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Here's a question I got as a homework assignment:

Let $\{A_i\}_{i=1,\ldots,\infty}$ a sequence of events in the probability space $(\Omega,F,\mathbb{P})$. Show that if $\mathbb{P}(A_i)=1$ for all $i$ then $\mathbb{P}(\bigcap_{i-1}^{\infty}A_i)=1$

So, as the equation is very obvious, I don't know how to prove it.

Any suggestions?

Thanks!

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  • $\begingroup$ I switched the tag from Probability to Probability theory $\endgroup$ – Jean-Sébastien Nov 4 '12 at 16:24
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    $\begingroup$ It may seem obvious, but it really does depend on the fact that you’re considering only countably many events. If there were uncountably many events, the probability of their intersection could even be $0$. $\endgroup$ – Brian M. Scott Nov 4 '12 at 16:38
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Look at the sequence of sets $B_{k+1} = A_{k+1} \bigcap B_k$, where $B_1 = A_1$. Prove that $\mathbb{P}(B_k) = 1$ and note that $$B_1 \supseteq B_2 \supseteq B_3 \supseteq \cdots$$ and $$\bigcap_{k=1}^{\infty} A_k = \bigcap_{k=1}^{\infty} B_k$$

Recall the following result you should have proved for sequence of nested decreasing sets.

Continuity from above: If $C_k \downarrow C$, that is, $C_1 \supseteq C_2 \supseteq C_3 \cdots$ and $\displaystyle \bigcap_{k=1}^{\infty} C_k = C$, then $\mathbb{P}(C_k) \downarrow \mathbb{C}$.

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  • $\begingroup$ Can you please clarify the last line? I'm not familier with the arrow pointing down... $\endgroup$ – Yotam Nov 4 '12 at 16:59
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$$ \bigcap_{n=1}^{+\infty}A_n=\Omega\setminus B \quad B=\bigcup_{n=1}^{+\infty}(\Omega\setminus A_n) \quad \mathbb P(\Omega\setminus A_n)=0 \quad \mathbb P(B)\leqslant\sum_{n=1}^{+\infty}\mathbb P(\Omega\setminus A_n) $$

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That really depends on what you already know about probability. If you know that the countable union of events of probability zero has probability zero then this is merely a case of applying DeMorgan laws:

Let $B_i$ be the complement of $A_i$, then $\Bbb P(B_i)=0$, then $\Bbb P(\bigcup B_i)=0$ as well, and by DeMorgan laws this is the same as saying $\Bbb P(\bigcap A_i)=1$.

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