The probability generating function (pgf) of $X\sim \text{Poisson}(\lambda)$ is $$G_x(t) = e^{-\lambda(1-t)}.$$

Find pgf of $X$ if $\lambda\sim \text{Unif}(0,2).$ Then find $\mathbb P(X=2).$

My solution:

$G_{x\mid \lambda\sim Unif(0,2)}(t)=\int_{0}^2 e^{-\lambda(1-t)} f_\lambda(t) dt = \int_{0}^2 e^{-\lambda(1-t)} \frac{1}{2} dt = \frac{1}{2} \int_{0}^2 e^{-\lambda(1-t)} dt = \frac{1}{2} e^{2\lambda} - \frac{1}{2} e^{-\lambda}$.

Then $\mathbb P(X=2) = \frac{G_x^{(2)}(0)}{2!}$, so here

$G_x^{(2)}(t) = 2e^{2\lambda} - \frac{1}{2}e^{-\lambda}$ and $G_x^{(2)}(0)= 2-\frac{1}{2}$.

Finally, $$\mathbb P(X=2) = \frac{3}{4}$$

Is it correct?

  • $\begingroup$ @KaviRamaMurthy Are you sure the integration should be w.r.t $t$ and not $\lambda$?, because we want the unconditional PGF and probability. $\endgroup$ Jun 20, 2019 at 9:02
  • $\begingroup$ Yes, I am interested in this question too. If I integrate w.r.t. $λ$, I get 0, that is also possible. $\endgroup$
    – Lucyy
    Jun 20, 2019 at 9:06
  • $\begingroup$ @Lucyy As pointed out by Vishaal Sudarsan you have mixed up the variables. You have to integrate w.r.t. $\lambda$ and get the moment generating function as a function of $t$. The differentiate twice w.r.t. $t$. $\endgroup$ Jun 20, 2019 at 9:08
  • $\begingroup$ @Kavi Rama Murthy but now I am getting the negative probability.. $\endgroup$
    – Lucyy
    Jun 20, 2019 at 9:30
  • $\begingroup$ \lambda is the proper syntax for $\lambda$. $\endgroup$ Jun 20, 2019 at 13:33

1 Answer 1


$\mathbb{E}\left[t^{X}\mid\lambda=u\right]=e^{-u\left(1-t\right)}$ so that $\mathbb{E}\left[t^{X}\mid\lambda\right]=e^{-\lambda\left(1-t\right)}$ and: $$G_{X}\left(t\right)=\mathbb{E}t^{X}=\mathbb{E}\left[\mathbb{E}\left[t^{X}\mid\lambda\right]\right]=\mathbb{E}e^{-\lambda\left(1-t\right)}=\frac{1}{2}\int_{0}^{2}e^{-\lambda\left(1-t\right)}d\lambda=\begin{cases} \frac{1-e^{-2\left(1-t\right)}}{2\left(1-t\right)} & \text{if }t\neq1\\ 1 & \text{otherwise} \end{cases}$$

Now $P(X=2)$ can be found on base of: $$P(X=2)=\frac{G_{X}^{\left(2\right)}\left(0\right)}{2!}$$

  • $\begingroup$ @StubbornAtom You are correct in stating that the word "independent" (I removed it) is not the right term to use here. It was meant to say that $\Lambda$ only determined the parameter of the distribution and had no other effect on $X$. $\endgroup$
    – drhab
    Jun 20, 2019 at 12:05

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