Take the 2-minute tour ×

I am thinking about the time-scaling of Cornish-Fisher VaR (see e.g. page 130 here for the formula).

It involves the skewness and the excess-kurtosis of returns. The formula is clear and well studied (and criticized) in various papers for a single time period (e.g. daily returns and a VaR with one-day holding period).

Does anybody know a reference on how to scale it with time? I would be looking for something like the square-root-of-time rule (e.g. daily returns and a VaR with d days holding period). But scaling skewness, kurtosis and volatility separately and plugging them back does not feel good. Any ideas?

share|improve this question
Have you tried to do the calculation yourself? In the case of IID returns it shouldn't be too complicated. Write log returns of aggredated returns as the sum of individual log returns. Then calculate the skewness and Kurtosis of the sum. –  Zarbouzou Jun 21 '12 at 12:12
I did the calculations. I think skewness scales with 1/sqrt(n) and kurtosis with 1/n. So both vanish as the law of large numbers already tells us. But if I just plug in those scalings and apply sqrt(n) for the volatility then a strange time-independent term pops up. This gives me the feeling that just scaling the inputs (skewness,kurtosis and volatility) does not yield a proper solution. –  Richard Jun 21 '12 at 13:24
This [paper][1] shows that your calculations are correct and mistakes are easily made ;) [1]: onlinelibrary.wiley.com/doi/10.1111/j.1540-6288.1989.tb00340.x/… –  Bob Jansen Jun 21 '12 at 18:28
Thanks for the Link ... I guess you don't have a link that I can read or at least preview for free? Something like a link to a preprint. –  Richard Jun 22 '12 at 9:08
add comment

3 Answers

up vote 3 down vote accepted

If $z_\alpha$ is the so-called standard normal $z$-score of the significance level $\alpha$ such that $$ \frac 1 {\sqrt{2\pi}}\int_{-\infty}^{z_\alpha} e^{-\xi^2/2}d\xi=\alpha $$ and we assume normality, (ignoring skewness and kurtosis,) then we can estimate the $\alpha$ quantile of a distribution with cdf $\Phi$ as $$\Phi^{-1}(\alpha)=\mu + \sigma z_\alpha.$$ The Cornish-Fisher expansion is an attempt to estimate this more accurately directly in terms of the first few cumulants as $$\Phi^{-1}(\alpha)=y_\alpha,$$ where (before we have applied any scaling) $$y_\alpha= \kappa_1 + \frac {z_\alpha}2 + \frac {z_\alpha \kappa_2 }2 + \frac {(z_\alpha^2-1) \kappa_3}6 + \frac {(z_\alpha^3-3z_\alpha) \kappa_4}{24} - \frac {(2z_\alpha - 5z_\alpha) \kappa_3^2}{36}. $$ (Note that $\mu=\kappa_1$ and $\sigma^2=\kappa_2$.) Expressed directly in terms of cumulants, let us try to scale this directly with time as the convolution of infinitely divisible, independent identically distributed random variables. Cumulants of all orders scale linearly with time in this case, since they are simply additive under convolution. $$y_\alpha[t] = \kappa_1t + \frac {z_\alpha}2 + \frac {z_\alpha \kappa_2 t}2 + \frac {(z_\alpha^2-1) \kappa_3 t}6 + \frac {(z_\alpha^3-3z_\alpha) \kappa_4 t}{24} - \frac {(2z_\alpha - 5z_\alpha) \kappa_3^2 t^2}{36}. $$ but we want $y_\alpha[t] = \mu t + (\sigma\sqrt t )x_\alpha[t]$ where $x_\alpha[t]$ is the quantile function of a random variable with zero mean and unit variance. First the term $\kappa_1t$ drops off as our $\mu t$ since all the other cumulants are shift-invariant. Second we need to divide each remaining cumulant $\kappa_kt$ by $(\sigma\sqrt t)^k=(\kappa_2t)^{k/2}$ (because the $k$th cumulant is homogeneous of order $k$.) So: $$y_\alpha[t] = \mu t + \sigma\sqrt t \left[ z_\alpha + \frac {(z_\alpha^2-1) \kappa_3 t}{6(\kappa_2t)^{3/2}} + \frac {(z_\alpha^3-3z_\alpha) \kappa_4 t}{24(\kappa_2t)^2} - \frac {(2z_\alpha - 5z_\alpha) \kappa_3^2 t^2}{36(\kappa_2t)^3}\right]. $$ $$y_\alpha[t] = \mu t + \sigma\sqrt t \left[ z_\alpha + \frac {(z_\alpha^2-1) \kappa_3}{6\sigma^3t^{1/2}} + \frac {(z_\alpha^3-3z_\alpha) \kappa_4}{24\sigma^4 t} - \frac {(2z_\alpha - 5z_\alpha) \kappa_3^2}{36\sigma^6t}\right]. $$ but generally we write $\gamma_1=\kappa_3/\sigma^3$ and $\gamma_2=\kappa_4/\sigma^4$ for the skewness and the kurtosis respectively, so that $$y_\alpha[t] = \mu t + \sigma\sqrt t \left[ z_\alpha + \frac {(z_\alpha^2-1) \gamma_1}{6\sqrt t} + \frac {(z_\alpha^3-3z_\alpha) \gamma_2}{24t} - \frac {(2z_\alpha - 5z_\alpha) \gamma_1^2}{36t}\right]. $$

