Poisson question

#1
Can anyone shed some light on this one?

When cycling home at night, I notice that sometimes my rear light is switched off when
I arrive home. Presumably the switch is loose and can flip from on to off or back again
when I go over bumps. I suppose that the number n of flippings per trip has a Poisson
distribution (e^−λ)(λ^n) /n!. If the probability that the light is still on when I arrive home is p, find λ.

Thanks.
 
#2
Hi, new to the forum and don't know how to use symbols here but I have the answer so I will hack at it will Latex. It is not that hard once you realize there must be a simplificaiton if you want to solve for λ analytically. Basically you are given the probability that the distribution is even.


p = P(even) = \Sum_{even} \frac{e^{-λ} λ^n}{n!}
= e^{-λ} \Sum_{even} \frac{ λ^n}{n!}
= e^{-λ} (1 + \frac{ λ^2}{2!} \frac{ λ^4}{4!} + .... )
= e^{-λ} cosh(λ)
= e^{-λ} *0.5 (e^{λ} +e^{-λ} )
= 0.5 (e^0 + e^{-2λ})

ln(2p-1)=-2λ

λ=-0.5 * ln(2p-1)

Let me know if I made any mistakes,
Keith