Question

We have k coins. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0, 1]. We assume that conditioned on Q = q, all coin tosses are independent. Let T; be the number of tosses of the ith coin until that coin results in Heads for the first time, for i = 1, 2,..., k. (T; includes the toss that results in the first Heads.)

You may find the following integral useful: For any non-negative integers k and m,
₁
∫ qᵏ(1-q)ᵐdq = k!m!/(k+m+1)
⁰
Find the PMF of T1. (Express your answer in terms of t using standard notation.)
Fort 1, 2,...,
PT₁ (t) =

Answer 1

The PMF of the time until the first heads appears, T1, for a sequence of coin tosses with a uniformly distributed probability of heads is 1/t!, where t is the number of tosses until the first heads occurs.

Step-by-step explanation:

The student is asking about the probability mass function (PMF) of the time until the first heads appears (T1) in a sequence of coin tosses, where the probability of heads q is a random variable that is uniformly distributed between 0 and 1. Given the scenario, we know that the coin tosses are independent, and the probability of T1 being equal to t is the probability that the first (t-1) tosses result in tails and the t-th toss results in heads.

First, let's denote the probability of heads as 'q' and the probability of tails as '1-q'. Then, the PMF of T1 can be calculated as:

• For the first (t-1) coin tosses, we have tails, which occurs with probability (1-q)^(t-1).
• On the t-th toss, we get heads, which occurs with probability q.

Since q is uniformly distributed, the expected value of q can be found by integrating over the possible values of q:

PT₁(t) = ∫ (1-q)^(t-1) * q dq

Using the given integral formula, we can solve this to get the PMF:

PT₁(t) = (t - 1)! * 0! / (t + 0 + 1) = 1/t!
Therefore, the PMF of T1, expressed in terms of t, is 1/t!