Mostly theoretical question. Are there any examples of non-normal distributions that has first four moment equal to those of normal? Could they exist in theory?

  • $\begingroup$ Considering even just a mixture of 2 normals (5 parameters--2 means, 2 variances, and the mixture probability), you can solve for a wide variety of first four moments. $\endgroup$ Dec 4, 2019 at 22:45
  • $\begingroup$ I gave a discrete example at a later near-duplicate of this question: stats.stackexchange.com/a/378923/225256 $\endgroup$
    – Matt F.
    May 28, 2021 at 16:21
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    $\begingroup$ @SheridanGrant, a mixture of two normals won't work. Let $X$ be a mixture with $p$ of $N(m,s)$ and $(1-p)$ of $N(n,t)$. Then forcing $X$ and $N(0,1)$ to have the same first, second and third moments leads to $p=n/(n-m)$, $s=\sqrt{(3-m^2+2mn)/3}$, $t=\sqrt{(3-n^2+2mn)/3}$. Under those conditions, the difference in fourth moments between $X$ and $N(0,1)$ is just $2mn(m^2-mn+n^2)/3$; if that is $0$ then $m=0$ or $n=0$, and $X$ is just a single standard normal. $\endgroup$
    – Matt F.
    May 29, 2021 at 12:13
  • $\begingroup$ @MattF.--huh, cool. I did a quick search for an analysis of "how many moments can you match with a mixture of k Gaussians" and didn't find anything, do you know of any results? $\endgroup$ Jun 10, 2021 at 6:10

2 Answers 2


Yes, examples with skewness and excess kurtosis both zero are relatively easy to construct. (Indeed examples (a) to (d) below also have Pearson mean-median skewness 0)

(a) For example, in this answer an example is given by taking a 50-50 mixture of a gamma variate, (which I call $X$), and the negative of a second one, which has a density that looks like this:

dgam 2.3

Clearly the result is symmetric and not normal. The scale parameter is unimportant here, so we can make it 1. Careful choice of the shape parameter of the gamma yields the required kurtosis:

  1. The variance of this double-gamma ($Y$) is easy to work out in terms of the gamma variate it's based on: $\text{Var}(Y)=E(X^2)=\text{Var}(X)+E(X)^2=\alpha+\alpha^2$.

  2. The fourth central moment of the variable $Y$ is the same as $E(X^4)$, which for a gamma($\alpha$) is $\alpha(\alpha+1)(\alpha+2)(\alpha+3)$

As a result the kurtosis is $\frac{\alpha(\alpha+1)(\alpha+2)(\alpha+3)}{\alpha^2(\alpha+1)^2}=\frac{(\alpha+2)(\alpha+3)}{\alpha(\alpha+1)}$. This is $3$ when $(\alpha+2)(\alpha+3)=3\alpha(\alpha+1)$, which happens when $\alpha=(\sqrt{13}+1)/2\approx 2.303$.

(b) We could also create an example as a scale mixture of two uniforms. Let $U_1\sim U(-1,1)$ and let $U_2\sim U(-a,a)$, and let $M=\frac12 U_1+\frac12 U_2$. Clearly by considering that $M$ is symmetric and has finite range, we must have $E(M)=0$; the skewness will also be 0 and central moments and raw moments will be the same.


Similarly, $E(M^4)=\frac{1}{10} (1+a^4)$ and so the kurtosis is $\frac{\frac{1}{10} (1+a^4)}{[\frac16 (1+a^2)]^2}=3.6\frac{1+a^4}{(1+a^2)^2}$

If we choose $a=\sqrt{5+\sqrt{24}}\approx 3.1463$, then kurtosis is 3, and the density looks like this:

enter image description here

(c) here's a fun example. Let $X_i\stackrel{_\text{iid}}{\sim}\text{Pois}(\lambda)$, for $i=1,2$.

Let $Y$ be a 50-50 mixture of $\sqrt{X_1}$ and $-\sqrt{X_2}$:

enter image description here

by symmetry $E(Y)=0$ (we also need $E(|Y|)$ to be finite but given $E(X_1)$ is finite, we have that)


by symmetry (and the fact that the absolute 3rd moment exists) skew=0

4th moment: $E(Y^4) = E(X_1^2) = \lambda+\lambda^2$

kurtosis = $\frac{\lambda+\lambda^2}{\lambda^2}= 1+1/\lambda$

so when $\lambda=\frac12$, kurtosis is 3. This is the case illustrated above.

