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Given that $R_i$ denotes a random amount (£1 or -£1), depending on the outcome of the coin $i$-th toss:

1) Why $E[R_i R_j]=0$. I can understand this conceptually, I think, or at least I thought I did before I have written that... Does that say expected value of $R_i$ multiplied by$R_j$? And how to show mathematically that it is zero? Would it be: $$\Sigma_{r_i,r_j:p(r_i,r_j)>0} \ r_i r_j p(r_i r_j)$$ No, that can't be true, because then we would have a non-zero probability for at least one combination, and since tosses are independent it would not make sense. Oh.. does it just mean that the tosses are independent?

2) Now if $S_i$ is the total amount of money accumulated up to and including the $i$-th toss, then why is $E[S_i^2]=E[R_1^2+2R_1R_2+...]=i$ ?

3) And a final question. Why is $E[S_6|R_1,R_2,...,R_5]=S_5$

And of course when I say why, I would ideally like to see some extended mathematical explanation.

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We have the definition of $E(X)$: $$ E(X) = \sum_{x\in X} xP(x) $$ That is $E(X) = \text{The expected (mean) value when looking at the variable X}$.
And you defined a variable $R_i$ to have the sample space $\{1, -1\}$ (pound). The probability of each sample is the same, because we are flipping coins. So you have $P(1)=0.5,\, P(-1)=0.5$. You also defined the variable $S_i$ to be the accumulated value of $R_j$ from $j = 1$ to $j = i$.

Answer to your first question: $$ E(R_iR_j) = P(r_i=1,r_j=1)\cdot 1\cdot 1 +\\ P(r_i=1,r_j=-1)\cdot 1\cdot (-1) +\\ P(r_i=-1,r_j=1)\cdot (-1)\cdot 1 +\\ P(r_i=-1,r_j=-1)\cdot (-1)\cdot (-1) =\\ \frac{1}{4}\cdot 1\,+\,\frac{1}{4}\cdot (-1)\,+\,\frac{1}{4}\cdot (-1)\,+\,\frac{1}{4}\cdot 1 = 0 $$

Answer to your thrid question: $$ E(S_6)=E(S_5+R_6)=E(S_5)+E(R_6)=S_5+E(R_6)\\\text{where }E(R_6)=1\cdot P(1)+(-1)\cdot P(-1) = 1\cdot 0.5+(-1)\cdot 0.5=0\\\text{So we get } E(S_6)=S_5 $$

The $E(...|...)$ just specifies the variables $R_1$ to $R_5$ is already chosen.

Finally, the hard one can be solved by using a rule for two independent variables: If $X$ is independent on $Y$ then $E(X\cdot Y)=E(X)\cdot E(Y)$. $$ E(S_i^2)=E\left({\left(\sum_{j=1}^i{R_j}\right)}^2\right)= E\left(\sum_{j=1}^i\left(R_j^2+2\sum_{j<k\le i}{R_j\cdot R_k}\right)\right) =\\ \sum_{j=1}^i\left(E(R_j^2)+2\sum_{j<k\le i}{E(R_j\cdot R_k)}\right) =\\ \sum_{j=1}^i\left(E(R_j^2)+2\sum_{j<k\le i}{E(R_j)\cdot E(R_k)}\right) =\\ \sum_{j=1}^i\left(E(R_j^2)+2\sum_{j<k\le i}{0\cdot 0}\right) = \sum_{j=1}^iE(R_j^2) = i\cdot E(R_1^2) = i\cdot\left(0.5\cdot 1^2\, +\, 0.5\cdot (-1)^2\right) = i $$ For those who are interested, this holds too: $$ E(S_i^n)= \begin{cases} 0\quad\;\ n\text{ is odd}\\ i^{\frac n2}\quad n\text{ is even} \end{cases} $$

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  • $\begingroup$ This is great! Thank you very much. For the very last line, I do not see how we got $i$ in front of the expected expression, and why did $E[R_j^2]$ change to $E[R_1^2]$? $\endgroup$ – i squared - Keep it Real Nov 4 '15 at 9:14
  • $\begingroup$ We know that $R_j$ does not depend on $j$ (in terms of probability) so we can choose $j$ to be what we want (in this case $1$). $\endgroup$ – Niklas Bäckström Nov 4 '15 at 15:31
  • $\begingroup$ This can be done because the expression inside $E$ does not already contain the variable $R_1$. So an example when this cannot be done would be: $\sum_{j=1}^i{E(R_j\cdot R_1)}$. $\endgroup$ – Niklas Bäckström Nov 4 '15 at 15:41

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