My question is related to this post: Linearity of expectation for infinite sums?
What makes the switch between expectancy and sum fails? One is indeed giving 0, the other one infinite.
How shall we then calculate $E \Sigma_1^∞ X_n$?
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There are two general theorems that allow exchange of integrals: Tonelli for positive measurable functions and Fubini for absolutely integrable functions. Tonelli's theorem simply states that the order between two integrations (summation is a form of integral) can be exchanged if the random variables are positive; Fubini's theorem assert the same when the random variables are no longer positive but if we add the hypothesis that their absolute values are summable (i.e. either $E \sum |X_i| < \infty$ or $\sum E|X_i| < \infty$).... – William M. Mar 11 '22 at 16:42
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...When you have i.i.d. variables, you will always have $\sum E|X_i| = \infty$ and therefore, Fubini's theorem will never apply or if the random variables are positive, then $E \sum X_i = \infty.$ – William M. Mar 11 '22 at 16:42
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Thanks for the reply. Therefore, you are saying that above expectancy for centered normal distributions will be ∞, which is counter intuitive and does not seem to apply through simulation. Can you please develop further this aspect please? – yeahman269 Mar 16 '22 at 13:59
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No, I am saying that Fubini's theorem will never apply. In fact, I believe $\sum\limits_{i = 0}^\infty X_i$ diverges almost surely for independent and identically distributed random variables. For if it were to converge, then we would have that $X_i \to 0$ almost surely, and we can show that $P(X_i > 1) = c$ and then use Borel-Cantelli lemma to show that $X_i > c$ for infinitely many indices, almost surely, therefore, the sum $\sum\limits_{i = 0}^\infty X_i$ diverges almost surely. In this way, your question is actually nonesense. – William M. Mar 16 '22 at 14:42