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Suppose $P([\frac1n, \frac12]) \leq \frac13$ for all $n= 1,2,3,...$

Must we have $P((0, \frac12)) \leq \frac13$?

This problem is from a Continuity (Boole's, Bonferroni's Theorems) section in a Statistics textbook. How do I prove this with a method that is relevant to the content taught in that section?

Thanks in advance!

Plus Twenty
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Shukie
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1 Answers1

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$(0,\frac 1 2) \subset \cup_n [\frac 1 n, \frac 1 2]$ and the events $[\frac 1 n, \frac 1 2]$ increase as $n$ increases. Hence $P(0,\frac 1 2) \leq (\lim P([\frac 1 n, \frac 1 2])\leq \frac 1 3$.

Kavi Rama Murthy
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  • Thank you very much! That was kind of my solution too, but mine doesn't really reference continuity. I just put: (0,1/2) is a subset of [1/n,1/2], so P(0,1/2) < P[1/n,1/2] <= 1/3. – Shukie Jun 11 '19 at 09:12
  • @Shukie for no particular $n$ do we have $(0,1/2)$ contained in $[1/n,1/2]$ so your argument is not correct. – Kavi Rama Murthy Jun 11 '19 at 09:14