the probability of X being even in a Poisson is higher than the probability of it being odd

Question

Ive came across this question and im not sure how to solve it : if X~Pois($\lambda$) , $\lambda >0$
with the identity

$$e^{\lambda}=\sum_{k=0}^\infty \frac{\lambda^k}{k!}, \space \forall \lambda \in \Bbb R$$

Prove that the probability of X being even is higher than the probability of it being odd

Answer 1

1) The sum of even X and odd X is equal to $1$.

2) The difference of even X and odd X is

$\sum_{k=0}^{\infty} (-1)^{k}\cdot e^{-\lambda}\cdot \frac{\lambda^k}{k!}= e^{-\lambda}\cdot \sum_{k=0}^{\infty} (-1)^{k} \frac{\lambda^k}{k!}=e^{-2\lambda}$

Now you can sum 1) and 2). I think you can take it from here.

Answer 2

You begin with the expressions: $$\mathsf P(E) = \sum_{k=0}^\infty \mathsf P(X=2k) $$

$$\mathsf P(O) = \sum_{k=0}^\infty \mathsf P(X=2k+1)$$

Then you apply the p.m.f. , $\mathsf P(X=n)= \lambda^n e^{-\lambda}/n!$

Finally: Carefully subtract one from the other and examine the result.