Independent coin tosses , double or halve current sum

(Quant job Interviews - Questions and Answers - Joshi et al, Question 3.5)

Suppose you have a fair coin. You start with 1 dollar, and if you toss a H your position doubles, if you toss a T your position halves. What is the expected value of the money you have if you toss the coin to infinity ?

Now the answer is stated as follows:

We work out what happens with one toss, then $n$ tosses and then let $n$ tend to infinity. Let X denote a toss then: $$\mathbb E (X) = \frac{1}{2} * 2 + \frac{1}{2} * 0.5= {5\over4}$$ Provided the tosses are independent, the product of expectations is the expectation of the product. Let $X_j$ be the effect of toss $j$. This means that $$\mathbb E (\prod_{j=1}^{n} X_j) = \prod_{j=1}^{n} \mathbb E (X_j) = ({5\over4})^n$$ this clearly tends to infinity as n tends to infinity

Now, I don't understand this answer :(

First, the way the answer is written out, surely the ${5\over4}$ is the expectation of the outcome of the first toss $X_1$ , not that of a toss $X_j , j \ge 1$ ?

Secondly, whilst I do understand that the tosses are independent, it would seem that the $X_{j+1}$ is actually quite heavily dependent on the $X_{j}$ before it ?

So then why is it so obvious that $\mathbb E ( X_{j+1} ) = \mathbb E ( X_{j})$ ?

• Let $X_i$ be the outcome of the $i$th toss translated into $\{-1,1\}$. Then $M_n=2^{X_1+\dots+X_n}$. The sum in the exponent is symmetric around $0$ and spreads out as $n\to \infty$, hence $E(M_n)\to \infty$. – A.S. Nov 11 '15 at 14:38

Let $S_{i}$ be the random variable that denotes your dollars after the toss $i$.
It's clear that $E(S_{1}) = \frac{1}{2}\times 2 + \frac{1}{2} \times \frac{1}{2} = \frac{5}{4}$.
Then, $S_{2}$ takes two "values" : $2S_{1}$ or $\frac{S_{1}}{2}$. Thus $E(S_{2}) = \frac{1}{2}\times 2E(S_{1}) + \frac{1}{2} \times \frac{E(S_{1})}{2} = \frac{5}{4} \times E(S_{1}) = (\frac{5}{4})^{2}$.
By recurrence $$E(S_{n}) = (\frac{5}{4})^{n}$$