# 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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# An inequality of the mean involving truncation

Let ${X_1,X_2,...}$ be i.i.d. r.vs with ${\mathbb{E}[|X_i|] < \infty}$ and ${Y_k = X_k \mathbb{I}_{(|X_k| \leq k)}}$. Then

$\displaystyle \boxed{ \mathbb{E}[X_1] \geq \sum_{k=1}^{\infty} \frac{var(Y_k)}{4 k^2 } }$

Proof:

First we proove the following useful result

If ${X \geq 0}$ and ${ a > 0}$ then

$\displaystyle \boxed{ \mathbb{E}[X^a] = \int_{0}^{\infty} a x^{a-1} P(X >x) dx }$

$\displaystyle \begin{array}{rcl} \int_{0}^{\infty} a x^{a-1} P(X >x) dx &=& \int_{0}^{\infty} \int_{\Omega}a x^{a-1} \mathbb{I}_{(X>x)} dP dx \\ &=& \int_{\Omega} \int_{0}^{\infty} a x^{a-1} \mathbb{I}_{(X>x)} dP dx \\ &=& \int_{\Omega} \int_{0}^{X} a x^{a-1} dP dx = \mathbb{E}[X^a] \end{array}$

Note you can find the same lemma on Feller Vol.2 (p. 150) as

$\displaystyle \mathbb{E}[X^a]= \int_{0}^{\infty} x^a F \{ dx \} = a \int_{0}^{\infty} x^{a-1} [ 1- F(x)] dx$