6
$\begingroup$

Tweedie distributions are a family of distributions from the exponential dispersion family that have power-law mean-variance relationship:

\begin{align} \mathbb E[X] &= \mu \\ \operatorname{Var}[X]&=\phi \mu^p \end{align}

What is the formula for skewness?


For small integer values of $p$, these are well-known distributions (Gaussian, Poisson, gamma, inverse Gaussian). The most interesting case is $1<p<2$ which corresponds to the compound Poisson-Gamma distribution that has point mass at zero and is continuous for $x>0$. The formulas for the density as well as for the parameters of Poisson and Gamma in terms of $\mu$ and $\phi$ can be found e.g. in Dunn & @GordonSmyth (2005) Series evaluation of Tweedie exponential dispersion model densities.

I couldn't find the skewness formula anywhere so I derived it myself, and am posting this Q&A to share the result.

$\endgroup$
2
  • $\begingroup$ In this question and your answer wherever you write "$0\lt p \lt 1$" it seems you should be writing "$1\lt p \lt 2$." Is this a correct impression? $\endgroup$
    – whuber
    Commented Jan 30, 2018 at 0:21
  • 1
    $\begingroup$ Thank you @whuber, this was a mistake. I fixed the parameter range in both Q and A. $\endgroup$
    – amoeba
    Commented Jan 30, 2018 at 8:01

1 Answer 1

6
$\begingroup$

Exponential dispersion family is a broad family of distributions allowed in GLMs. The general form of the PDF can be written as follows:

$$f(x;\theta,\phi)=a(x,\phi)\exp\Big[\frac{1}{\phi}\big(x\theta-\kappa(\theta)\big)\Big].$$

The term $\kappa(\theta)$ is denoted with kappa because it is intimately related to the cumulants. Specifically, cumulant generating function (CGF) is given by

$$K(t;\theta,\lambda)=\frac{1}{\phi}\big(\kappa(\theta + t\phi)-\kappa(\theta)\big)$$

(see Wikipedia or Eq 2.6 in Jørgensen 1987, or Jørgensen's The Theory of Dispersion Models, 1997. Note that with $\phi=1$ the family reduces to the natural exponential family, see Wikipedia for its CGF.)

It follows that the first three cumulants are given by:

\begin{align} \kappa_1 &= \kappa'(\theta)\\ \kappa_2 &= \phi\kappa''(\theta)\\ \kappa_3 &= \phi^2\kappa'''(\theta) \end{align}

(Again note that for the natural exponential family cumulants are simply derivatives of $\kappa(\theta)$.)

For Tweedie distribution it must hold that

\begin{align} \kappa_1 &= \kappa'(\theta) = \mu\\ \kappa_2 &= \phi\kappa''(\theta) = \phi\mu^p \end{align}

so it follows that $$\kappa_3=\phi^2\kappa'''(\theta)=\phi^2(\kappa''(\theta))'=\phi^2(\mu^p)'=\phi^2p\mu^{p-1}\mu'=\phi^2p\mu^{p-1}\mu^p=\phi^2p\mu^{2p-1}.$$

Now we can compute skewness:

$$\operatorname{Skewness}[X]=\frac{\kappa_3}{\kappa_2^{3/2}}=\frac{\phi^2p\mu^{2p-1}}{(\phi\mu^p)^{3/2}}=\phi^{1/2}p\mu^{p/2-1}.$$

As a sanity check, this formula yields correct values for $p=0$, $p=1$, and $p=2$; these are skewness formulas for the Gaussian, Poisson, and gamma.

Let's verify that it works correctly for $1<p<2$:

# Tweedie random generation, using compound Poisson-Gamma representation
def tweediernd(n=1, p=1.5, phi=10, mu=1):
    # See Dunn & Smyth paper linked above for these formulas
    lambd = mu**(2-p)/(2-p)/phi   # Poisson rate
    alpha = -(2-p)/(1-p)          # gamma shape
    beta = phi*(p-1)*mu**(p-1)    # gamma scale

    x = np.zeros(n)
    for i in range(n):
        x[i] = np.sum(np.random.gamma(alpha, scale=beta, 
                      size=np.random.poisson(lambd)))
    return x

np.random.seed(42)
x = tweediernd(n=10000)
print('Mean:    ', np.mean(x))            # 1
print('Variance:', np.var(x))             # 10
print('Skewness:', scipy.stats.skew(x))   # sqrt(10)*1.5 = 4.74

This yields:

Mean:     0.996421833721
Variance: 9.86859188577
Skewness: 4.763172234662853
$\endgroup$
5
  • 2
    $\begingroup$ Clark and Thayer give skewness and kurtosis but I bet they were not the first; I'd expect someone like Tweedie or Jorgensen. But anyway: Clark, David R. and Charles A. Thayer. 2004. “A Primer on the Exponential Family of Distributions.” CAS Discussion Paper Program, 117-148. Their skewness agrees with yours. $\endgroup$
    – Glen_b
    Commented Nov 14, 2017 at 9:17
  • $\begingroup$ Oh, I should have mentioned ... its on the last page. $\endgroup$
    – Glen_b
    Commented Nov 14, 2017 at 9:31
  • $\begingroup$ Yes, took me some time to find :-) I was scrolling downwards. $\endgroup$
    – amoeba
    Commented Nov 14, 2017 at 9:33
  • $\begingroup$ I just searched for "Tweedie" in it. There's only about 4 hits $\endgroup$
    – Glen_b
    Commented Nov 14, 2017 at 9:36
  • $\begingroup$ Oh, it's probably in Johnson & Kotz or Encyclopedia of statistical sciences. $\endgroup$
    – Glen_b
    Commented Nov 14, 2017 at 9:37

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.