Does the likelihood function for the Poisson distribution integrate to 1

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?

• 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 9 '17 at 16:21
• Likelihood functions don't sum (/ integrate) to 1 in general. – Glen_b Nov 9 '17 at 16:49
• 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 9 '17 at 17:01
• @JohnDoe, yes, exactly!!! – MikeP Nov 9 '17 at 17:39
• 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 9 '17 at 18:17

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 $k \in \mathbb{N}$.
• 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