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Imagine we have a seqence of (positive) absolutely continuous random variables $X_n$ and we know $\sup_n X_n$ is a measurable random variable and $\sup_n X_n < \infty$, i.e. there is a maximum one. Is the following true?

$$P\left( \sup_{n} X_n \geq a \right) = P\left( \bigcup_n \left\{ X_n \geq a\right\} \right).$$

If not, is then the following true? $$P\left( \sup_{n} X_n \geq a \right) \leq P\left( \bigcup_n \left\{ X_n \geq a\right\} \right).$$

Thanks for the help! :)

1 Answers 1

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\begin{align*} P\biggl(\sup X_n \geq a\biggr) &= 1 - P\biggl(\sup X_n < a \biggr) \\ &= 1 - P\biggl(\bigcap_n\{X_n < a\} \biggr) \\ &= 1 - \biggl[1 - P\biggl(\bigcup_n\{X_n \geq a\} \biggr)\biggr] \\ &= P\biggl(\bigcup_n\{X_n \geq a\} \biggr) \end{align*}

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    It's generally not true that $\sup X_n$X_n=a-1/n$. – 2017-02-08
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    In the equation we have $\sup_n X_n \geq a$, not $<$, in that case $\sup_n X_n < a$ is equivalent to $\bigcap_n \{X_n < a\}$. But you mean that the above argument is wrong then? I can't find the mistake. – 2017-02-08
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    That is true. But these random variables are *absolutely continuous* by assumption. That certainly excludes the example you have given above. – 2017-02-09