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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

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  • $\begingroup$ PS: $\displaystyle e^x = \sum\limits_{k=0}^\infty \dfrac{x^k}{k!}$ , or $\displaystyle e^\lambda = \sum\limits_{k=0}^\infty \dfrac{\lambda^k}{k!}$ .. Don't jumble them together $\endgroup$
    – Graham Kemp
    Nov 9, 2016 at 4:47
  • $\begingroup$ @grahamKemp I think I got it I have to add up $e^x$ and $e^{-x}$ in order to get an even sequence. After that, I'm not sure why I have to divide it by 2. And thank you ! $\endgroup$
    – user387171
    Nov 9, 2016 at 4:51

2 Answers 2

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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.

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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.

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  • $\begingroup$ you know how I try substracting but I have some trouble Here goes: $e^{-\lambda} (\sum_{k=0}^{\infty} \frac{\lambda^{2k}}{(2k)!} - \sum_{k=0}^{\infty} \frac{\lambda^{2k+1}}{(2k+1)!})$ to $e^{-\lambda} (\frac{e^\lambda}{2\lambda^{\frac{1}{2}}} - \sum_{k=0}^{\infty} \frac{\lambda^{2k+1}}{(2k+1)!})$ I dont know how to do the second part $\endgroup$
    – user387171
    Nov 9, 2016 at 5:38

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