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I am currently self-studying with Casella & Berger and there is a problem that has me stumped. Please note I am NOT asking for a solution, I just need clarification on the terms used.

Here is the exercise in question:

C&B exercise

The way I understand the definition of the set Ek, it is not generally true that P(E1) = P(A1) + ... + P(An).

To illustrate, with A1 = {1,2} and A2 = {1,2,3}, E1 would denote elements contained in exactly one of A1 and A2, namely E1 = {3}. The statement in problem (b) would therefore be equivalent to P({3}) = P({1,2}) + P({1,2,3}).

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  • $\begingroup$ "Exactly" in (a) means what it says. Thus $E_1=\{3\}$ and $E_2=\{1,2\}$. Your example should read $$P\left(\{1,2\}\cup\{1,2,3\}\right)=P(\{3\})+P(\{1,2\}).$$ $\endgroup$
    – whuber ♦
    Commented Oct 18, 2016 at 21:19
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    $\begingroup$ One note that might be relevant to the question you attempt to solve (but not necessarily the question you pose here) is that there is an errata sheet (stat.ufl.edu/archived/casella/class/errata7.pdf) for Casella & Berger and some of the parts of question 1.42 have errors in the text. $\endgroup$
    – Matt Brems
    Commented Oct 18, 2016 at 21:24
  • $\begingroup$ Thank you both for your replies. I have posted an answer based on the information provided, do let me know if this goes against site etiquette. $\endgroup$
    – overdisperse
    Commented Oct 18, 2016 at 21:36
  • $\begingroup$ @whuber You might have missed that the question is about (b), not (a)! OP's example seems to indeed be a counterexample against the statement in the original (b) (when $P(\{1,2\})>0$), and in the errata (b) is replaced with something completely different. $\endgroup$
    – Juho Kokkala
    Commented Oct 19, 2016 at 6:35
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    $\begingroup$ @whuber Let us assume $n=2$, $A_1=\{1,2\},~A_2=\{1,2,3\}$ (is something already incorrect?). By substituting $k=1$ into "Let $E_k$ denote the set..." we get that $E_1$ is the set containing outcomes that are contained in exactly $1$ of $A_1,A_2$. $1$ and $2$ are contained in $A_1$ and $A_2$, so 2 sets. $3$ is only in $A_2$, possible other outcomes are in $0$ $A_i$s. Thus we conclude $E_1=\{3\}$. Then, the left hand side of the equation in (b) becomes $P(E_1)=P(\{3\})$. The rhs is $P(A_1)+P(A_2) = \sum_{i=1}^2 P(A_i) = P(A_1)+P(A_2) = P(\{1,2\})+P(\{1,2,3\})$ $\endgroup$
    – Juho Kokkala
    Commented Oct 19, 2016 at 18:24

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Piecing together the replies by whuber and Matt Brems, it seems I did understand question 1.42(b) correctly, it's just that there's a mistake in the book.

(I'm still quite unsure of what proper etiquette mandates on this site but my observations lead me to conclude self-replying with an answer is appropriate)

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  • $\begingroup$ Your understanding of the site is correct--but this does not appear to answer your question! Indeed, upon re-reading your original post I am left wondering what it is you were trying to ask. $\endgroup$
    – whuber ♦
    Commented Oct 18, 2016 at 22:02

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