# Is Poisson–Lindley an exponential family? If not, why? [closed]

\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.

• 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 ... Commented Apr 14, 2021 at 3:34
• I don't understand why you keep destroying the equations. Is there something you would like to change about them?
– whuber
Commented Apr 14, 2021 at 17:13
• I have no idea what $thi$ is supposed to be. Commented Apr 24, 2021 at 12:00
• This question isn't answerable without explaining what $thi$ is.
– Sycorax
Commented Apr 24, 2021 at 17:09
• Most probably the $thi$ is supposed to be $\phi$ which is consistent with notation used in an exponential dispersion family. Commented May 18, 2021 at 17:50

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.

• 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? Commented Apr 14, 2021 at 18:47
• 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 Commented Apr 14, 2021 at 18:48
• (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."
– whuber
Commented Apr 14, 2021 at 19:48
• I meant in this form: f(y)=exp{ [y(theta) - b(theta)]/ a(thi) + c(y, thi)} Commented Apr 15, 2021 at 6:16
• Based on the pdf linked by OP, looks like they were interested in an exponential dispersion family instead. Commented May 18, 2021 at 18:02