How Products of Normal Variables Change Return Kurtosis

# Normally Distributed Returns Become Leptokurtic Due to Compounding


# Normally Distributed Returns Become Leptokurtic Due to Compounding












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??

## Answer by pbr142 (score 3, accepted)

https://quant.stackexchange.com/a/10583

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

and

https://math.stackexchange.com/questions/133938/what-is-the-density-of-the-product-of-k-i-i-d-normal-random-variables?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.

## Answer by Richi Wa (score 2)

https://quant.stackexchange.com/a/10571

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?