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Suppose that $\{X_{n}, n\ge 1\}$ are arbitrary random variables such that $\sum_{n}\pm X_{n}$ converges almost surely for any choice of signs $\pm 1$. Show that $\sum X_{n}^2<\infty$ almost surely. [Hint: Consider $\sum_{n} B_{n}(t)X_{n}(\omega)$ where the random variables $\{B_{n},n\ge 1\}$ are Bernoulli. Apply Fubini on the space of $(t,\omega)$.]

Remarks

  1. This is problem 7.7.4(b) of Resnick's A Probability Path.
  2. We know that if $\sum \pm a_{i}$ converges almost surely, then $\sum a_{i}^2$ converges if the $a_{i}$ are fixed (non-random). This is discussed here, for example. It's also Lemma 7.6.1 of Resnick.
  3. The point of the problem seems to be to pass from knowledge of the result for fixed values of the $X_{n}$ to random values; and presumably that's the point of the hint about Fubini. But I don't understand exactly what's needed to draw this conclusion.
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    Note $\sum a_i^2$ converges almost surely if and only if $\sum a_i^2$ converges, because it isn't a random variable. Maybe you realized this, but the phrase "$\sum a_i^2$ converges almost surely" is kind of weird when you realize you are talking about a fixed quantity. – 2017-01-11
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    You're right, I will edit the question. – 2017-01-11

1 Answers 1

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Let $A = \{(t,w):\sum B_n(t) X_n(w)\}$ converges. For each $w$, let $A_w = \{t : (t,w) \in A\}$, and for each $t$, let $A^t = \{w:(t,w) \in A\}$.

By hypothesis, for almost every $t$, the set $A^t$ has measure 1. By Fubini, $$ \int \text{measure}(A_w) \, dw = \int \text{measure}(A^t) \, dt ,$$ since both sides equal $\int \!\!\int \chi_A(t,w) \, dt \, dw$. Hence for almost every $w$, the set $A_w$ has measure 1. But by your point 2, for any $w$, the set $A_w$ has measure 1 only if $\sum |X_n(w)|^2 < \infty$.

Thus for almost every $w$, we have $\sum |X_n(w)|^2 < \infty$.

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    Here I believe $A$ is a subset of the measure space $B\times X$ where $B$ is the countable product of copies of $\mathbf{R}$ with the discrete measure with mass $1/2$ at $-1$ and $1$ and $X$ is the countable product of $\mathbf{R}$ with Lebesgue measure; these countable products have the product measure. Does Fubini hold in that generality? – 2017-01-11
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    Perhaps the answer to my previous question is clear because we are dealing with probability measures, so the spaces are $\sigma$-finite (actually finite) and therefore Fubini holds. – 2017-01-11
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    Also the space $\{\pm 1\}^{\mathbb N}$ is measure equivalent to $[0,1]$. (Think about writing the numbers in base 2, or google "Radermacher functions.") – 2017-01-11