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Hi,

(Please forgive me if my question is vague or trivial)

Let a normal distribution $P(\mu, \sigma)$ and a poisson distribution $Q(\lambda)$. I want to find a distribution $Q'$ that is :

  • a poisson with parameter $\lambda'$
  • contains the elements of $P$, which altogether form $Q'$

Or in other words: $Q'$ is like a projection (or a reduction) of $P$ with respect to $Q$.

Do I have to use bayesian inference (posterior, prior, likelihood...) ?

Thanks a lot !

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I don't understand this question at all. What are the "elements of $P$"? What does it mean for $Q'$ to "contain" those elements? You say you want to find a distribution $Q'$ which is Poisson with parameter $\lambda'$; saying that the distribution is Poisson with parameter $\lambda'$ is already specifying the distribution completely; what more is there to "find"? –  James Martin Apr 20 '12 at 20:47
    
Forgive me. Here is another explanation: $p, q, q'$ are probability distributions over a finite sample space $\Omega=1,2,3...n$ The elements of $p$ (resp. $q$) are $p_i=p(i)$ (resp. $q_i=q(i)$)$\ \ \forall i \in [1,n]$ We have $p \sim \mathcal{N}(\mu, \sigma),\ q \sim \mathcal{Pois}(\lambda)$, and$\ q' \sim \mathcal{Pois}(\lambda')$ What I meant is that $q'$ is built by sampling $p$ according to the distribution $q$ ($q'\sim p|q$). If $q$ is a poisson then $q'$ will contain the $p_k$ that fit the most a poisson distribution ($q$). –  cybgeex Apr 21 '12 at 9:43
    
It's like finding the best fit $q'$ to the distribution $p$ such that $q'$ has the same distribution law of $q$. –  cybgeex Apr 21 '12 at 9:44
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