At time $0$, a coin that lands on heads with probability $p$ is flipped and lands on heads. At times chosen with a Poisson process of rate $\lambda$, the coin is flipped again. What is the probability that the coin is on its head at time $t$?
Note: Flip means I toss the coin again.
I'm having trouble with parsing this question, but also I don't know how I'd solve it any way I think about it.
Am I renewing the Poisson process every time I flip? As in, flip, generate Poisson random number, flip again, generate Poison random number, flip again, etc.
Or is the number generated by the Poisson process the regular interval between flips?
And either way, how would I solve it?
Best Answer
$$N\sim\mathrm{Poisson}(\lambda t).$$
$$ \Pr( N\text{ is even}) = e^{-\lambda t} \sum_{n=0}^\infty \frac{(\lambda t)^{2n}}{(2n)!}= e^{-\lambda t}\cosh(\lambda t)\to\frac 1 2\text{ as }t\to\infty. $$
OK, I misconstrued the word "flipped" (see comments below). And I also treated it as a "fair" coin above. I'll edit further,
unless I don't.Construing "flip" in the more usual sense (not just turning the coin upside down manually, but randomly getting "heads" or "tails"), the problem is simpler: The conditional probability given $N>0$ is $p$, and the conditional probability given $N=0$ is $1$. So we get \begin{align} & 1\cdot\Pr(N=0) + p \Pr(N>0) \\[8pt] = {} & e^{-\lambda t} + p (1-e^{-\lambda t}) \\[8pt] = {} & (1-p)e^{-\lambda t} + p\to p\text{ as }t\to\infty. \end{align}
The process is indeed renewed at each flip, but what you randomly generate at each renewal is the exponentially distributed time $T$ until the next renewal, satisfying $$ \Pr(T>t) = \Pr(N= 0) = e^{-\lambda t} $$ since the event $T>t$ is the same as the event $N=0$ (i.e. not just the probabilities, but the events themselves are the same, in the sense that either event occurs if and only if the other does). Here I'm construing $N$ as the number of later renewals between the time of renewal and and that time plus $t$.