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I am trying to solve the following question from Sheldon Ross, Introduction to Probability Models.

chapter 1 question 47 sheldon ross

I am struggling to prove the first of the 3 conditions,

the 3 conditions for a probability

I thought of using the inclusion exclusion theorem, but I can't think of how to prove these bounds for $ P(A_i \cup B)$ . I was thinking that showing $P(\phi | B) = 0$ and $P(\Omega |B) = 1 $ should be enough since $\phi \subseteq A_i \subseteq \Omega $. But I want rigor in my proof.

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Now I think my proof for condition 2 is correct and straightforward. For condition 3, I used $n=2$ as the base case and proved it via induction. But I am not sure it is a valid proof since $\infty$ is not a natural number and stackexchange seems to agree that it is not valid. I was wondering if induction is enough to show that countable additivity holds for any countably infinite sequence of events ${A_1, A_2, \cdots }$ which are mutually exclusive. When can I use induction to prove that it holds for an infinite of cases? What other tools are there?

My attempts -

page 2 of my work page 3 of my work

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    $\begingroup$ Maybe I have misunderstood your question, but it seems much simpler than you are making it. As I understand it, you are trying to prove that $0 \le P(A \mid B) \le 1$ for all events $A,B$. Well, just write the definition: $P(A \mid B) = P(A \cap B) / P(B)$. This is a ratio of two nonnegative numbers, so it is nonnegative. And $A \cap B \subset B$, so the numerator is less than or equal to the denominator. Done. $\endgroup$ – Nate Eldredge Aug 12 '19 at 14:34
  • $\begingroup$ @NateEldredge is there any proof for the fact that if A is a subset of B then P(A) must be less than P(B)? It seems obvious, but I thought the proof must be only depending on axioms and logic $\endgroup$ – Aditya P Aug 12 '19 at 14:43
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    $\begingroup$ Sure. Write $B = A \cup (B \setminus A)$. This is a disjoint union so by axiom (iii) we have $P(B) = P(A) + P(B \setminus A)$. But by axiom (i) we have $P(B \setminus A) \ge 0$. This implies $P(B) \ge P(A)$. $\endgroup$ – Nate Eldredge Aug 12 '19 at 15:19
  • $\begingroup$ 😂😂 thanks I can't believe I overcomplicated that. What about condition 3? Do you think the induction proof is good enough? $\endgroup$ – Aditya P Aug 12 '19 at 15:21
  • $\begingroup$ No, induction can prove your claim for finite unions, but not for infinite unions. $\endgroup$ – Nate Eldredge Aug 12 '19 at 16:15
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Since $0\leq P(A\cap B)\leq P(B)$, it follows that

$$0\leq P(A\mid B)= \frac{P(A\cap B)}{P(B)}\leq 1$$

Here is a proof for (3):

$$P(\bigcup_n A_n\mid B) =\frac{P(\bigcup_n A_n\cap B)}{P(B)}$$ $$= \sum_n \frac{P(A_n\cap B)}{P(B)} = \sum_n P(A_n\mid B)$$

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  • $\begingroup$ Any idea about if induction can be used to prove propery 3? $\endgroup$ – Aditya P Aug 12 '19 at 15:02
  • $\begingroup$ Induction on what? There is nothing to induct on. This is rather a straightforward calculation, which I added to my answer. $\endgroup$ – ε-δ Aug 12 '19 at 15:53
  • $\begingroup$ I mean that proof only holds for finite unions right? But the actual condition needs it to be proven for infinite number of unions. $\endgroup$ – Aditya P Aug 12 '19 at 16:17
  • $\begingroup$ The statement $B\cap \bigcup_{i\in I} A_i=\bigcup_{i\in I}(B\cap A_i)$ holds for every index set $I$, so not only finite ones. This is an easy set theory exercise. $\endgroup$ – ε-δ Aug 12 '19 at 16:19

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