TITLE: What is the probability of getting an even number from a Poisson random draw?
QUESTION [12 upvotes]: Below is a graph showing the probability of drawing an odd number (y-axis) from a Poisson distribution with a given expected value (x-axis)
x = seq(0,1e4,1)         // range of values to explore 
lambdas = seq(0,4,0.01)  // expected value of the Poison distribution

FracOdd = numeric(n+1)       // response variable
for (i in 1:length(lambdas)) 
{
   FracOdd[i] = sum(dpois(x,lambdas[i])[seq(2,length(x),2)])  // calculate probability of drawing an odd number
}

plot(y=FracOdd,x=lambdas, type="l", lwd=3, xlab="Expected value", ylab="Probability of drawing an odd number")   // plot the data


It seems that the probability of drawing an odd number from a Poisson distribution with non-infinite mean is always lower than the probability of drawing an even number.

What is the intuition for why the probability of drawing an odd number from a Poisson distribution always below 0.5?
Is the function I drew numerically, easy to derive analytically?

REPLY [8 votes]: Yeah, it's reasonably well known.
$$\begin{align}\mathsf P(\text{Even}) ~&=~ \sum_{k=0}^\infty \dfrac{\lambda^{2k} e^{-\lambda}}{(2k)!}\\[1ex] ~&=~ e^{-\lambda}\cosh(\lambda)\\[1ex] ~&=~ e^{-\lambda}\tfrac 12(e^{\lambda}+e^{-\lambda})\\[1ex]~&=~ \tfrac12(1+e^{-2\lambda})
\\[2ex]\mathsf P(\text{Odd}) ~&=~ \sum_{k=0}^\infty \dfrac{\lambda^{2k+1} e^{-\lambda}}{(2k+1)!} \\[1ex]~&=~ e^{-\lambda}\sinh(\lambda)\\[1ex] ~&=~ e^{-\lambda}\tfrac 12(e^{\lambda}-e^{-\lambda})\\[1ex]~&=~ \tfrac12(1-e^{-2\lambda})\\[2ex]e^{-\lambda}(\cosh\lambda-\sinh\lambda)~&=~e^{-2\lambda}\end{align}$$