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Returns possess non-zero skewness and excess kurtosis. If these assets are temporally aggregated both will disappear due to the law of large numbers. To be exact, if we assume IID returns skewness scales with $\frac{1}{\sqrt{n}}$ and kurtosis with $\frac{1}{n}$.

I'm interested in a concise, clear and openly accessible proof for the above statement preferably for all higher moments.

This question is inspired by this question by Richard which deals with, among other things, the behaviour of the higher moments of returns under temporal aggregation. I know about two papers that answer this question. Hawawini (1980) is wrong and Hon-Shiang and Wingender (1989) is behind a paywall and a bit inscrutable.

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    $\begingroup$ Might want to apply the procedure described here: papers.ssrn.com/sol3/papers.cfm?abstract_id=1635484 $\endgroup$
    – John
    Jun 24, 2012 at 19:59
  • $\begingroup$ Thanks, Bob, for continuing the discussion. I wonder why there is so much work published for one period and hardly anything for temporal aggregation. $\endgroup$
    – Richi Wa
    Jun 26, 2012 at 11:33
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    $\begingroup$ The results by @JL344 are very nice but only apply to the IID case. If you drop that assumption you have to use another model for the returns again and derive these properties under that model. I don't think it is possible to make any general statements on the higher moments under temporal aggregation which would explain the lack of literature: interesting results are not possible. $\endgroup$
    – Bob Jansen
    Jun 26, 2012 at 20:28
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    $\begingroup$ I am very aware that returns are not iid (e.g. volatility clustering). But as a starting point the iid case is just fine to me. If this case is clear it makes sense to move on. $\endgroup$
    – Richi Wa
    Jun 27, 2012 at 13:06

1 Answer 1

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Just to be painfully clear, it only seems to make sense to consider the logarithm of returns, i.e. $X=\log (1+\frac r{100})$ for a simple return of $r\%$ in an arbitrary period because this is what sums when returns are temporally aggregated. A basic property of cumulants is that cumulants of all orders are additive under convolution, for which a proof can be found here here.

So if $X_1$, $X_2$, ... $X_n$ are i.i.d., then all the cumulants of $$Y_n = \sum_{i=1}^nX_i$$ scale linearly with $n$, i.e. $$\kappa_k(Y_n)=n\kappa_k(Y_1).$$ However, I suspect that you are normalizing this sum so that the variance (or volatility) remains constant with increasing $n$. So instead let us consider $$Z_n=\frac{Y_n}{\sqrt n}= \frac 1 {\sqrt n} \sum_{i=1}^nX_i.$$ Another basic property of cumulants is that the $k$th cumulant is homogeneous of order $k$ as to scale. Using both properties together we have $$\kappa_k(Z_n)=\left(\frac 1 {\sqrt n}\right)^k\kappa_k(Y_n)=\left(\frac 1 {\sqrt n}\right)^kn\kappa_k(Y_1)=\frac {\kappa_k(Z_1)}{n^{(k-2)/2}}.$$

(Don't forget that $Z_1=Y_1=X_1$.) Now we can show that the statistics scale as you have described: $$\textrm{variance}=\kappa_2(Z_n)=\kappa_2(Z_1)\propto 1;$$ $$\textrm{skewness} =\frac{\kappa_3(Z_n)}{\kappa_2(Z_n)^{3/2}}=\frac{\frac{1}{n^{1/2}}\kappa_3(Z_1)}{\kappa_2(Z_1)^{3/2}}\propto \frac 1{\sqrt n};$$ $$\textrm{ex. kurtosis}=\frac{\kappa_4(Z_n)}{\kappa_2(Z_n)^2}=\frac{\frac{1}{n}\kappa_4(Z_1)}{\kappa_2(Z_1)^{2}}\propto \frac 1 n.$$

There is no reason this cannot be extended to higher orders, although it works out more directly in terms of cumulants than of moments.

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    $\begingroup$ Great answer! Do you know how to apply these findings to the Cornish-Fisher VaR? Just scaling skewness, kurtosis and variance seems wrong to me. For one period Cornish-Fisher VaR reads $$ \mu-\sigma z^*$$ where $$z^* = z +\frac {S (z^2-1)}{6}+\frac {(z^3 - 3z)K}{24}-\frac {(2z^3 - 5z)S^2}{36}$$ and $z$ is the desired standard normal quantile, $S$ is skewness and $K$ is excess-kurtosis. If we apply the scalings found we get $$ \sigma \sqrt{n} z^* = \sigma (z \sqrt{n} +\frac{S (z^2-1)}{6}+\frac{(z^3 - 3z)K}{24 \sqrt{n}}-\frac{(2z^3 - 5z)S^2}{36 \sqrt{n}}$$, thus one term independent of time. $\endgroup$
    – Richi Wa
    Jun 26, 2012 at 21:31
  • $\begingroup$ Above you see the standard term for the volatility scaling with $\sqrt{n}$ and 2 terms vanishing with $\frac{1}{\sqrt{n}}$ as we expect, but one term persists. This makes me think that "naive" plugging in must be wrong. As far as I know Cornish-Fisher Var is based on a cumulant expansions. Thus your results for the cumulant could be the path to go. $\endgroup$
    – Richi Wa
    Jun 26, 2012 at 21:36
  • $\begingroup$ @Richard You are correct in that the Cornish-Fisher expansion is based on the cumulants.[1] engineering.asu.edu/sites/default/files/shared/… $\endgroup$
    – JL344
    Jun 26, 2012 at 23:32
  • $\begingroup$ @Richard -- Sorry I hit enter too soon. I believe that at some point you will need to just naively plug things in, but it would be nice to find an elegant way to do so just to make sure everything is in fact scaled correctly. $\endgroup$
    – JL344
    Jun 26, 2012 at 23:53
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    $\begingroup$ sorry to bother you again. I am also sure, that the calculations are correct. These are: step one: replace $S $ by $\frac{S}{\sqrt{n}}$ and $K $ by $\frac{K}{n}$ (analogously for $S^2$) step two: multiply by $\sqrt{n}$ coming from the volatility term. Then we arrive at the equation that you find above. But there is a term independent of $n$. This means no matter how much data I aggregate over time there is a constant "distance" from Gaussian behaviour (which would be just $\sigma \sqrt{n}$). This is still strange to me. Do I make myself clear? $\endgroup$
    – Richi Wa
    Jun 27, 2012 at 11:32

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