The values are more clumped in the middle. The range of extreme values is much larger. To Josh's response, yea I am trying to model the returns as they compounded in real markets. $100, up 2%, down 1.6 -->100(1+.02)(1+.016). $\endgroup$
    – jessica
    Commented Mar 13, 2014 at 14:14
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    $\begingroup$ @jessica if $r$ is normally distributed with mean $0$ and variance $\sigma$ then $1+r$ is normally distributed with mean $1$ and variance $\sigma$. The distribution of $\prod_{i=1}^n (1+r_i)$ is not normal. If you look at log-returns then you can just sum up for accumulation over time. Then you stay in the log-return world. $\endgroup$
    – Richi Wa
    Commented Mar 13, 2014 at 17:24
  • $\begingroup$ @jessica: NORM.INV(RAND(),0,1) yields numbers roughly between -3/+3. That's not -3% to +3%; it's -300% to +300%. Hence my comment that your "returns" are huge. $\endgroup$ Commented Mar 13, 2014 at 19:52

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