The Value-at-Risk is then $$\mathrm{VaR} = K_0 \left( 1 - {\exp (y_{\alpha}[t]-rt})\right),$$ where $K_0$ is the initial capital, $\alpha$ is some level of significance, say 1 to 5% or so, and $r$ is some instantaneous risk-free rate, appropriate discount rate, or required rate of return, however one chooses to define it. (This expression ought to become negative for a long enough time, because in the long run one will almost surely make money if $\mu>0$.)

share|improve this answer
Thank you for the rigorous answer. Thus the final result is that plugging in the separate scaling rules for volatility, skewness and kurtosis is correct. And I have to live with the term $$\sigma \frac{(z^2-1)\gamma_1 }{6} $$ which is not affected as time is aggregated. Thank you! –  Richard Jun 28 '12 at 12:23
add comment

Skewness decays with time, but the rate of that skewness decay will vary based on the instruments and how they are traded, so a simple estimator such as the square root of time rule is not appropriate.

I typically recommend that to scale VaR or ES it makes more sense to lower your confidence level (raise the alpha parameter) to one that makes sense for your holding period.

So, for example, assume that we are working with daily returns, as in your question. Now assume that I have a one-month holding period, and I want a 'monthly VaR'.

I would argue that a rational confidence level for this is 95%, or 1 in 20, corresponding approximately to the loss that will be exceeded about 1 day a month.

For monthly returns, as in a hedge fund portfolio, a confidence of 92% may be most appropriate, to specify the VaR that will be exceeded, on average, once a year.

I think that this is a much more rational approach than asking 'what loss level will be exceeded once in 10000 years?', as many papers and standards bodies recommend. These numbers aren't very useful, as many other authors have pointed out.

Also, extension to Cornish Fisher Expected Shortfall (also called CVaR or Expected Tail Loss) with the same approach as above, helps scale these numbers in a rational way, asking what the mean loss is when the loss exceeds the VaR.

More information on this is available in our published work including this paper from the Journal of Risk which also covers additive/coherent portfolio decomposition of Cornish Fisher VaR and Expected Shortfall.

share|improve this answer
Thank you Brian. I am totally aware of the shortcomings of VaR and modified VaR. But my client asks for 90 days Modified VaR. The questions is still how to mathematically correctly calculate Modified VaR for holding periods greater than one day. You are right that Expected Shortall is much better anyways. Personally, I even prefer Weighted-VaR as developed in A.S. Cherny's paper. I will also go through the paper whose link you gave me. –  Richard Jun 21 '12 at 13:35
Something like the alpha-weighting of VaR described for VaR derived from extreme value theory described in Tsay could be derived for the Cornish Fisher Distribution. As another alternative, square root of time sigma scaling will get most of the structure, unless the series is highly skewed or kurtotic. I don't think there is a precise or complete answer in the literature for this, and I haven't done the work to develop a precise answer. –  Brian G. Peterson Jun 22 '12 at 14:45
add comment

The time scaling of higher moments for ordinary (discrete) returns as per the Wingender paper is illustrated in Excel and VBA in the following spreadsheet demonstration files:

Terminal-Wealth-Time-Horizon-Calcs-Normal-and-Modified-VBA and;


Available here

For more on the weaknesses of the Cornish Fisher expansion see the presentation on Why-Distributions-Matter-16-Jan-2012

share|improve this answer
Thanks for the link. Searching the web for papers on the topic I have already visited your webpage. –  Richard Sep 7 '12 at 8:44
add comment

Your Answer


