$$\begin{aligned}f_Y(y_i)&=\frac{{\theta_i}^2\left(y_i+\theta_i+2\right)}{\left(1+\theta_i\right)^{y_i+3}}\\ &=\exp\ \log\left[\frac{{\theta_i}^2\left(y_i+\theta_i+2\right)}{\left(1+\theta_i\right)^{y_i+3}}\right]\\ &=\exp\left[\log\left(\frac{{\theta_i}^2\left(y_i+\theta_i+2\right)}{\left(1+\theta_i\right)^{y_i+3}}\right)\right]\\ &=\exp\left[\log\left({\theta_i}^2\right)+\log\left(y_i+\theta_i+2\right)-\log\left(1+\theta_i\right)^{y_i+3}\right] \end{aligned}$$

Need to write in the form

$$f(y) = \exp{ [y(\theta) - b(\theta)]/ a(thi) + c(y, thi)}$$

I am actually stuck here and I am not sure about whether it can be written as canonical form or not. I have read somewhere that mixed distributions are not members of the exponential family.

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    $\begingroup$ Hint: You get a term $\log( y+\theta-2 )$ (where $-2$ has the wrong sign ...). There is no way that can be written as a separated product ... $\endgroup$ Commented Apr 14, 2021 at 3:34
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    $\begingroup$ I don't understand why you keep destroying the equations. Is there something you would like to change about them? $\endgroup$
    – whuber
    Commented Apr 14, 2021 at 17:13
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    $\begingroup$ I have no idea what $thi$ is supposed to be. $\endgroup$ Commented Apr 24, 2021 at 12:00
  • $\begingroup$ This question isn't answerable without explaining what $thi$ is. $\endgroup$
    – Sycorax
    Commented Apr 24, 2021 at 17:09
  • $\begingroup$ Most probably the $thi$ is supposed to be $\phi$ which is consistent with notation used in an exponential dispersion family. $\endgroup$ Commented May 18, 2021 at 17:50

1 Answer 1


In many cases there's a relatively simple, mindless test you can apply.

Recall that an Exponential family of distributions has densities of the form

$$f(x,\theta) = \exp(\eta(\theta)T(x) + A(\theta)+B(x)).$$

Suppose there is a region of values of $(x,\theta)$ in which $\eta,$ $T,$ $A,$ and $B$ are differentiable. Applying the logarithm and taking derivatives shows

$$\frac{\partial^2}{\partial x\partial \theta}\log f(x,\theta) = \eta^\prime(\theta)T^\prime(x),\tag{*}$$

effectively "killing off" the $A$ and $B$ terms.

If you can restrict this region to one where $\eta^\prime$ and $T^\prime$ each remain positive or negative, without equaling zero, and each is differentiable, you can repeat this process (after taking absolute values, if necessary, to assure the log can be applied):

$$\frac{\partial^2}{\partial x\partial \theta} \log|\frac{\partial^2}{\partial x\partial \theta}\log f(x,\theta)| = 0.$$

The result is zero because taking the log splits $(*)$ into a sum of a function of $\theta$ and a function of $x;$ just as at the outset, the mixed partial derivative kills both terms.

The Lindley-Poisson distribution is a discrete distribution on the values $x\in\{0,1,2,\ldots\}$ with probabilities

$$f(x,\theta) = \frac{\theta^2(x+\theta+2)}{(\theta+1)^{x+3}}$$

for $\theta\gt 0,$ giving

$$\log f(x,\theta) = 2\log\theta + \log(x+\theta+2) - (x+3)\log(\theta+1).$$

Its mixed partial derivative can be mechanically computed using basic laws of differentiation as

$$\frac{\partial^2}{\partial x\partial \theta}\log f(x,\theta) = 0 - \frac{1}{(x+\theta+2)^2} - \frac{1}{\theta+1}.$$

This is constantly negative. Repeating this operation on its absolute value gives (again purely mechanically)

$$\frac{\partial^2}{\partial x\partial \theta} \log|\frac{\partial^2}{\partial x\partial \theta}\log f(x,\theta)| = \frac{6}{(x+\theta+2)^4}\ne 0,$$

proving this is not an Exponential family.

  • $\begingroup$ Hello @Whuber, thank you for the response. But can you please clarify why is that for distribution to belong to the exponential family the above differentiation should equal to zero? $\endgroup$
    Commented Apr 14, 2021 at 18:47
  • $\begingroup$ In fact, I wanted to write it in the form which is in the link below. stat.purdue.edu/~ovitek/STAT526-Spring11_files/pdfs/hw8-sol.pdf $\endgroup$
    Commented Apr 14, 2021 at 18:48
  • $\begingroup$ (1) I searched that pdf file for "Lindley" but found nothing, then searched for "Poisson" but still found nothing about the Poisson-Lindley distribution. (2) Re your request for clarification, I gave the full derivation in the first half of this answer between the lines "Recall that" and "kills both terms." $\endgroup$
    – whuber
    Commented Apr 14, 2021 at 19:48
  • $\begingroup$ I meant in this form: f(y)=exp{ [y(theta) - b(theta)]/ a(thi) + c(y, thi)} $\endgroup$
    Commented Apr 15, 2021 at 6:16
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    $\begingroup$ Based on the pdf linked by OP, looks like they were interested in an exponential dispersion family instead. $\endgroup$ Commented May 18, 2021 at 18:02

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