# Partial Converse to Borel-Cantelli Lemma

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```					                        Partial Converse to Borel-Cantelli Lemma
Matt Rosenzweig

Lemma 1. (Paley-Zygmund Inequality) If X is a nonnegative random variable not almost every zero with
E[X 2 ] < ∞, then

E[X]2
P {X > 0} ≥
E[X 2 ]

Proof. By Cauchy-Schwarz,
1
2                             1
E[X] = E[X1{X>0} ] ≤ E[X 2 ]E[12
{X>0} ]                       = E[X 2 ]P {X > 0}       2

Squaring both sides and dividing by E[X 2 ] completes the proof.
Proposition 2. Suppose (En )∞ is a collection of events with
n=1

∞                                        k        k
m=1      n=1   P(En ∩ Em )
P(En ) = ∞,          lim inf               k
<∞
n=1
k→∞              (   m=1   P(En ))2

Then P(lim supn→∞ En ) > 0.

Proof. For each j, let 1Ej denote the indicator function of the event Ej . For each n ∈ N, deﬁne a random
n                                              2
variable Xn by Xn := j=1 1Ej . Clearly, Xn is nonnegative and E[Xn ] ≤ n2 < ∞. Also, it is evident
2                                         ∞
that E[Xn ] > 0 for all n suﬃciently large since n=1 P(En ) = ∞ ⇒ P(En ) > 0 for some n. Applying the
Paley-Zygmund inequality, we obtain
2                                       2
n                                      n
Xn              E[Xn ]2                 j=1 E[1Ej ]                            j=1 P(Ej )
P            >0      ≥     2
=           n     n                      =    n         n
E[Xn ]           E[Xn ]              j=1   i=1 E[1Ei 1Ej ]             j=1       k=1 P(Ej    ∩ Ek )

Note that
2
n       n                                                      n
j=1     k=1   P(Ej ∩ Ek )                                      j=1 P(Ej )
lim inf                          2      < ∞ ⇒ lim sup              n      n                    >0
n→∞               n                                    n→∞         j=1    k=1 P(Ej    ∩ Ek )
j=1   P(Ej )

By Fatou’s lemma,
2
n
Xn                                Xn                                        j=1 P(Ej )
P lim sup          >0      ≥ lim sup P               >0           ≥ lim sup        n      n                    >0
n→∞      E[Xn ]               n→∞          E[Xn ]                 n→∞           j=1    k=1 P(Ej    ∩ Ek )
∞
Since lim supn→∞ E[Xn ] = ∞ by hypothesis that k=1 P(Ek ) = ∞, we obtain that, with positive probability,
lim supn→∞ Xn = ∞, which is equivalent to P {lim supn→∞ 1En = 1} = P(lim supn→∞ En ) > 0.

1

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