Marginal Distribution of X w/ Lambda Parameter: Probability Help

AI Thread Summary
The discussion revolves around finding the marginal distribution of a Poisson random variable X with a parameter λ that follows an exponential distribution. The user expresses uncertainty in applying the equations for dependent distributions and seeks clarification on the correct approach, specifically mentioning the need for Bayes' rule. They attempt to compute the marginal distribution using an integral but encounter difficulties with complex outputs from Mathematica and a calculator. The conversation highlights the challenges of integrating probability distributions and the importance of proper manipulation for solvable integrals. Overall, the thread emphasizes the need for guidance in applying theoretical concepts to practical problems in probability.
ryzeg
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I am a little shaky on my probability, so bear with me if this is a dumb question...

Anyway, these two random variables are given:

X : Poisson (\lambda)
\lambda : Exponential (\theta)

And I simply need the marginal distribution of X and the conditional density for \lambda given a value for X

I have all the equations for dependent distributions, but do not know how to apply them to this ostensibly easy problem...

Any help?
 
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You'll need Bayes' rule for this. What results have you got so far?
 


I was doing this, but I think it is wrong:

<br /> f_X(x) = \int^{\lambda=\infty}_{\lambda=0} \frac{\lambda^{x}}{x!} e^{-\lambda} \times \theta e^{-\theta \lambda} d \lambda<br />

Plugging this integral into Mathematica gives a really nasty output with a incomplete gamma function, and my TI-89T cannot evaluate it.
 


I take that back; the integral is doable with a little manipulation. Damn machines...
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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