(d) all my examples so far have been symmetric, since symmetric answers are easier to create -- but asymmetric solutions are also possible. Here's a discrete example.

enter image description here

(e) Now, here's an asymmetric continuous family. It will perhaps be the most surprising for some readers, so I'll describe it in detail. I'll begin by describing a discrete example and then build a continuous example from it (indeed I could have started with the one in (d), and it would have been simpler to play with, but I didn't, so we also have another discrete example for free).

$\:\,$ (i) At $x=-2,1$ and $m=\frac12 (5+\sqrt{33})$ ($\approx 5.3723$) place probabilities of $p_{-2}= \frac{1}{36}(7+\sqrt{33})$, $p_1=\frac{1}{36}(17+\sqrt{33})$, and $p_m=\frac{1}{36}(12-2\sqrt{33})$ (approximately 35.402%, 63.179% and 1.419%), respectively. This asymmetric three-point discrete distribution has zero skewness and zero excess kurtosis (as with all the above examples, it also has mean zero, which simplifies the calculations).

$\:$ (ii) Now, let's make a continuous mixture. Centered at each of the ordinates above (-2,1,m), place a Gaussian kernel with common standard deviation $\sigma$, and probability-weight given by the probabilities above (i.e. $w=(p_{-2},p_1,p_m)$). Phrased another way, take a mixture of three Gaussians with means at $-2,1$ and $m$ each with standard deviation $\sigma$ in the proportions $(p_{-2},p_1,p_m)$ respectively. For any choice of $\sigma$ the resulting continuous distribution has skewness 0 and excess kurtosis 0.

Here's one example (here the common $\sigma$ for the normal components is 1.25):

enter image description here

(The marks below the density show the locations of the centers of the Gaussian components.)

As you see, none of these examples look particularly "normal". It would be a simple matter to make any number of discrete, continuous or mixed variables with the same properties. While most of my examples were constructed as mixtures, there's nothing special about mixtures, other than they're often a convenient way to make distributions with properties the way you want, a bit like building things with Lego.

This answer gives some additional details on kurtosis that should make some of the considerations involved in constructing other examples a little clearer.

You could match more moments in similar fashion, though it requires more effort to do so. However, because the MGF of the normal exists, you can't match all integer moments of a normal with some non-normal distribution, since that would mean their MGFs match, implying the second distribution was normal as well.


Good points are made by Glen_b. I would only add consideration of the Dirac Delta function as additional grist for the mill. As Wikipedia notes, "The DDF is a generalized function, or distribution, on the real number line that is zero everywhere except at zero, with an integral of one over the entire real line" with the consequence that all higher moments of the DDF are zero.

Paul Dirac applies it to quantum mechanics in his 1931 book The Principles of Quantum Mechanics but it's origins date back to Fourier, Lesbesgue, Cauchy and others. The DDF also has physical analogues in modeling the distribution, e.g., of the crack of a bat hitting a baseball.

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    $\begingroup$ What has this to do with the question? $\endgroup$ Jun 1, 2015 at 13:25
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    $\begingroup$ The question is explicit about making the "first four moment[s] equal to those of [a] normal [distribution]". You haven't a hope of even matching the second central moment when you use a delta distribution. $\endgroup$
    – whuber
    Jun 1, 2015 at 15:58
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    $\begingroup$ Perhaps you can give an example where you match moments of a standard normal (mean 0, variance 1, $E[(X-\mu)^3]=E(X^3)=0$ and $E[(X-\mu)^4]=E(X^4)=3$). If you do that, it will answer the questions being raised and clarify your point. $\endgroup$
    – Glen_b
    Jun 1, 2015 at 16:45
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    $\begingroup$ @A. Donda: Excess kurtosis is the 4th standardized moment about the mean minus 3, i.e. $\mathrm{E}(X - \mathrm{E} X)^4 / (\mathrm{E}(X - \mathrm{E} X)^2)^2$, so I don't think you can say it's -3 in the case of Dirac's delta function - rather it's undefined, as the variance is zero. $\endgroup$ Jun 1, 2015 at 18:18
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    $\begingroup$ @Mike Hunter: I think the questions in the title & body are equivalent: once you have a distribution with defined skewness & excess kurtosis both equal to zero, matching mean & variance to any Gaussian you want is just shifting & stretching. I stress defined because both skewness & kurtosis are standardized moments, so the Dirac delta function doesn't have them. $\endgroup$ Jun 1, 2015 at 18:51

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