Calculating the expected value of a transformation

When confronted with an apparent contradiction in a mathematical derivation, consider developing the answer in a completely different fashion. With statistical problems, one method that is often available is to simulate the problem.

This R code flips the coin as many times as you specify--I use a million (1e7) flips here (which keeps the run time under one second)--identifies the heads, computes the numbers of flips $Y$ preceding each head, and estimates the expectation by using the average of $\exp(-Y).$ It prints out the "z-score," defined as the difference between that and your estimator as expressed as a multiple of the standard error of estimate. Typically, a z-score between $-2$ and $2$ would be taken as evidence your formula is correct.

set.seed(17)
k <- which(rbinom(1e6, 1, 1/2) == 1)
Y <- diff(c(0, k)) - 1
Z <- exp(-Y)
s <- sqrt(var(Z) / length(Z))              # Standard error of estimate
(mean(Z) - exp(1) / (2 * exp(1) - 1)) / s  # The Z-score

The output is

> [1] 0.2558145

This example is not unusual insofar as (a) it's quick and easy to code and (b) takes almost no computational time. As such, it illustrates why it's a good policy always to check your probability calculations with a simulation.

Assuming the description of the question is accurate, the book's claimed result for the expectation cannot possibly be correct because $$\frac{2e^2}{2e-1} > \frac{2e^2}{2e} = e > 2.7,$$ whereas $$\begin{align} \operatorname{E}[Z] &= \operatorname{E}[e^{-Y}] \\ &= e^0 \Pr[Y = 0] + e^{-1} \Pr[Y = 1] + \cdots \\ &< \Pr[Y = 0] + \Pr[Y = 1] + \cdots \\ &= 1. \end{align}$$

There is no need for explicit calculation of the expectation--the above is easy to mentally deduce.

The answer by whuber gives excellent general advice for how to deal with situations of this kind. For completeness, I will show you a derivation of the expected value at issue, which is a particular value of the moment generating function of the geometric distribution. Taking $Y \sim \text{Geom}(p)$ and applying the law of the unconscious statistician allows us to derive the moment generating function:

$$\begin{align} m_Y(t) \equiv \mathbb{E}(e^{tY}) &= \sum_y e^{ty} \cdot \text{Geom}(y|p) \\[6pt] &= \sum_{y=0}^\infty e^{ty} \cdot p (1-p)^y \\[6pt] &= p \sum_{y=0}^\infty (e^t (1-p))^y \\[6pt] &= \frac{p}{1-(1-p)e^t}. \\[6pt] \end{align}$$

In this particular case you have $p=\tfrac{1}{2}$ and $t=-1$ which gives:

$$\mathbb{E}(e^{-Y}) = \frac{\tfrac{1}{2}}{1-\tfrac{1}{2} e^{-1}} = \frac{1}{2-e^{-1}} = \frac{e}{2e-1} = 0.6126998.$$

The book’s answer,

$$\frac{2 e^2}{2 e-1},$$

cannot be right here, because $\mathbb Z = e^{-\mathbb Y}$ always satisfies $0 < \mathbb Z \le 1$, so its expectation must also be at most $1$.

Your calculation is correct. We have, by definition of expectation,

$$E[\mathbb Z] = E\left[e^{- \mathbb Y}\right] = \displaystyle \sum_{y=0}^\infty e^{-y} \left(\frac{1}2 \right)^{y+1} = \frac{1}2 \displaystyle \sum_{y=0}^\infty \left(\frac{1}{2 e} \right)^y.$$

This is a geometric series with ratio $1/(2e)$, so

$$E[\mathbb Z] = \frac{1}2 \cdot \frac{1}{1 - \frac{1}{2 e}} = \frac{e}{2 e-1}.$$

The random variable $\exp(-Y)$ is never more than $1$ and half the time is less than $1,$ so its average value cannot be $2e^2/(2e-1)\approx 3.33098.$

The way you did it is correct.

To check that, I used the following R commands:

e <- exp(1)
n <- seq(0,30)
sum(exp(-n)/2^(n+1))

This gave me $0.6126998.$

The command

e/(2*e-1)

gave me the same number.