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Kurtosis/4th central moment in terms of mean and variance

  1. Jul 5, 2012 #1
    Hi All,

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

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

    Thanks in advance!


    P.S.: Some comments on my attempts.

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

    The term [itex]E(X^2)[/itex] can be expressed as [itex]E(X^2) = \mu^2 + \sigma^2[/itex] but I didn't manage to find the way to express [itex]E(X^3)[/itex] and [itex]E(X^4)[/itex] in terms of [itex]\mu[/itex] and [itex]\sigma^2[/itex].
    Last edited: Jul 5, 2012
  2. jcsd
  3. Jul 5, 2012 #2


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    In general you can't express higher order moments in terms of mean and variance, although you can for a normal distribution.
  4. Jul 6, 2012 #3
    Thanks! Unfortunately the underlying distribution in my problem isn't Gaussian. But I think I could use approximations to the mean of a function of a random variable: [itex]E[g(X)] \approx g(\mu) + \frac{1}{2}\sigma^2g''(\mu)[/itex]
  5. Jul 6, 2012 #4


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    Hey fchopin and welcome to the forums.

    What distribution are you using? Is it a standard one?
  6. Jul 6, 2012 #5
    No, the underlying distribution is much more complex. It is the result of the convolution of a generalized Pareto distribution with a uniform distribution, and with a uniform distribution again. The random variable is the sum of a generalized Pareto + uniform + uniform. I managed to obtain the expression of the pdf/cdf, but it is huge and computing [itex]\mu_4[/itex] by integration would definitely be a nightmare... I'm trying to find approximations based on the delta method. So far I have obtained [itex]\mu_4 \approx \sigma^4[/itex], but need to check its accuracy for my particular problem... Thank you all for you help and interest.
  7. Jul 6, 2012 #6


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    Can you calculate the Moment Generating Function either analytically as an exact expression or even as a good enough approximation?
  8. Jul 6, 2012 #7
    The PDF of a sum of several random variables is a convolution only if the variables are independent.

    On the other hand, by the convolution theorem, the moment generating function is a product of the corresponding moment generating functions. Taking the logarithm, the cumulant generating function is a sum of the corresponding cumulant generating function.

    However, the 4th cumulant is not identical to the 4th central moment.
  9. Jul 7, 2012 #8
    I'm afraid any attemp to obtain anything from the pdf/cdf analytically is a real nightmare as the resulting PDF is composed of four subequations, really huge... I spent one week computing these convolutions (yes, the involved variables are independent).

    Basically, my random variable Y is Y = X + A + B, where X is generalized Pareto, and A,B are uniform (independent of each other the 3 random variables). I was interested in the 4th central moment of Y, [itex]\mu_4(Y)[/itex], but I have finally found an alternative solution to my problem by expressing [itex]\mu_4(Y)[/itex] as a function of [itex]\mu_4(X)[/itex], [itex]\mu_4(C)[/itex] and the variance of X and C, where C=A+B is a triangular distribution. This is enough for my problem.

    By the way, the approximation [itex]\mu_4\approx\sigma^4[/itex] I was trying to use is really bad in this case since in practice [itex]\mu_4\gg\sigma^4[/itex].

    Thank you all for your comments.
  10. Jul 7, 2012 #9
    P.S.: The relation is

    [itex]\mu_4(Y) = \mu_4(X) + \mu_4(C) + 6Var(X)Var(C) [/itex]
  11. Jul 7, 2012 #10
    This is only true for statistically independent normally distributed random variables X and C.
  12. Jul 16, 2012 #11
    Hi Dick, I'm both learning and refreshing my memory on many of these topics...

    and I'm not certain what you mean; Don't normally distributed variables have an excess kurtosis of 0? / or kurtosis of 3?, so can his formula produce anything "but" 3, when applied to true normal distributions?

    And second, I'm quite interested in the convolution approach for addition...
    You mentioned cummulants, where the OP is asking about (central?) moments.
    I thought that there is a bijection between cummulants and moments for a given distribution.

    If that relationship holds for sums of distributions, such that the sum of the logs may be used for cummulants -- then it would seem to me that the OP might be able to generate a series of empirical random generator test cases for his distributions; and curve fit the bijection between moments and cummulants in the final distribution... ? Then he would only need to do interpolation on the curve fit as a very good approximation to the statistics he wishes to extract.

    Thanks for your comments. :smile:

    Oh, and if fchopin would comment, which generalized Pareto are you concerned with?


    Type I, Type II, Lomax, Type III, or Type IV ?
  13. Sep 17, 2013 #12
    4th moment and Kurtosis in convolution

    With some hesitation (with mathematics nor physics being my core business) I would like to comment on the previous discussion:

    The formula given by fchopin for the 4th moment of the convolution of two functions (y = x*h)

    y4= x4 + h4 + 6 x2 h2

    is valid for all regular functions, not only for a normal distribution.
    While the first, second and third moments are nicely additive in convolution, the fourth (and higher?) is apparently not. I have no simple explanation for this.
    With the Kurtosis defined as x4 / x2^2, a Guassian has K = 3.
    You will find that the K-value of the convolution of 2 Gaussians, calculated with fchopins formula, is again 3, as it should be.
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