Conditional Expectation Question (Probability Theory)

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SUMMARY

The discussion centers on calculating the conditional expectation E(e-Λ|X=1) where X follows a Poisson distribution with PMF Pλ{X=x} = λxe-λ/x! and Λ follows an Exponential distribution with PDF f(λ)=e-λ for λ>0. The solution involves finding the marginal distribution of X and applying Bayes' theorem to derive the conditional distribution fΛ|X(λ|x). The final answer is confirmed to be 4/9, demonstrating the successful integration of discrete and continuous probability concepts.

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  • Ability to perform integration involving probability density functions
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Homework Statement



(Question is #6 on p.171 in An Introduction to Probability and Statistics by Ruhatgi & Saleh)

Let X have PMF Pλ{X=x} = λxe/x!, x=0,1,2...
and suppose that λ is a realization of a RV Λ with PDF
f(λ)=e, λ>0.

Find E(e|X=1)


The Attempt at a Solution



The answer according to the solutions in the back of the text is 4/9.

I'm really not sure where to begin. My problem is that one of these distributions is continuous and one is discrete and there are no conditional probability examples that are like that.

What I know is that X~Poisson and Λ~Exponential.

I believe that E(e|X=1)=integral{efΛ|X(λ|x)dλ,from λ=0 to infinity}. However, I do not see an easy way to calculate fΛ|X(λ|x).

Also, would the first distribution be considered the conditional probability mass function of X given λ? (That's how I was treating it in my attempts).

Thanks in advance for any tips.
 
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Yay. Nevermind. I was somehow able to solve it. I found the marginal distribution of X, and replaced the fλ|X with fX|λ(x|λ)*f(λ)/f(x) and it worked. :)
 

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