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I am having trouble understanding what Exercise 1.11 is asking.

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So, what I assumed from the previous exercise is when m = 2 and k = 1 that would mean that there were only two sets A and B and we're taking the cases of either A or B.

The symmetric difference is $A \triangle B = (A \backslash B) \cup (B \backslash A)$

So based on the previous exercise I got $A \triangle B$ to be in the event space which by definition is

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So $A \triangle B = (A \backslash B) \cup (B \backslash A)$ becomes $A \triangle B = (A \cap B^{c}) \cup (B \cap A^{c})$

Taking the complement

$A \triangle B = (A^{c} \cup B) \cap (B^{c} \cup A)$ and again becomes

$A \triangle B = (A^{c} \cup B)^{c} \cup (B^{c} \cup A)^{c} = (A \cap B^{c}) \cup (B \cap A^{c})$

Now 1.11 is asking that I need to show that the sets are also in an event space. At first, when I asked about this chat, it was suggested that if m = 3 and k =2 then I would have three sets A,B,C and that there are only two events because

$(A \cap B \cap C^{c}) \cup (A \cap B^{c} \cap C) \cup (A^{c} \cap B \cap C)$ holds because it's closed under complements, unions, and intersections. But still...I can't just write let m and k be specific numbers because this needs to hold true for every case.

The pattern also works for m = 4 and k =3, but that's like taking the intersections and unions of A, B, C, and D and either A, B , C , or D won't be counted as in the probability of one of those events doesn't occur.

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To begin, we'll use induction on $m$. Suppose this proposition is true for $k=1$ and $m=q$ where $q \geq 1$. Given sets $A_1, \ldots A_{q+1} \in \mathcal{F}$, the set of points $X$ that belong to exactly 1 of the $A_i$ ($1 \leq i \leq q)$ is also in $\mathcal{F}$ by the induction hypothesis. Notice that since $X, A_{q+1} \in \mathcal{F}$, the symmetric difference of $X$ and $A_{q+1}$ is in $\mathcal{F}$; this is exactly the set of points that are in exactly one of $A_1, \ldots , A_{q+1}$. Since we proved the base case $m=2$, we know this is true for any $m$.

We know $1 \leq k \leq m$, and the case $k=m$ is trivial. We can use a similar strategy by inducting on $k$. We can simply show that if $A_1, \ldots A_m \in \mathcal{F}$ and the set of points of $\Omega$ that only belong to $A_1, \ldots A_{\ell}$ ($1 \leq \ell < m$) is in $\mathcal{F}$, then the set of points that belong to $A_1, \ldots , A_{\ell + 1}$ also belongs to $\mathcal{F}$ by a similar (but slightly more complicated) argument as above. Can you finish from here?

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  • $\begingroup$ Ah so it's like induction where we consider the base case, k case, and then using our k case we can get into the k+1 case to prove that this is indeed in an event space. I have been thinking about this problem for a while...so I'm thinking that the end result we would have like (l+1)+1= l+2 ? $\endgroup$ – usukidoll Aug 28 '18 at 8:52

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