# 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 ... – kjetil b halvorsen Apr 14 at 3:34
• I don't understand why you keep destroying the equations. Is there something you would like to change about them? – whuber Apr 14 at 17:13
• I have no idea what $thi$ is supposed to be. – The Pointer Apr 24 at 12:00
• This question isn't answerable without explaining what $thi$ is. – Sycorax Apr 24 at 17:09

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? – CORNICHON Apr 14 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 – CORNICHON Apr 14 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 Apr 14 at 19:48
• I meant in this form: f(y)=exp{ [y(theta) - b(theta)]/ a(thi) + c(y, thi)} – CORNICHON Apr 15 at 6:16
• That form might be reminiscent of an Exponential family but it is not the same. – whuber Apr 15 at 15:48