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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?

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    $\begingroup$ Depending on the sort of student you are, this might be helpful to get some insight into Poisson processes. $\endgroup$ – Pedro Mar 11 '15 at 19:20
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$$N\sim\mathrm{Poisson}(\lambda t).$$

$$ \Pr( N\text{ is even}) = e^{-\lambda t} \sum_{n=0}^\infty \frac{(\lambda t)^{2n}}{(2n)!}= e^{-\lambda t}\cosh(\lambda t)\to\frac 1 2\text{ as }t\to\infty. $$

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 $$ 1\cdot\Pr(N=0) + p \Pr(N>0) = e^{-\lambda t} + p (1-e^{-\lambda t}) = (1-p)e^{-\lambda t} + p\to p\text{ as }t\to\infty. $$

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 $t$.

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  • $\begingroup$ I guess by 'flipped', he means tossed and not simply turned upside down. Not sure though, but the first sentence says "is flipped and lands on...". In that case, the answer would remain $p$. $\endgroup$ – Bravo Mar 11 '15 at 18:47
  • $\begingroup$ Yes, this is correct, I will clarify. Also, I'm looking for probability of heads at time t, not as t goes to infty. $\endgroup$ – ok_ Mar 11 '15 at 19:00
  • $\begingroup$ I'm having difficult grasping why we want the Poisson mean to be $\lambda t$. Sorry! $\endgroup$ – ok_ Mar 11 '15 at 19:18
  • $\begingroup$ @ok_ : That's what "rate" means. It can be shown (and it's not hard) that if the average of the exponentially distributed time until the next renewal is $1/\lambda$ then the number of renewals in a time interval of lentgth $t$ is Poisson-distributed with average $\lambda t$. The parameter $\lambda$ is in units of frequency, i.e. some number of arrivals per unit of time, so $1/\lambda$ is in units of time and $\lambda t$ is unitless. ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 11 '15 at 19:31
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    $\begingroup$ Someone posted an answer here and then deleted it, and linked to this video and said something relevant is in the last 15 minues, but I haven't looked at that part yet. $\endgroup$ – Michael Hardy Mar 11 '15 at 19:34
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Let $A_t$ be shorthand for "Something is true at time $t$". In your case, $A_t$ is "the coin shows heads at time $t$". Then the method for solving is:

\begin{align} P(A_t) &= \sum_{N = 0}^{\infty} P(A_t \mid N \text{ events have occurred by time }t)\cdot P(N \text{ events have occurred by time }t) \\ &= \sum_{N = 0}^{\infty} P(A_t \mid N \text{ events have occurred by time }t) \cdot e^{-\lambda t}\frac{(\lambda t)^{N}}{N!}\end{align}

[In your case, however, the problem is much simpler: Whether the coin is heads or not just depends on the most recent flip. It does not depend on how many flips have occurred. So the answer is simply $p$.]

EDIT: This is wrong: When $N = 0$ we have $P(A_t \mid N) = 1$ as mentioned below. Otherwise we get $p$. So the probability is:

$$P(A_t) = e^{-\lambda t} + \sum_{N = 1}^{\infty}p \cdot e^{-\lambda t}\frac{(\lambda t)^{N}}{N!} = e^{-\lambda t} + p(1 - e^{-\lambda t})$$

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  • $\begingroup$ I don't think this is correct. You start of with a head, so the probability of getting a head is bigger at times close to 0. If you take t equal to infinity what you say will be correct. $\endgroup$ – Pedro Mar 11 '15 at 18:53
  • $\begingroup$ Yeah you're right. I missed that the coin landed on heads at time 0. $\endgroup$ – Alex Zorn Mar 12 '15 at 0:20

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