NO it means that, we just know lambda...
I'm sure I can solve this, but before I attempt it I want to know how you write
this density Exp .
I'm glad I asked, since that's not the exponential density I use.
I just solved this. The answer looks weird, but is a valid distribution.
It's a geo with
I hate using f(x) when talking about discrete distributions, but this is a mixture of discete and continuous.
Multiplying the conditional of Y given lambda with the distribution of lambda gives you the joint distribution of Y and lambda.
That being said, this joint density is
So, the marginal of Y is
NOW, what I do instead of making a substitution I just recognize the GAMMA density and note that ALL densities integrate to one.
(you could make a substitutin and then use the gamma function, but I prefer...)
the integral is one, NOT how I write my gamma, but I did it your way
also since y is an integer.
This reduces to
y=0,1,2,... which is that geo.
This is a good example of what I am trying to master, therefore, I'd like to try to summarise the approach used here. We are given a conditional distribution for Y on Lambda and told the (marginal?) distribution for Lambda. We are asked to find the (marginal?) distribution of Y.
1. First, if Y is conditioned on Lambda distributed according to some distribution (here Poisson), that means that Y is the random variable in the Poisson formula, as here:
3. And finally to find we integrate (or summate) the joint density over the range of Lambda. Whether this is integration or summation will be determined by the type of Lambda variable, right? Here Lambda is exponentially distributed, therefore, continous, therefore, we integrate.
Does it make sense?