Question

At time $0,$ a coin that comes up heads with probability p is flipped and falls to the ground. Suppose it lands on heads. At times chosen according to a Poisson process with rate $\lambda$, the coin is picked up and flipped. (Between these times, the coin remains on the ground.) What is the probability that the coin is on its head side at time$t$? Hint: What would be the conditional probability if there were no additional flips by time $t$, and what would it be if there were additional flips by time $t$?

Short answer

The required probability is$e^{-\lambda t}+p·\left(1-e^{-\lambda t}\right)$

Step-by-step solution

Step 1:Given information

At time $0$, a coin that comes up heads with probability $p$ is flipped and falls to the ground. Suppose it lands on heads. At times chosen according to a Poisson process with rate $\lambda$, the coin is picked up and flipped. (Between these times, the coin remains on the ground.)

Step 2:Explanation

Since we select our times of flipping a coin according to the Poisson process with the rate $\lambda$, the number of times that we flip a coin before has distribution Pois $\left(\lambda t\right)$. Name that random variable $Y$. If $Y=0$, by the definition, the probability that we will finish with Head on is 1 since we do not flip a coin and we are told that we begin with Head. If $Y\ge 1$, the probability that we finish with Heads is simply $p$since it only depends on the final throw. Hence, the probability is

$p=P($Head$\mid Y=0\left)P\right(Y=0)+P($Head $\mid Y\ge 1\left)P\right(Y\ge 1)$

$=e^{-\lambda t}+p·\left(1-e^{-\lambda t}\right)$

Step 3:Final answer

The required probability is$e^{-\lambda t}+p·\left(1-e^{-\lambda t}\right)$