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Suppose we play a game at a casino. There is a \$5 stake and three possible outcomes: with probability $1/3$ you lose your stake, with with probability $1/3$ the bank returns your stake plus \$5, and with probability $1/3$ the bank simply returns your stake.

Let $X$ denote your winning in one play. And you play 1000 times. What would you expect the magnitude of your win or loss to be approximately?


Is this question asking the same as to find the $Var(S)$ where $S=X_1+\cdots X_{1000}$, so the final answer should be $1000\cdot Var(X)$?

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Well, no, the question is asking about an expected value.... –  user5137 Dec 9 '11 at 23:43
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The variance measures the expected squared win or loss. So the standard deviation (square root of the variance) is a reasonable answer to the question. As @JackManey points out, this isn't exactly the same as the expected value of the absolute win or loss, but for the normal distribution they're the same order of magnitude. –  mjqxxxx Dec 9 '11 at 23:48
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Hard to know, but my guess is that it is sloppy language for $E(\sum X_i)$. "What would you expect" is kind of a giveaway, too inexact. –  André Nicolas Dec 10 '11 at 0:34
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I'd interpret the question as looking for either $SD(S)$ or $E(|S|)$, which it has been pointed out are similar in magnitude if $E(S) = 0$. –  Michael Lugo Dec 10 '11 at 0:47
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2 Answers 2

up vote 1 down vote accepted

The key word here is probably "expect". With a large number of trials, you would win on average $\langle X \rangle$ dollars per trial and you have 1000 trials, so you should expect to win about 1000$\langle X \rangle$ dollars overall. Note that angle brackets denote expectation of the random variable.

Var$(S)$, on the other hand, would give you a measure of how close to this mean you might actually get.

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If I understand correctly, you are interested in the expected value of the magnitude of the win (or) loss (and not the magnitude of the expected value of the win (or) loss). Hence, you are interested in computing $\mathbb{E}(S_n)$ of the underlying random variable where $$S_n = \left| X_1 + X_2 + \cdots + X_n \right|.$$ Note that $S_n$ can take values in the set $\{0,5,10,\ldots,5n\}$.

Now let us find out the probability that $S_n = 5m$ for some $m \in \{0,1,2,\ldots,n\}$.

This means we are interested in the event $X_1 + X_2 + \cdots + X_n = \pm 5m$.

Let us first evaluate the probability of the event $X_1 + X_2 + \cdots + X_n = 5m$.

For this event to occur, if you lose $k$ times, you need to win $m+k$ times and get back your stake the remaining $n-m-2k$ times where $k \in \left \{0,1,2,\ldots, \left \lfloor \frac{n-m}{2} \right \rfloor \right\}$.

Hence, the desired probability of the event $X_1 + X_2 + \cdots + X_n = 5m$ is given by $$\sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac1{3^n}$$

Hence, the desired probability of the event $S_n = 5m$ is given by $$P_n(m) = \sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac2{3^n}$$

Hence, the expected value of $S_n$ is given by $$\sum_{m=0}^{n} 5m P_n(m) = \sum_{m=0}^{n} \sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac{10m}{3^n}$$

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$S_n$ does not converge to zero almost surely. –  mjqxxxx Dec 10 '11 at 0:04
    
I do not understand the formula for $E(|S_n|)$ (did you spot that $X_k$ takes the values $-5$, $+5$, AND $0$?). The assertion that $E(|S_n|)\to0$ is incorrect (to begin with, $n\mapsto E(|S_n|)$ is increasing by strict convexity). –  Did Dec 10 '11 at 1:00
    
@DidierPiau: I think I have now corrected it. Thanks! I am not sure how to go about the asymptotics (Stirling??), but does $\lim_{n \rightarrow \infty} \mathbb{E}(S_n)$ exists? –  user17762 Dec 10 '11 at 7:01
    
@DidierPiau: My $S_n = |X_1 + X_2 + \cdots +X_n|$ –  user17762 Dec 10 '11 at 22:22
    
The answer is that $E(|X_1+X_2+\cdots+X_n|)/\sqrt{n}$ converges to an explicit finite nonzero limit. –  Did Dec 10 '11 at 22:24
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