Is there any non-gaussian distribution has skewness $0$ and kurtosis $3$? Thanks!

  • 4
    $\begingroup$ I don't think the downvotes and votes to close are justified, although the formulation of the question could be improved. $\endgroup$
    – Abdelmalek Abdesselam
    Nov 26, 2018 at 15:35

4 Answers 4


The discrete distribution with probabilities \begin{align} p(-2)&=1/12\\ p(-1)&=\ 1/6\\ p(0)&=\ 1/2\\ p(1)&= \ 1/6\\ p(2)&=1/12 \end{align} has the same mean, variance, skewness and kurtosis as the Gaussian.


Note that in terms of cumulants $\kappa_n$, $n\ge 1$ one has $$ {\rm Mean}=\kappa_1 $$ $$ {\rm Variance}=\kappa_2 $$ $$ {\rm Skewness}=\frac{\kappa_3}{\kappa_{2}^{\frac{3}{2}}} $$ $$ {\rm Kurtosis}=3+\frac{\kappa_4}{\kappa_2^2} $$ My understanding of the OP's question is whether there are RVs other than the $N(0,1)$ which satisfies $\kappa_1=0,\kappa_2=1,\kappa_3=0,\kappa_4=0$. This is the truncated Hamburger moment problem for the sequence of moments $$ (m_0,m_1,m_2,m_3,m_4)=(1,0,1,0,3)\ . $$ The corresponding Hankel matrix is $$ H_2=\left( \begin{array}{ccc} 1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 3 \end{array} \right) $$ which is nonsingular and thus has Hankel rank $3=n+1$ for a sequence of moments $(m_0,\ldots m_{2n})$. This implies that there are infinitely many atomic measures (with $n+1$ atoms) that have the moments up to order four as the standard Gaussian. So the answer to the OP's question is yes. The above is a particular case of a result by Curto and Fialkow in Houston J. Math. 1991. A good account is in chapter 9 of the book "The Moment Problem" by Konrad Schmüdgen.

A perhaps more interesting question is under what extra condition does the answer become no, i.e., one has uniqueness. Being in fixed wiener chaos comes to mind because of the fourth moment theorem. There is a similar result by Newman for RV's that satisfy the Lee-Yang theorem as in ferromagnetic spin systems (see Theorem 3 in this article). Approaches to triviality of the $\phi^4$ model in dimension $\ge 4$ also use the fourth moment to show the Gaussian property.


Yes, there is. The Laplace$(0,\frac{1}{2})$ distribution has the required properties. Its probability density function is given by $f(x)=\exp(-2\lvert x\rvert)$ over the real line.

  • 2
    $\begingroup$ ...and therefore any convex combination of the density functions of this and the Gaussian will have this property. $\endgroup$
    – Igor Rivin
    Nov 25, 2018 at 1:32
  • $\begingroup$ I was reluctant to propose mixture-based answers with weighted normality but thank you for your comment! $\endgroup$
    – Nicéphore Bayekula
    Nov 25, 2018 at 7:32
  • 2
    $\begingroup$ Lapalce$(0, \frac{1}{2})$ has skewness 0 and excess kurtosis 3. Sorry for not making clear. By kurtosis $3$ I mean excess kurtosis $= 0$, like Gaussian distribution. $\endgroup$
    – pockeystar
    Nov 25, 2018 at 17:54

Skewness and Kurtosis, opposite to the popular belief, are non-uniquely defined concepts. There are general definitions of Skewness and Kurtosis:

Convex transformations of random variables

and the definitions based on the third and fourth moments are just one of them. In fact, these definitions are not even defined for all distributions, as the Cauchy distribution, a clearly symmetric distribution, has undefined skewness and kurtosis.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.