limsup and liminf of sequence of random variables and Borel Cantelli

Question

For a sequence of independent random variables $X_n$, if we have $P( \limsup [X_n > a] ) = 1$, then $P( [ \limsup X_n ] > a ) = 1$. Does this automatically mean $P( \liminf [X_n > a] ) = 1$ as well?

Since $P(([\limsup X_n] > a)^c)=P( \liminf(X_n > a)^c)=P([\liminf X_n]\le a) = 0$, we have $P( \liminf [X_n > a] ) = 1$.

Please let me know if I have made stupid mistakes. Thank you in advance.

Answer 1

Important inequalities

Williams - Probability with Martingales

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Deduced similarly:

(iii) If $\liminf x_n > z$, then

$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n < z $, then

$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)

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$$P( \limsup [X_n > a] ) = 1 \to P( [ \limsup X_n ] > a) = 1$$

No. Consider $X_n = a + \frac{1}{n}$.

The converse, which is,

$$P( \limsup [X_n > a] ) = 1 \leftarrow P( [ \limsup X_n ] > a) = 1$$

is true (see above).

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$$P( [ \limsup X_n ] > a) = 1 \to P( \liminf [X_n > a] ) = 1$$

No.

$$P( [ \limsup X_n ] > a) = 1 \to P( \limsup (X_n > a)) = 1$$

$$P( \limsup (X_n > a)) = 1 \nrightarrow P( \liminf [X_n > a] ) = 1$$

The converse of the latter, which is,

$$P( \liminf [X_n > a] ) = 1 \to P( \limsup (X_n > a)) = 1$$

is true (obviously?).

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$$P(([\limsup X_n] > a)^c)=P( \liminf(X_n > a)^c)$$

No.

$$P(([\limsup X_n] > a)^c) = P([\limsup X_n] \le a)$$

$$P( \liminf[(X_n > a)^c]) = P( \liminf(X_n \le a))$$

$$P( [\liminf(X_n > a)]^c) = P( \limsup(X_n \le a))$$

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$$P( \liminf(X_n > a)^c)=P([\liminf X_n]\le a)$$

No.

$$P( \liminf[(X_n > a)^c]) = P( \liminf(X_n \le a))$$

$$P( [\liminf(X_n > a)]^c) = P( \limsup(X_n \le a))$$

$$P([\liminf X_n]\le a) = P(([\liminf X_n] > a)^C)$$

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$$P([\liminf X_n]\le a) = 0 \to P( \liminf [X_n > a] ) = 1$$

Yes. Contrapositive of 3