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A discrete stochastic variable $X$ is defined as the number of eggs laid by a bird at a certain time. Depending on the specie of the bird, the Poisson constant for this distribution is $\lambda = \lambda_1$ or $\lambda_2 $. What is $\text {Pr}(X = k)$?

Am I crazy to think that it is the sum of the Poisson functions on both constants?

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up vote 2 down vote accepted

The question requires interpretation. One reasonable thing is to suppose that the bird laying the eggs is of Species $1$ with probability $p$, and of Species $2$ with probability $1-p$. Perhaps, quite unreasonably, you are expected to assume that $p=\frac{1}{2}$.

Then if $X$ is the number of eggs, $$\Pr(X=k)=pe^{-\lambda_1}\frac{\lambda_1^k}{k!}+(1-p)e^{-\lambda_2}\frac{\lambda_2^k}{k!}.$$ If $\lambda_1\ne \lambda_2$, this is not Poisson for any "mixed" $\lambda$, except in the extreme cases $p=0$ or $p=1$.

Remark: I am not sure what you meant by the sum of the Poisson functions. Maybe you meant $\Pr(X=k)=e^{-\lambda_1}\frac{\lambda_1^k}{k!}+e^{-\lambda_2}$. If so, that is not quite right, though if we take $p=\frac{1}{2}$, as you may be intended to do, it is very close. But we need to divide by $2$. In general $\Pr(X=k)$ is a weighted average of the two mass functions.

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a bird can't be of two different species – user31280 Oct 27 '12 at 4:05
True, I changed the wording. The bird that laid the eggs might be of Species $1$ or it might be of Species $2$. – André Nicolas Oct 27 '12 at 4:08
Thanks you very much, again! – user31280 Oct 27 '12 at 4:24

In a general case, you can always write the interested probability as follows $${\rm{Pr}}(X=k) = {\rm{Pr}}(X=k|\lambda_1){\rm{Pr}}(\lambda=\lambda_1)+{\rm{Pr}}(X=k|\lambda_2){\rm{Pr}}(\lambda=\lambda_2),$$ which uses the total probability formula.

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