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I and many of my class mates are struggling very hard on this problem:

Let $X$ be a random variable with Poisson distribution, with parameter $\lambda$, where $\lambda$ itself is a random variable with exponential distribution of mean $1/\theta$, that is, $X \sim \text{Poisson}(\lambda)$, $\lambda \sim \exp(\theta)$. Find the marginal distribution of $X$.

I tried:

P(X=x) = (w.r.t λ from 0 to inf) ∫ P(X=x, λ=θ) = ∫ P(X=x | λ=θ)* P(λ=θ)

And then got stuck a.k.a. something not integrate-able

Help... T.T

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  • $\begingroup$ In your attempt you're conditioning on the event $\{\Lambda=\theta\}$. You should be conditioning on the event $\{\Lambda=\lambda\}$ and using the fact that $f_{\Lambda}(\lambda)=\theta e^{-\theta \lambda}$ for $\lambda \geq 0$ and $f_{\Lambda}(\lambda)=0$ otherwise. $\endgroup$ – Matthew Pilling Sep 13 '20 at 16:21
  • $\begingroup$ Oh wow... I see! But how was I supposed to know that it was conditioning on {Λ=λ} and not {Λ=θ}??? $\endgroup$ – Drew Sep 13 '20 at 17:31
  • $\begingroup$ $\Lambda$ is a random variables that can take on ANY non negative real number. $\Lambda$ does not have to equal $\theta$. $\endgroup$ – Matthew Pilling Sep 13 '20 at 17:33
  • $\begingroup$ Use MathJax for formatting math. $\endgroup$ – StubbornAtom Sep 13 '20 at 18:47
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Hint

Using total probability yield

$$\mathbb P\{X=k\}=\int_0^{\infty } \mathbb P\left\{X=k\mid \lambda =t\right\}f_\lambda (t)\,\mathrm d t=\frac{\theta }{k!}\int_0^\infty t^ke^{-(1+\theta )t}\,\mathrm d t, $$ which is integrable.

Several possibilities :

  • Brute force, $k$ integration by part is requiert.

  • Nevertheless, an antiderivative of $t^ke^{-t}$ is of the form $$(a_0+a_1t+...+a_kt^k)e^{-t}.$$ So, you can easily find the $a_i$.

  • An other way : do the substitution $u=(1+\theta )t$, and write the integral using Gamma function.

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    $\begingroup$ Regardless of approach, it simplifies to $\Bbb P\{X=k\}=\theta/(1+\theta)^{k+1}$, a zero-indexed geometric distribution with $p=\theta/(1+\theta)$. $\endgroup$ – J.G. Sep 13 '20 at 15:38
  • $\begingroup$ Why did you change "an antiderivative" to "the antiderivative"? There are multiple antiderivatives, exactly one of which has the stated form. You could say "the antiderivative vanishing at $\infty$". $\endgroup$ – J.G. Sep 13 '20 at 15:40
  • $\begingroup$ To discuss plurals you'd say something like, "the antiderivatives of $t^ke^{-t}$ are of the form $(a_0+a_1t+\cdots+a_kt^k)e^{-t}+C$". Your correction still doesn't make clear a $+C$ is needed for multiple antiderivatives to take the specified form. $\endgroup$ – J.G. Sep 13 '20 at 15:46
  • $\begingroup$ No, don't erase it. $\endgroup$ – J.G. Sep 13 '20 at 15:47
  • $\begingroup$ In your attempt you're conditioning on the event $\{\Lambda=\theta\}$. You should be conditioning on the event $\{\Lambda=\lambda\}$ and using the fact that $f_{\Lambda}(\lambda)=\theta e^{-\theta\lambda}$ for $\lambda\geq0$ and $f_{\Lambda}(\lambda)=0$ otherwise. $\endgroup$ – Matthew Pilling Sep 13 '20 at 16:01

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