I am working on finding higher order moments for a given theoretical function, to be used in modelling of daily log-returns. The PDF is,

$f_r(x) =$ $\begin{cases} \quad \frac{1}{2}ae^{a\left(x-\mu\right)} & \text{if x } < \mu \\ \quad \frac{1}{2}be^{-b\left(x-\mu\right)} & \text{if x } \geq \mu \\ \end{cases} $

I believe the mean and variance are correct as,

E[X] =$-\frac{1}{2 a}+\frac{1}{2 b}+\mu$ and

Var[X] = $\frac{3}{4 a^2}+\frac{1}{2 a b}+\frac{3}{4 b^2}$

In computing skewness I have for $\mu_3 = \acute{\mu_3} - 3\acute{\mu_1}\acute{\mu_2} +2\acute{\mu_1}^3 $,

$= -\frac{7}{4 a^3}-\frac{3}{4 a^2 b}+\frac{3}{4 a b^2}+\frac{7}{4 b^3}$.

And for the kurtosis, $\mu_4 = \acute{\mu_4} - 4\acute{\mu_1}\acute{\mu_3} +6\acute{\mu_1}^2\acute{\mu_2} -3\acute{\mu_1}^4 $,

$= \frac{117}{16 a^4}+\frac{15}{4 a^3 b}+\frac{15}{8 a^2 b^2}+\frac{15}{4 a b^3}+\frac{117}{16 b^4}$

My first questions, are the $\mu_3$ and $\mu_4$ the correct moments for Skew and Kurt? When I am trying to estimate the parameters to fit the theoretical and sample moments, it does not seem so.. I am happy to provide more details etc if deemed necessary to clarify the issue. Thanks.

  • $\begingroup$ You can view this is a mixture of a (shifted) Exponential and the negative of another shifted Exponential. Raw moments (around zero) of mixtures are easy to compute, because they are comparable linear combinations of the raw moments of the component distributions. But must you? I like @Juan Rulfo's reply which directs you to an alternative model. $\endgroup$
    – whuber
    Mar 15, 2013 at 18:01
  • $\begingroup$ Yes, its a given exercise so I am not able to choose which model to use, unfortunately. When you say they are easy to compute, and linear combinations of the component distributions, its what I have suggested right? $\endgroup$
    – Geoffrey
    Mar 15, 2013 at 19:35

1 Answer 1


Note that $f_r$ is not continuous at $\mu$. This feature has to be taken into consideration since this makes the model more difficult to interpret.

An alternative continuous model is the skew-Laplace or two-piece Laplace distribution for which there are many equivalent expressions. These models have the same tail behaviour as the one you mention but they are continuous. One of them is studied in the following document


They present expressions for the moments as well as skewness and kurtosis.

The sort of distributions you are working on is known as a "split-distribution". They were popular in the 70s but now it is preferred to use two-piece distributions since they provide an easier interpretation. See for example the paper

Galbraith and Zhu (2011). Modeling and forecasting expected shortfall with the generalized asymmetric Student-t and asymmetric exponential power distributions. Journal of Empirical Finance, Volume 18, Issue 4, Pages 765–778.

See also the book

S. Kotz, T.J. Kozubowski and K. Podgorski (2001). The Laplace Distribution and Generalizations: A Revisit with Applications to Communications, Economics, Engineering, and Finance. Birkhauser, Boston.

  • $\begingroup$ Thanks for this, @JuanRulfo! I will definitively take a look. Hopefully, I have access to the cited paper. The exercise is related to finance and risk management, and later I am asked to calculate a Value at Risk and Expected Shortfall expression for the function, so I would like to be comfortable with f before calculating these values. $\endgroup$
    – Geoffrey
    Mar 15, 2013 at 19:44

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