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Let $T_1, T_2, \ldots, T_n$ be $n$ iid. exponential variables with common mean $a$. After a period of time $T_1$ has elapsed a biassed coin, with heads probability $p$, is tossed. If a tail appears the coin is tossed again after a period $T_2$ has elapsed, and so on until a head appears.

What is the distribution of the time until the first head appears?

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The time it takes for the first head to appear is a random variable equal to:

$$Y = \begin{cases} T_1 & \text{with probability $p$} \\ T_1 + T_2 & \text{with probability $(1-p)p$}\\\ \dots \\ \sum_{i=1}^n T_i & \text{with probability $(1-p)^{n-1}p$}\\ \dots \end{cases}$$

With this in mind, how can we compute $P(Y \le y)$? There are several possibilities for this event to happen:

  • $T_1 \le y$ and the first coin comes up head
  • $T_1 + T_2 \le y$ and the first coin comes up tail, and the second comes up head

and so on. Note that these events are disjoint (the coin toss do not match) so this means that

$$P(Y \le y) = \sum_{i=1}^\infty P\left(\sum_{k=1}^iT_k \le y, \{\text{first $i-1$ coin toss are tail, the $i$-th one is head}\} \right) $$

Now, the coin toss are independent on the exponential random variables, and the sum of independent exponential random variables with parameter $\alpha$ is a gamma with parameter $n$ and $\alpha$, i.e. $\sum_{i=1}^n T_i \sim \Gamma(n, \alpha)$. Call $F_{n, \alpha}$ the cdf of a $\Gamma(n, \alpha)$ distributed random variable. We can therefore write

$$P\left(\sum_{k=1}^iT_k \le y, \{\text{first $i-1$ coin toss are tail, the $i$-th one is head}\} \right) = F_{i, \alpha}(y)(1-p)^{i-1}p$$

Substituting in we get $$P(Y \le y) = p\sum_{i=1}^\infty F_{i, \alpha}(y)(1-p)^{i-1}$$

We know the cdf of the gamma distribution in terms of the lower incomplete gamma function, so that

$$P(Y \le y) = p\sum_{i=1}^\infty \frac {(1-p)^{i-1}}{(i-1)!}\gamma(i, \alpha y)$$

Using the recurrence relation for $\gamma$ ($\gamma(s+1, x) = s\gamma(s,x)-x^se^{-x}$) we could potentially simplify that expression, but it doesn't seem so easy.

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