I want to calculate the unconditional distribution of $x$

$$p(x|p) = \int_0^\infty p(x,\theta|p) d\theta = \int_0^\infty p(x|\theta)p(\theta|p) d\theta$$

where $x|\theta \sim Poisson(\lambda)$ and $\lambda$ is a random variable that follows the Lindley distribution $$p(\theta|p) = \frac{p^2}{p+1}(\theta+1)e^{(-\theta p)}$$

The proven result is $$p(x|p) = \frac{p^{2}(p+2+x)}{(p+1)^{(x+3)}}$$

Doing so I have :

\begin{align*} p(x|\lambda) &= \int_0^\infty p(x,\theta|\lambda) d\theta\\ &= \int_0^\infty p(x|\theta)p(\theta|\lambda) d\theta\\ &= \int_0^\infty \frac{e^{-\theta}\theta^x}{x!} \frac{p^2}{p+1}(\theta+1)e^{(-\theta p)}d\theta\\ &= \frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1) d\theta\\ \end{align*}

So far so good. But I split the integral (am I allowed to do that?) into :

$$ \frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1) d\theta = \frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x+1} d\theta +\int_0^\infty e^{-\theta(1+p)}\theta^{x} d\theta $$


Given that : \begin{align*} \int_0^\infty \theta^x e^{-\theta (p+1)} d\theta &=\frac{\Gamma(x+1)}{(p + 1)^{(x+1)}} \int_0^\infty \frac{(p + 1)^{(x+1)}}{\Gamma(x+1)}\theta^x e^{-\theta (p+1)} d\theta\\ &=\frac{\Gamma(x+1)}{(p + 1)^{(x+1)}} \int_0^\infty \text{Gamma}(\theta; x+1, p+1) d\theta\\ &=\frac{\Gamma(x+1)}{(p + 1)^{(x+1)}} \end{align*}

then becomes : $$\frac{p^2}{(p+1)\Gamma(x+1)} \left(\frac{\Gamma(x+1)}{(p+1)^{x+1}} + \frac{\Gamma(x)}{(p+1)^x} \right)$$

Am I right ?


1 Answer 1


Assuming your work up to the last line (before edit) is correct we have with a little algebraic manipulation \begin{align} f_X(x|p) &=\frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1)\,\mathrm d\theta\\ &=\frac{p^2}{(p+1)^{x+2}}\int_0^\infty (1+\theta)\frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-\theta(p+1)}\,\mathrm d\theta, \end{align} which can be written as the expected value $$ f_X(x|p)=\frac{p^2}{(p+1)^{x+2}}\mathsf E(1+X),\quad X\sim\operatorname{Gamma}(x+1,p+1). $$ By linearity of the expected value $$ \mathsf E(1+X)=1+\mathsf EX=1+\frac{x+1}{p+1}; $$ hence, $$ f_X(x|p)=\frac{p^2}{(p+1)^{x+2}}\left(1+\frac{x+1}{p+1}\right). $$


If you need to show the steps in evaluating $\mathsf EX$ then write $$ \mathsf EX=\int_0^\infty \theta\frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-\theta(p+1)}\,\mathrm d\theta. $$ Substituting $u=(p+1)\theta$ and making use of the integral definition of the gamma function will get you the final result.


As requested by the OP: \begin{align} f_X(x|p) &=\frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1)\,\mathrm d\theta\\ &=\frac{p^2}{(p+1)}\int_0^\infty(1+\theta) \frac{1}{\Gamma(x+1)}e^{-\theta(1+p)}\theta^{x}\,\mathrm d\theta\\ &=\frac{p^2}{(p+1)}\int_0^\infty(1+\theta) \frac{1}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-(p+1)\theta}\,\mathrm d\theta\\ &=\frac{p^2(p+1)^{-(x+1)}}{(p+1)}\int_0^\infty(1+\theta) \frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-(p+1)\theta}\,\mathrm d\theta\\ &=\frac{p^2}{(p+1)^{x+2}}\int_0^\infty(1+\theta) \frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-(p+1)\theta}\,\mathrm d\theta\\ &=\frac{p^2}{(p+1)^{x+2}}\int_0^\infty(1+\theta) \operatorname{Gamma}(\theta|x+1,p+1)\,\mathrm d\theta\\ \end{align}

  • $\begingroup$ How the $x!$ become the power of $x+2$ in the first line ? $\endgroup$ Commented Jan 26, 2022 at 15:57
  • $\begingroup$ Note that $x!=\Gamma(x+1)$. Also note that $1/(p+1)=(p+1)^{x+1}/(p+1)^{x+2}$ $\endgroup$ Commented Jan 26, 2022 at 16:00
  • $\begingroup$ It seems correct but I think between the first 2 lines there are a lot of calculations that might help me more.If you please can edit your answer with these calculations I would appreciate it.Nonetheless thank you for your effort is helpful $\endgroup$ Commented Jan 26, 2022 at 16:37
  • $\begingroup$ @HungryHomer See if the edit helps you. $\endgroup$ Commented Jan 26, 2022 at 17:11

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