Is there a known distribution, $f(x|\theta_1,\theta_2,\theta_3,\theta_4)$, with the following properties:

  1. $E(X^n)=\theta_n$ for $n \in \{1, 2, 3, 4\}$.
  2. If $\theta_3=0$ and $\theta_4=3\theta_2^2$, then $f$ is the normal density with mean $\theta_1$ and variance $\theta_2-\theta_1^2$.
  3. $f$ is a continuous function in each parameter.
  4. (Any other regularity conditions that preclude contrived solutions.)

Essentially, I'm looking for something akin to Johnson's SU distribution but with parameters that are more interpretable.

EDIT: I consider the following to be an example of a contrived solution.

If $\theta_3=0$ and $\theta_4=3\theta_2^2$, then $$f(x|\theta_1,\theta_2,\theta_3,\theta_4)=\dfrac 1 {\sqrt{2\pi}}\exp(\dfrac 1 2 \dfrac {(x-\theta_1)^2} {\theta_2})$$ otherwise $$f(x|\theta_1,\theta_2,\theta_3,\theta_4)=p(x|x_1,x_2,x_3,x_4,x_5)$$ where $p$ is the unique PMF with support over $\{x_1,x_2,x_3,x_4,x_5\}$ which satisfies Property 1 (provided a solution exists for the given $\theta's$). Obviously, this solution does not provide any modeling utility and it also doesn't satisfy Property 3. It's possible that other contrived solutions may exist which do satisfy Property 3. However, I'm only interested in ones which have utility for modeling - hence Property 4.

  • 4
    $\begingroup$ Your question appears to desire to treat each parameter $\theta_i$ as an independent parameter. But one would expect that changing the skewness $\theta_3$ would cause the mean $\theta_1$ to change. Similarly, one would expect that changing the kurtosis $\theta_4$ would typically cause the variance $\theta_2$ to change. Indeed, if you look at the formal definitions for skewness and kurtosis, they are functions of $\theta_1$ and $\theta_2$. $\endgroup$
    – wolfies
    Commented Mar 25, 2018 at 17:44
  • $\begingroup$ Great point. I hadn't considered that. I'll modify the properties to refer to raw moments instead. $\endgroup$
    – jjet
    Commented Mar 25, 2018 at 17:59
  • $\begingroup$ If $\theta_4=0$, then $x$ must be identically $0$ and therefore there are no (other) distributions satisfying both $(1)$ and $(2)$. $\endgroup$
    – whuber
    Commented Mar 25, 2018 at 18:32
  • 1
    $\begingroup$ I forgot to modify property 2 after I changed property 1. It's corrected now. $\endgroup$
    – jjet
    Commented Mar 25, 2018 at 18:46
  • $\begingroup$ Thank you; now it makes sense. It's unclear what you might mean by "contrived solutions," though. After all, the Normal family constitutes a 2D manifold which is a submanifold of arbitrarily many 4D manifolds of distributions. The restrictions in (2), (3), and (4) don't reduce the possibilities by any meaningful amount. If you would like to refine that family of solutions, then you will need to provide more specific properties associated with the additional parameters $\theta_3$ and $\theta_4.$ $\endgroup$
    – whuber
    Commented Mar 25, 2018 at 19:00

1 Answer 1


One option are Lambert W random variables (skewed, heavy-tailed), which can be parameterized either as $f(y \mid \mu_x, \sigma_x, \gamma)$ or $f(y \mid \mu_x, \sigma_x, \delta_{\ell}, \delta_r, \alpha)$, respectively (Disclaimer: I am the author of these, so I am biased on whether they are interpretable or not -- I find them def much more interpretable than a asinh() function ;) ).

As you care about 3rd and 4th moments, double heavy-tailed Lambert W x Gaussian (or Tukey's h / hh as special case) might be useful to look at. They arise as a non-linear transformation of $N(\mu_x, \sigma_x^2)$ random variable $X$ to (setting $\alpha = 1$ for simplicity)

$$ Y = \mu_x + \sigma_x \cdot \left( U \exp\left(\frac{\delta}{2} \cdot U^{2}\right) \right), \quad U := \frac{X - \mu_x}{\sigma_x} \sim N(0, 1) $$

It can be extended to a skewed version, by allowing $\delta$ to be different for the left side ($X < \mu_x$) vs the right side ($X > \mu_x$); hence $\delta \rightarrow (\delta_l, \delta_r)$. Clearly, $Y \sim N(\mu_x, \sigma_x^2)$ if $\delta = 0$.

The interpretation is that there is a latent process $X$ that is Gaussian; however, we only observe & measure the extreme skewed / heavy-tailed version of it through $Y$. As an example take the stock market: here you could think of $X$ as "news" occurring in the world (Gaussian), but we can only observe / measure them through the lens of collective market actions -- and as we know people freak out over unlikely events (adding heavy-tails); and people react more extreme to negative news than to positive ones (adding skewness). This collective response is captured via $\delta_l$ and $\delta_r$ parameters, which push events far from the mean even further away (generating heavy tails). Obviously, this should not be taken as a literal explanation of the market, but as (one) interpretation (see Table 4 & Figure 7 for an illustration on SP500 returns).

The distribution of $Y$, $f(y \mid \mu_x, \sigma_x, \delta_l, \delta_r)$, has the properties you request in 1. & 2. (set $\delta \equiv 0$) and 3. (see Eq. (23) here); re 4.: I assume you mean that you want to exclude pathological cases that are theoretically interesting, but practically useless. For that matter several applications in the original papers as well as several posts here illustrating applications of it with simulations and real world examples (How to transform data to normality?, How to transform leptokurtic distribution to normality?, Transformations to approximate normality with high kurtosis data) should suffice.

  • $\begingroup$ This is great! I'm going to have to dig into your papers to really digest this. Probably sometime after work today. I'll get back to you once I've gotten a chance to think about this some. $\endgroup$
    – jjet
    Commented Mar 26, 2018 at 12:56

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