**Question:** Suppose X is a Poisson random variable with mean λ. The parameter λ is itself a random variable whose distribution is exponential with mean 1. Show that P{X =n} = (^{1}⁄_{2})^{n+1}

**Analytical Solution**

There is not much to it, just set up the conditional expression and that should be it

P{X=n} = Integral( P{ X=n | Λ = λ } P{ Λ = λ }) ∂λ with ( λ ∈ [0, ∞) ) = ^{Γ(n+1)}⁄_{n! 2n+1} = (^{1}⁄_{2})^{n+1}

*Note, P{ X=n | Λ = λ } would the PMF of the Poisson Distribution*

**Simulation Solution**

The first image below is showing the analytical solution and the second one is showing the experiment. The code for the both is provided below in the code section.

**Code**

Module[
{\[CapitalLambda] := RandomVariate[ExponentialDistribution[1]], data},
data =
KeySort@
Counts[
Table[RandomVariate[PoissonDistribution[\[CapitalLambda]]], 10000]
];
BarChart[
data,
ImageSize -> 788,
Frame -> True,
ChartLabels -> Placed[{Keys@data, Values@data}, {Below, Above}],
ChartStyle -> Directive[EdgeForm[Red], LightBlue],
FrameLabel -> (Style[#, 20] & /@ {"Outcome", "Count"})]
]
Plot[1 / 2^(n + 1), {n, 0, 10}, PlotRange -> All, Frame -> True,
ImageSize -> 788,
GridLines -> {Range[0, 10, 0.5], Range[0, 0.5, 0.05]},
AspectRatio -> 0.5,
FrameLabel -> (Style[#, 20] & /@ {"n",
"\!\(\*FractionBox[\(1\), SuperscriptBox[\(2\), \(n + 1\)]]\)"})]

End of the post 😉

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