These paragraphs are from p. 82 of Introduction to Probability, 2nd Edition by Dimitri Bertsekas and John Tsitsiklis. I don't quite understand why the expectation in the last example is undefined; why is it not equal to zero?
Random variables with undefined expectation
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probability
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0The authors say that expectation well defined means that the series $\sum_x xp_X(x)$ converges absolutely (their definition). The series $\sum_k 2^k p_X(2^k)=\sum_k 2^k\cdot 2^{-k}=\sum_k 1$ diverges; thus, the series $\sum_x |x|p_X(x)$ in their second example, which equals twice $\sum_k 2^k p_X(2^k)$, also diverges. – 2017-02-01
2 Answers
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$\sum_x\lvert x\rvert p_X(x)=\sum_{k\geq2}\lvert2^k\rvert2^{-k}+\sum_{k\geq2}\lvert-2^k\rvert2^{-k}=\infty.$
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Well, in the last example the expectation is $$ \sum_{k\geq2}P(X=-2^k)\times (-2^k)+ \sum_{k\geq2}P(X=2^k)\times 2^k = \sum_{k\geq2}\frac1{2^k}\,(-2^k)+ \sum_{k\geq2}\frac1{2^k}\, 2^k = -\sum_{k\geq2}1+ \sum_{k\geq2}1 $$ which is ill-defined as a difference of two infinite sums. Of course you can regroup terms and obtain an infinite sum of zeros, but infinite sums are tricky (in a similar fashion you could show that the difference of infinite sums is $1$, $-1$ of any integer value).