Almost sure convergence via pairwise independence

If {A_1,A_2,...} are pairwise independent and {\sum_{n=1}^{\infty}P(A_n)=\infty} then as {n \rightarrow \infty}

\displaystyle \boxed{ \frac{\sum_{m=1}^{n}\mathbb{I}_{A_m}}{\sum_{m=1}^{n}P(A_m)} \xrightarrow{a.s.} 1 }

Proof:

Let {X_m = \mathbb{I}_{A_m}} and {S_n = X_1+...+Xn}. Since {A_m} are pairwise independent, the {X_m} are uncorrelated and thus

\displaystyle var(S_n) = var(X_1) + ... + var(X_n)

Since {X_m \in \{0,1 \}}

\displaystyle var(X_m) \leq \mathbb{E}[X_m^2] = \mathbb{E}[X_m] \Rightarrow var(S_n) \leq \mathbb{E} [S_n]

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