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I am currently reading a probability book (Marek Fisz Probability theory and Mathematical statistics), in the problems section I have encountered the following exercise:

$(a)\space \space prove: A_1 + A_2 + \dots + A_n = A_1 + (A_2 - A_1A_2) + (A_3 - A_1A_3 - A_2A_3) + \dots + (A_n - A_1A_n - A_2A_n - \dots - A_{n-1}A_n)$

$(b)\space \space prove: \bigcup_{i=1}^{\infty} A_i= A_1 + (A_2 - A_1A_2) + (A_3 - A_1A_3 - A_2A_3) + \dots $

Where $A_i$ are random events.

TLDR: why the author treats these to cases as different ?

MY REASONING:

I showed the first using induction, which is simple once you notice (and prove) that $(A_k - A_1A_k - A2A_k - \dots - A_{k-1}A_k) = (A_k - A_k(A_1 + \dots + A_{k-1}))$.

I suspect that induction is not applicable to the second so I used the following technique - I supposed that $x \in \bigcup_{i=1}^{\infty} A_i $ then there exists a set on natural numbers $I$ such that $i\in I \rightarrow x \in A_i$. Let $ i = \inf I$ then $j<i \rightarrow x \notin A_j$ so $x \in (A_i - A_1A_i - A_2A_i - \dots - A_{i-1}A_i)$, so $\bigcup_{i=1}^{\infty} A_i \subset A_1 + (A_2 - A_1A_2) + (A_3 - A_1A_3 - A_2A_3) + \dots $

Showing that $ A_1 + (A_2 - A_1A_2) + (A_3 - A_1A_3 - A_2A_3) + \dots \subset \bigcup_{i=1}^{\infty} A_i$ might be done applying the same principle as before.

Again, quite straightforward.

I noticed the prove for $(b)$ is valid for $(a)$.The question is: why the author divided the two cases ? Or I made some mistakes while proving and everything is not that straightforward ?

Actually, I was not very suspicious until I encountered the following exercise (in thesame chapter)

$show: P( \bigcup_{i=1}^{n} A_i ) \le \sum_{i=1}^{n} P( A_i )$

$show: P( \bigcup_{i=1}^{\infty} A_i ) \le \sum_{i=1}^{\infty} P( A_i )$

Which can be easily shown using the aforementioned exercise (as hinted by the author, since the right side of the equation is made of independent events), the probability axioms and the fact $P(A) \le P(B) $ if $A \subset B$.

Again the question is the same: why the author treats these to cases as different ?

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Of course any proof of (b) is also going to be a valid proof of (a) since you can just reduce (a) to (b) by taking $A_i = \emptyset$ for $i \geq n$.

But, yes, your proof is fine. The reason they're separated into two cases is presumably that the first can be handled with induction alone, whereas the second requires a bit more mathematical maturity in dealing with infinite sets.

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