Assume a policyholder is four times more likely to file two claims as to file exactly three claims. Assume also that the number X of claims of this policyholder is Poisson. What is $E(X^2)$?
Here's what I have so far:
$ 4(P(X=3))=P(X=2)=(\lambda^2*e^{-2})/2!\\ P(X=3)=(\lambda^3*e^{-3})/3!$
Solving this algebra gives us that $\lambda=2.039$
Since we're asked to find the 2nd moment about X, isn't that just $\lambda$? The answer to$ E(X^2)$ is 21/16. why is that and what procedure do you use to even get a fraction here?