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

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    $\begingroup$ 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 $E(M_n)\to \infty$. $\endgroup$
    – A.S.
    Nov 11, 2015 at 14:38

2 Answers 2

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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 $$E(S_{n}) = (\frac{5}{4})^{n}$$

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  • $\begingroup$ Also solved it by recursion, official answer seemed fuzzy. I found it easier to formulate as $$\Pi_{t}=\Pi_{t-1} \cdot B$$ $$\text{E}(\Pi_{t})=\text{E}(\Pi_{t-1} \cdot B)$$ by independence $$\text{E}(\Pi_{t})=\text{E}(\Pi_{t-1}) \cdot \text{E}(B)$$ $$\text{E}(\Pi_{t})=\text{E}(\Pi_{t-1}) \cdot 1.25$$ Which then leaves me with a system of equations $$\begin{cases} \text{E}(\Pi_{t})=\text{E}(\Pi_{t-1}) \cdot 1.25 \\ \text{E}(\Pi_{0})=1 \end{cases}$$ to solve, yielding $$\text{E}(\Pi_{t})=1.25^{t}$$ $\endgroup$ Apr 28, 2020 at 15:23
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$$ X_j = \begin{cases} 2 & \text{with probability } 1/2, \\ 1/2 & \text{with probability } 1/2. \end{cases} $$ 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 $Y_j.$

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