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So like let $Y | \mu \sim \operatorname{Poisson}(\mu $. $P(Y =y | \mu ) = \frac{e^{-\mu}\mu^y}{y!}$. Let $\mu \sim \Gamma(\alpha, \beta )$. I'm looking to find the marginal distribution of Y i.e. $P(Y = y)$

I don't have a textbook but found online that the equation is

$$P(Y = y) = \int_{x}^{y} P(Y = y | \mu )f(\mu )du$$

How do I find the limits for x and y ?

amWhy
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1 Answers1

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By the Law of total probability, $P(Y=y) = \int\limits_R P(Y=y|X=x)f_X(x)dx$, where $X$ is some continuous random variable and $R$ is the support of $X$. Letting $X$~$\mathrm{Gamma}(\alpha,\beta), R = (0,\infty).$ Thus the limits of your integration should be $0$ and $\infty$. Hope that helps.

jeremy909
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