Poisson with its parameter as an exponential random variable

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

• 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. – Matthew Pilling Sep 13 '20 at 16:21
• Oh wow... I see! But how was I supposed to know that it was conditioning on {Λ=λ} and not {Λ=θ}??? – Drew Sep 13 '20 at 17:31
• $\Lambda$ is a random variables that can take on ANY non negative real number. $\Lambda$ does not have to equal $\theta$. – Matthew Pilling Sep 13 '20 at 17:33
• Use MathJax for formatting math. – StubbornAtom Sep 13 '20 at 18:47

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.

• Regardless of approach, it simplifies to $\Bbb P\{X=k\}=\theta/(1+\theta)^{k+1}$, a zero-indexed geometric distribution with $p=\theta/(1+\theta)$. – J.G. Sep 13 '20 at 15:38
• 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 $\infty$". – J.G. Sep 13 '20 at 15:40
• 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. – J.G. Sep 13 '20 at 15:46
• No, don't erase it. – J.G. Sep 13 '20 at 15:47
• 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. – Matthew Pilling Sep 13 '20 at 16:01