Does the following fact have a name?

Theorem 1 Suppose {X_1,\ldots,X_n,\ldots} are a countable collection of 0/1 random variables over a probability space {(\Omega, {\cal B}, \mathop{\mathbb P})} such that for every integer {n} and bits {b_1,\ldots,b_n} the event

\displaystyle  X_1 = b_1 \wedge \cdots \wedge X_n = b_n

is measurable.

Suppose that

\displaystyle  \sum_{i=1}^\infty \mathop{\mathbb E} X_i \ \ \ {\rm converges}

Then

\displaystyle  \mathop{\mathbb P} \left[ \sum_i X_i = \infty \right] = 0

Proof: Intuitively, we want to say that by linearity of expectation we have {\mathop{\mathbb E} [ \sum_{i=1}^\infty X_i ] = O(1)} and so by Markov’s inequality

\displaystyle  \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i = \infty\right] \leq \frac{O(1)}{\infty} = 0

Just to make sure that this can be made rigorous, let’s belabor the proof step by step, doing everything completely from first principles.

Let {c:= \sum_{i=1}^\infty \mathop{\mathbb E} X_i}.

First of all, the event

\displaystyle  \sum_{i=1}^\infty X = \infty

is measurable, because it is the countable intersection over all {t} of the events {\sum_{i=1}^\infty X \geq t}, which are measurable because each of them is the countable union over all {n} of the events {\sum_{i=1}^n X_i \geq t}. So suppose towards a contradiction that its probability is not zero, then there is {\epsilon >0} such that

\displaystyle  \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i = \infty\right] = \epsilon

In particular, for every {t},

\displaystyle  \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] \geq \epsilon

But, by linearity of expectation, Markov’s inequality, and our assumption, we have that for every {n} and every {t}

\displaystyle  \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] \leq \frac {\mathop{\mathbb E} \sum_{i=1}^n X_i}{t} = \frac{\sum_{i=1}^n \mathop{\mathbb E} X_i}{t} \leq \frac ct

Now,

\displaystyle  \epsilon \leq \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] \leq \frac ct

which is a contradiction if we choose {t > c/\epsilon}.

Maybe we should also justify {\mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right]}. Define the disjoint events {E_1,\ldots,E_n,\ldots} as

\displaystyle  E_n := \left( \sum_{i=1}^n X_i \geq t \right) \wedge \left( \sum_{i=1}^{n-1} X_i < t \right)

Then

\displaystyle  \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \mathop{\mathbb P} \left [ \bigcup_{i=1}^\infty E_i \right] = \sum_{i=1}^\infty \mathop{\mathbb P} [ E_i ]

and

\displaystyle  \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \bigcup_{i=1}^n E_i \right]= \lim_{n\rightarrow \infty} \sum_{i=1}^n \mathop{\mathbb P}[ E_i] = \sum_{i=1}^\infty \mathop{\mathbb P}[E_i]

\Box

About these ads