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:
$$\mathbb E (X) = \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot 0.5= {5\over4} $$

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
$$ \mathbb E \left(\prod_{j=1}^n X_j\right) = \prod_{j=1}^n \mathbb E (X_j) = \left({5\over4}\right)^n$$
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 $\mathbb E ( X_{j+1} ) = \mathbb E ( X_j)$ ?
 A: $$
X_j = \begin{cases} 2 & \text{with probability } 1/2, \\ 1/2 & \text{with probability } 1/2. \end{cases}
$$
You said "Let $X_j$ be the effect of toss $j.$" If you think these are not independent, I wonder if you thought this meant "Let $X_j$ be your position after toss $j.$" or something like that. But your position after toss $j$ is
$$
Y_j = \prod_{k=1}^j X_k.
$$
Certainly $Y_{j+1}$ is, in your words, "quite heavily dependent" on $Y_j.$
A: You're right, the given answer is not very clear. 
I'll try to provide a proof, but I warn you I'm far from being a probabilist so it might be not perfectly correct. 
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 
$$E(S_{n}) = (\frac{5}{4})^{n}$$
