Is it possible to express the kurtosis $\kappa$, or the 4th central moment $\mu_4$, of a random variable $X$ in terms of its mean $\mu = E(X)$ and variance $\sigma^2 = Var(X)$ only, without having to particularize to any distribution?

I mean, an expression like $\kappa = f(\mu, \sigma^2)$ or $\mu_4 = g(\mu, \sigma^2)$, valid for any distribution, where $f(\mu, \sigma^2)$ and $g(\mu, \sigma^2)$ are functions of the mean $\mu$ and variance $\sigma^2$.

P.S.: Some comments on my attempts.

$\kappa$ is related to $\mu_4$, and $\mu_4 = E(X^4) - 4\mu E(X^3) + 6\mu^2 E(X^2) - 3\mu^4$.

The term $E(X^2)$ can be expressed as $E(X^2) = \mu^2 + \sigma^2$ but I didn't manage to find the way to express $E(X^3)$ and $E(X^4)$ in terms of $\mu$ and $\sigma^2$.

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    $\begingroup$ $E(X^4)$ is not uniquely determined by the mean and variance. An ${\rm exponential}(1)$ random variable has mean and variance equal to 1, but has fourth moment equal to $4!=24$ while a normally distributed variable with mean $\mu=1$ and variance $\sigma^2 = 1$ has fourth moment equal to $10$. $\endgroup$
    – Macro
    Commented Jul 5, 2012 at 16:21
  • $\begingroup$ In the case of the normal distribution the complete distribution is determined by the mean and variance. So for a normal distribution the foruth central moment and all moments of the normal distribution can be expressed in terms of their mean and variance. @Macro This makes me puzzled why you would bring up the nromal distribution in your comment. $\endgroup$ Commented Jul 5, 2012 at 17:08
  • $\begingroup$ Thanks! I'm not an expert in statistics but need to deal with it for some problem. Well, in particular, I'm interested in having the variance of the sample variance expressed in terms of the mean and variance, and the variance of the sample variance depends on the excess kurtosis/4th central moment. That's why I need this particular parameter, and I'm trying to avoid fighting against the pdf to compute this moment as the pdf is the convolution of generalized Pareto with Uniform with Uniform... I've understood what I need isn't independent on the underlying distribution. Thanks. $\endgroup$
    – fchopin
    Commented Jul 5, 2012 at 17:21
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    $\begingroup$ @Macro When someone says that they want to know if the mean and variance determine the fourth moment for any distribution, I interpret that to mean in a given family will the mean and variance determine the fourth moment of the distribution? You apparently interpret it to be if you do not know the form of the distribution but know the mean and the variance will that determine the fourth moment? I think the answer to both questions is no. As IO reread the question i think your interpretation is correct. $\endgroup$ Commented Jul 5, 2012 at 17:24

1 Answer 1


What you think about here is something like a philosopher's stone of statistics.

The strict answer is:

No, it is impossible to express skewness or kurtosis via the mean and variance.

@Macro gave a counterexample of distributions with different skewness and kurtosis. A question of coming up with distributions for the given set of moments has entertained statisticians since the very early ages, and Pearson's system of frequency curves is one of the examples of how one could come up with a continuous distribution for the numeric values of the first four moments. You could also look at the moment generating function $m(t)={\rm E}[\exp(Xt)]$, a characteristic function $\phi(t)={\rm E}[\exp(iXt)]$, or a cumulant generating function $\psi(t) = \ln \phi(t)$. With some luck, you can try putting your four moments into them and invert these functions to obtain explicit expression of the densities. Finally, you can always find a distribution with discrete support on five points to satisfy the five equations for the moments of order 0 through 4 by solving a corresponding system of nonlinear equations.

To express the higher order moments via the lower order moments, you need to know the shape of the distribution and its parameters. For one-parameter (Poisson, exponential, geometric) or two-parameter (normal, gamma, binomial) distributions, you can express the higher order moments via the natural parameters of these distributions; e.g., for a Poisson with rate $\lambda$, skewness is $\lambda^{-1/2}$, and kurtosis is $\lambda^{-1}$ (sanity check: both going to zero as $\lambda \to \infty$, providing a normal approximation for Poisson for large $\lambda$). But these exceptions should not fool you; for more interesting distributions, including anything from the real world, you can just forget about doing anything meaningful with the kurtosis.

  • $\begingroup$ Thanks! Well, I'm studying something related to the variance of the sample variance estimator, and it depends on the excess kurtosis/4th central moment, so I need this particular parameter for my problem. I know the exact pdf of the underlying distribution, it is the result of the convolution of generalized Pareto with Uniform with Uniform (yes, Uniform twice), and the expression is so huge that I'm trying to avoid computing the 4th moment by integration. I think I could use approximations for the mean of a function of a random variable $E[g(X)] \approx g(\mu) + \frac{1}{2}\sigma^2 g''(\mu)$. $\endgroup$
    – fchopin
    Commented Jul 5, 2012 at 17:35
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    $\begingroup$ You are probably getting this idea from the delta-method, right? This is a first order approximation, and it only works well when the distribution of $X$ gets tighter as $n\to\infty$. If things remain $O(1)$, this expression fails miserably; think about $X\sim N(0,10)$ and $g(X)=\exp[X]$. The correct expression is $E[g(X)] = \exp( \mu + \sigma^2/2 ) = \exp( 5 ) = 148.41$; your first order approximation gives $\exp(0) + \frac12 \cdot 10 \cdot \exp(0) = 1 + 5 = 6$. $\endgroup$
    – StasK
    Commented Jul 5, 2012 at 18:42
  • $\begingroup$ Yes, delta method. This result can be found in the books of Benjamin & Cornell (1970), Papoulis (1984) and Blumenfeld (2001). I guess my problem meets the conditions for it to be a reasonable approximation, but that's an approximation after all. Will try... $\endgroup$
    – fchopin
    Commented Jul 6, 2012 at 7:52
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    $\begingroup$ By the way, the delta method, introducing various terms in the Taylor series, yields $E[g(X)] \approx g(\mu) + \frac{1}{2}\sigma^2g''(\mu) + \frac{1}{6}\mu_3g'''(\mu) + \frac{1}{24}\mu_4g^{iv}(\mu)$. Considering the expansion up to the 2nd or 3rd terms yields the result $\mu_4 = 0$, and considering all the terms up to the 4th one yields the result $\mu_4 = \mu_4$. So this technique doesn't yield any result here... $\endgroup$
    – fchopin
    Commented Jul 6, 2012 at 15:56

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