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$X$ has Poisson distribution with parameter $λ$, and $λ$ is a random variable uniformly distributed over the interval $[0,2].$

Find the probability mass function of X with parameter λ. Find the probability that the value of variable is 2.

My solution:

I found the probability mass function of $X$, then $λ$ is not a random variable, and it is: $G_x(t)=e^{{-λ}(1-t)}.$

Lucyy
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  • Do you want a solution verification? – tonychow0929 Jun 19 '19 at 06:54
  • I want to understand how to solve this. I know it is easy, but I do not know how to do that. – Lucyy Jun 19 '19 at 06:57
  • In other words, the random variable $X$ conditioned on $[\Lambda=\lambda]$ has a Poisson$(\lambda)$ distribution, where $\Lambda$ itself is uniform over $[0,2]$. You are asked to find the distribution of $X$. Use total probability theorem. – StubbornAtom Jun 19 '19 at 07:17

2 Answers2

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$$\mathbb{P}(X=t) = \int_0^2 \frac{e^{-\lambda}{\lambda}^t}{t!} \cdot \frac{1}{2} d\lambda = \frac{1}{2t!} \int_0^2 e^{-\lambda}{\lambda}^t d\lambda.$$

Can you complete the remaining steps?

(This involves the incomplete Gamma function. I doubt your answer is correct because $\sum_{t=0}^\infty e^{-\lambda(1-t)}) \neq 1$.)

tonychow0929
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$P[X=2]= \int_{0}^{2}{P[\Lambda =\lambda].P[X=2|\Lambda =\lambda]}d\lambda = \frac{1}{2\Gamma{3}}\int_{0}^{2}{e^{-\lambda}\lambda^{3-1}}d\lambda$

Now using the following theorem

$\frac{1}{2\Gamma{3}}\int_{0}^{2}{e^{-\lambda}\lambda^{3-1}}d\lambda = \frac{1-5e^{-2}}{2}$

Vishaal Sudarsan
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