Marlene Brooks

2022-10-12

Recursive proof of expectation of geometric distribution?

If $T\sim \mathsf{G}\mathsf{e}\mathsf{o}(p)$ ($0<p<1$) then $\mathbf{E}T=1/p$ is well known.

Suppose that I toss a sequence of coins that come up heads with probability p. On average, how many coins do I have to toss before I see a heads?

If $T\sim \mathsf{G}\mathsf{e}\mathsf{o}(p)$ ($0<p<1$) then $\mathbf{E}T=1/p$ is well known.

Suppose that I toss a sequence of coins that come up heads with probability p. On average, how many coins do I have to toss before I see a heads?

Jovanni Salinas

Beginner2022-10-13Added 18 answers

Step 1

We can condition on the first toss, let H denote the event that the first toss is a head.

$\begin{array}{rl}E[T]& =E[T|H]E[H]+E[T|{H}^{c}]E[{H}^{c}]\\ & =p+(1+E[T])(1-p)\\ & =1+(1-p)E[T]\end{array}$

Step 2

Solving for E[T] gives us $\frac{1}{p}$.

Notice that $E[T|{H}^{c}]=1+E[T]$ as the first toss and the remaining tosses are independent.

We can condition on the first toss, let H denote the event that the first toss is a head.

$\begin{array}{rl}E[T]& =E[T|H]E[H]+E[T|{H}^{c}]E[{H}^{c}]\\ & =p+(1+E[T])(1-p)\\ & =1+(1-p)E[T]\end{array}$

Step 2

Solving for E[T] gives us $\frac{1}{p}$.

Notice that $E[T|{H}^{c}]=1+E[T]$ as the first toss and the remaining tosses are independent.

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