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Here's a question I'm sturglling with:

Show that for an increasing sequence of events $$A_1\subset A_2\subset A_3\subset ...$$ the next equation holds $$P\big(\bigcup_nA_n\big)=\lim_{n\rightarrow\infty}P(A_n)$$

Here's what I have so far: let $(B_n)_{n=1}^{\infty}$ be a sequence of events such that $$B_{1}=A_{1},B_{2}=A_{2}\backslash A_{1},B_{3}=A_{3}\backslash A_{2}\cap A_{1},...,B_{n}=A_{n}\backslash(\cap_{i=1}^{n-1}A_{i})$$ So obviously every two elements in $B_n$ are disjoint and $\bigcup B_n = \bigcup A_n$ so $$P\big(\bigcup A_n\big) = P\big(\bigcup B_n \big) = \sum_n P(B_n)$$

And I'm stuck here. Really I'm not sure what a limit of a sequence of events here so I was just following the little I do know. I know sequence limits, but what does it mean when the elements are sets? Does the sequence converges to a set? What is that set?

Would appreciate any motivation of the limit issue, and on solving the question.

Thanks!

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  • $\begingroup$ I think you must have a bounded event space , ie no events happen outside this event space. Otherwise it limit doesn't seem to make sense. But i am not sure. $\endgroup$
    – Theorem
    Nov 15, 2012 at 9:44
  • $\begingroup$ From the question it looks like we have countably many events, so I'm not sure if it is bounded $\endgroup$
    – Yotam
    Nov 15, 2012 at 9:55
  • $\begingroup$ Note that you're not considering a limit of a sequence of events here at all. You're considering the limit of an ordinary sequence of numbers - $P(A_n)$ is a real number for each $n$. $\endgroup$ Nov 15, 2012 at 9:59
  • $\begingroup$ Yes if it is countably many events then you can at least cover all the events by countably infinit disjoint sets . so that makes sense to me . $\endgroup$
    – Theorem
    Nov 15, 2012 at 10:01
  • $\begingroup$ Also, the sets $B_n$ you've described are not pairwise disjoint. You may want to define $B_3 \setminus A_1 \cup A_2 = B_3 \setminus A_2$ instead (note that $A_2 \cap A_1 = A_1$, and so on...) $\endgroup$ Nov 15, 2012 at 10:14

3 Answers 3

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If $$A_1\subset A_2\subset A_3\subset ...$$ then $$\bigcup_n A_n=\lim_{n\rightarrow \infty }A_n$$

Try with $$B_{1}=A_{1},B_{2}=A_{2}\backslash A_{1},B_{3}=A_{3}\backslash A_2,...,B_{n}=A_{n}\backslash(A_{n-1})$$

$$P\big(\bigcup_n A_n\big) = P\big(\bigcup_n B_n \big) = \sum_{n=1}^\infty P(B_n)=\lim_{n\rightarrow \infty} P(A_n)$$

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I answer the question in the title. I see that an answer to the question in the body arrived already, so I leave that aside.

The only sensible definition of “limit” for a sequence of sets is their union, if the sequence is increasing, and the intersection, if it is decreasing. If it is neither, then it is not so clear what you would mean by the limit. But a reasonable definition goes as follows: Define $$\begin{aligned} \limsup_{n\to\infty} A_n&=\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k,\\ \liminf_{n\to\infty} A_n&=\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k,\\ \end{aligned}$$ and let the limit be the common value of these two if they coincide. Otherwise, the limit does not exist.

In words: The limit exists if every item that belongs to infinitely many $A_k$ belongs to all but a finite number of them, and then the limit consists of all the items that do.

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$P(\cup_{i=1}^{\infty} A_i) = P(\lim_{n \to \infty} \cup_{i=1}^{n} A_i) = P(\lim_{n \to \infty} A_n) = \lim_{n \to \infty} P(A_n)$, where the last equality is by the monotone convergence theorem: https://en.wikipedia.org/wiki/Monotone_convergence_theorem#Lebesgue.27s_monotone_convergence_theorem

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  • $\begingroup$ We do not need the MCT to show that the probability $\mathrm{P}$ is continuous (in the sense $\mathrm{P}(\lim_n A_n) = \lim_n \mathrm{P}(A_n)$ for nondecreasing sequence of events). $\endgroup$ Aug 12, 2017 at 22:25

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