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Approaching the following problem:

Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let $W$ denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win. Find $E[W]$

Realizing that the expectation is not just $E[ W] = \sum\limits_{x} i \left(\frac{ 1}{ 2}\right)$ I am unsure of how to approach the problem.

Letting $W_L$ be the value of $W$ accumulated by loses and $W_W$ be the value of $W$ accumulated by wins, I am inclined to believe we are looking at $E[ W_L] = \sum\limits_{i = 0}^\infty -i \left(\frac{ 1}{ 2}\right)^i$ and the $E[ W_W] = 1$.

However, 1) I do not know if this is the correct approach and 2) should this indeed be the correct approach, I do not understand [conceptually ] how to merge these two summations?

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How is $X$ related to $W$? –  sai Jul 27 '12 at 1:38
    
Typo, I apologize! Also since you mention it, I should probably go ahead and change that summation too. –  rudolph9 Jul 27 '12 at 1:50
    
Your approach is good except for the power of 1/2 in $E[ W_L] = \sum\limits_{i = 0}^\infty -i \left(\frac{ 1}{ 2}\right)^{i+1}$. For the overall winnings add $E[W_L]$ and $E[W_W]$ to get zero. –  Byron Schmuland Jul 27 '12 at 2:41

3 Answers 3

up vote 2 down vote accepted

Let $W$ be the "winnings," and $X$ the number of trials until the first head. Then $W=-(X-1)+1=2-X$.

Now use the fact that $E(X)=\frac{1}{1/2}$ and the fact that $E(2-X)=2-E(X)$.

Here we used the fact that if $p$ is the probability of success on any trial, and $p\ne 0$, then $E(X)=\frac{1}{p}$. (The random variable $X$ has geometric distribution.)

There are various ways to prove the required result. For example, when $p=1/2$, the probability that $X=n$ is equal to $\frac{1}{2^n}$. So $$E(X)=1\cdot \frac{1}{2}+2\cdot \frac{1}{^2}+3\cdot \frac{1}{2^3}+\cdots.\tag{$1$}$$ Multiply by $2$. We get $$2E(X)=1+2\cdot \frac{1}{2}+3\cdot \frac{1}{2^2}+4\cdot\frac{1}{2^3}+\cdots.\tag{$2$}.$$ Subtract $(1)$ from $(2)$. We get $$E(X)=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots.$$ The series on the left is an infinite geometric series with sum $2$.

Another way: We find $E(W)$ directly. With probability $\frac{1}{2}$ we win immediately, ending up with $1$ dollar. With probability $\frac{1}{2}$ we "win" $-1$ dollar on the first throw, and in effect the game starts all over again, and our expected net gain is $E(W)$. It follows that $$E(W)=\frac{1}{2}\cdot 1+\frac{1}{2}\cdot(-1)+\frac{1}{2}E(W)).$$ On the assumption that $E(W)$ is finite, we can now solve the above equation for $E(W)$, and find that $E(W)=0$.

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How are you able to assert $E[ X] = \frac{ 1}{ 1/2}$ and how does all fit together to get the $E[W] = 0$? –  rudolph9 Jul 27 '12 at 1:55
    
I do see how pluggin the value you got back into what you defined for $W$ would give you zero but conceptually I do not understand how you are able to simply plug back an expected value for the variable $X$ when solving for the the expected value of $W$ which is dependent on $X$? –  rudolph9 Jul 27 '12 at 1:59
    
I have added a lot of stuff, including many details. –  André Nicolas Jul 27 '12 at 5:43
    
Thanks for clarifying! –  rudolph9 Jul 27 '12 at 14:29

Let $X_1,X_2,\dots$ be the winnings on the individual bets, and $N$ the stopping time representing the gambler's strategy. Since it is a fair game, we have $E(X)=(-1){1\over 2}+(1){1\over 2}=0$. The gambler's total winnings are $W=X_1+X_2+\cdots+ X_N$ and by Wald's equation we get $\mathbb{E}(W)=\mathbb{E}(N)\mathbb{E}(X)=0.$

The moral of the story is that the average winnings of a fair game is zero, no matter what stopping strategy you adopt.

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Let $X$ denotes the number of trial of which winning occurs. Then we have:

$W=1+(-1)(X-1)=2-X$

Hence the expectation would be:

$E[W]\\=E[E[W|X]]\\=\sum\limits_{x} E[W|X=x]P\{X=x\}\\=\sum\limits_{x=1}^\infty \frac{(2-x)}{2^x}\\=0$

Note: The only thing that is little tricky here is to find $\sum\limits_{x=1}^\infty \frac{x}{2^x}$, however we have:

$\frac{1}{2}\sum\limits_{x=1}^\infty \frac{x}{2^x}\\=\sum\limits_{x=2}^\infty \frac{x-1}{2^x}\\=(\sum\limits_{x=1}^\infty \frac{x}{2^x}-\frac{1}{2})-\frac{1}{2}$

So that $\sum\limits_{x=1}^\infty \frac{x}{2^x}=2$.

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I don't see where your getting $\frac{ 1}{ 2} \sum\limits_{x = 1}^\infty \frac{ x}{ 2^x}$ from? –  rudolph9 Jul 27 '12 at 14:29

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