(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 $\mathbb E ( X_{j+1} ) = \mathbb E ( X_j)$ ?