- #1
Bazzinga
- 45
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Hey guys, I'm trying to find a conditional distribution based on the following information:
##Y|u Poisson(u \lambda)##, where ##u~Gamma( \phi)## and ##Y~NegBinomial(\frac{\lambda \phi}{1+ \lambda \phi}, \phi^{-1})##
I want to find the conditional distribution ##u|Y##
Here's what I've got so far:
##f(u|y)= \frac{f(u, y)}{p(y)} = \frac{p(y|u)}{p(y)} f(u)##
##=\frac{(u \lambda)^{y}e^{-u \lambda}}{y!} \frac{u^{ \alpha - 1}e^{-u/ \beta}}{ \Gamma ( \alpha) \beta^{ \alpha}} \frac{y! \Gamma ( \alpha)}{\Gamma (y+ \alpha)} ( \frac{1+\lambda \beta}{ \lambda \beta})^{y} (1+ \lambda \beta)^{ \alpha}##
where ##\beta = \phi## and ## \alpha = \phi^{-1}##
(I'm using ##\beta## and ##\alpha## right now because that's what it does in the notes)
I'm not sure where to go from here. I can cancel some terms out, but then I'm not sure what distribution I'm supposed to end up with. Could anyone give me a push in the right direction?
##Y|u Poisson(u \lambda)##, where ##u~Gamma( \phi)## and ##Y~NegBinomial(\frac{\lambda \phi}{1+ \lambda \phi}, \phi^{-1})##
I want to find the conditional distribution ##u|Y##
Here's what I've got so far:
##f(u|y)= \frac{f(u, y)}{p(y)} = \frac{p(y|u)}{p(y)} f(u)##
##=\frac{(u \lambda)^{y}e^{-u \lambda}}{y!} \frac{u^{ \alpha - 1}e^{-u/ \beta}}{ \Gamma ( \alpha) \beta^{ \alpha}} \frac{y! \Gamma ( \alpha)}{\Gamma (y+ \alpha)} ( \frac{1+\lambda \beta}{ \lambda \beta})^{y} (1+ \lambda \beta)^{ \alpha}##
where ##\beta = \phi## and ## \alpha = \phi^{-1}##
(I'm using ##\beta## and ##\alpha## right now because that's what it does in the notes)
I'm not sure where to go from here. I can cancel some terms out, but then I'm not sure what distribution I'm supposed to end up with. Could anyone give me a push in the right direction?