Breaking The linearity Of Expectation

In the previous problems, if we let X X denote the value of Scrooge’s winnings, and let Xn X_n denote the value of Scrooge’s winnings during each round, we have

X=X1+X2+X3+X4+ X = X_1 + X_2 + X_3 + X_4 + \ldots

Even though there are infinitely many terms (countable), only finitely many of them are non-zero. Hence, this sum makes sense under every scenario. In each game, E[Xi]=0 E[X_i] = 0 , since it is a fair value.

The linearity of expectation tells us that if X=X1+X2 X = X_1 + X_2 , then we have E[X]=E[X1]+E[X2] E[X] = E[X_1] + E[X_2] . As such, we would be very tempted to claim that

E[X]=E[X1]+E[X2]+E[X3]+=0 E[X] = E[X_1] + E[X_2] + E[X_3] + \ldots = 0

However, as we have seen, that is not the case. Only in the second problem, did we have E[X]=0 E[X] = 0 .

What’s the reason for this?

Note by Calvin Lin
5 years, 5 months ago

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1 vote

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What kind of a loser gambles when it is not possible to leave the casino with positive winnings!

Siam Habib - 5 years, 5 months ago

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Of course, I also simplified the scenarios so as to make the calculations easier. You could modify them by using the condition that "no one can make more than $100", and then analyze what happens.

When it comes to gambling, you will find that there are numerous motivations behind why people engage in it. Some play just for the fun / thrill of it, and are not looking at it as a way of making money.

Calvin Lin Staff - 5 years, 5 months ago

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I'm not sure but I think that the infinite series-es that we find in those questions are conditionally convergent and therefore they have different sums for different arrangements. It is true that the expected value of winnings of any particular game is always zero but when we add all of them and put them in the correct order they might provide us with an answer that is not zero.

Siam Habib - 5 years, 5 months ago

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Yes, we cannot simply interchange the order of summation. That is the basic error made here.

The detailed explanation is related to Fubini's Theorem, which asks when can we interchange the order of integration. For the combinatorial / probabilistic version, it is known as Wald's Lemma.

Calvin Lin Staff - 5 years, 5 months ago

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