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The above image shows a Poisson distribution for different values of the mean $\lambda$. If I was to fix a value of $k$ (say $k=3$) and sum over the different $\lambda$, is it true that $\sum_{\lambda}P(X=k| \lambda) = 1$?

If not, is there an obvious way to adapt this distribution so that this property is satisfied?

kjetil b halvorsen
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John Doe
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    since $\lambda \in \mathbb{R}_+$ your summation is done for inifinitely many $\lambda$. since you are summing up only positive terms, you can easily create a counterexample to your assertion. you have $\sum_k P(X=k|\lambda) = 1$ for all $\lambda$ but $\sum_\lambda P(X=k|\lambda) \to \infty $ for all $k$ – chRrr Nov 09 '17 at 16:21
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    Likelihood functions don't sum (/ integrate) to 1 in general. – Glen_b Nov 09 '17 at 16:49
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    In general not, but in this case, they appear to. I ran a quick matlab script and it appears this is true, but the mathematical proof of why isn't obvious to me at all. – MikeP Nov 09 '17 at 17:01
  • @MikeP That's interesting. I was really expecting it not to be true. What do you mean by 'in general not'? What case are you referring to? Thanks. – John Doe Nov 09 '17 at 17:05
  • I was agreeing with Glen's comment that these would not in general sum to 1. – MikeP Nov 09 '17 at 17:07
  • @MikeP Oh okay. But it seems to be true for particular case of the Poisson distribution? – John Doe Nov 09 '17 at 17:09
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    @JohnDoe, yes, exactly!!! – MikeP Nov 09 '17 at 17:39
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    A more useful and specific title would be mention the property here, like "Does the likelihood function for the Poisson distribution integrate to 1?" (As @glen_b this doesn't happen in general) – Silverfish Nov 09 '17 at 18:17

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Besides my comment, the claim is true if you replace the sum with an integral (which makes more sense). Indeed, one can show that for all $k \in \mathbb{N}$: $$\int_0^\infty P(X=k|\lambda)\,d\lambda = \int_0^\infty \frac{\lambda^kexp(-\lambda)}{k!}\,d\lambda = \frac{\Gamma(k+1)}{k!} = \frac{k!}{k!} =1.$$ In fact, the gamma function is already defined by $\Gamma(x) = \int_0^\infty \lambda^{x-1}\exp(-\lambda)\, d\lambda$. Moreover it is well known that $\Gamma(k+1) = k!$ if $k \in \mathbb{N}$.

chRrr
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  • Do you maybe know of other probability distributions which satisfy the condition that the likelihood function integrates to $1$? – John Doe Jan 22 '18 at 10:36