Exercise 1.2: This exercise derives the probability of an arbitrary (non-disjoint) union of events, derives the union bound, and derives some useful limit expressions.
a) For 2 arbitrary events $A_1$ and $A_2$, show that
$$A_1 \bigcup A_2=A_1 \bigcup\left(A_2-A_1\right)$$
where $A_2-A_1=A_2 A_1^c$. Show that $A_1$ and $A_2-A_1$ are disjoint. Hint: This is what Venn diagrams were invented for.

Solution: Let $B_1=A_1$. From (a) $B_1$ and $B_2$ are disjoint and (from (A.1)), $A_1 \cup A_2=$ $B_1 \cup B_2$. Let $C_n=\bigcup_{i=1}^n A_i$. We use induction to prove that $C_n=\bigcup_{i=1}^n B_i$ and that the $B_n$ are disjoint. We have seen that $C_2=B_1 \cup B_2$, which forms the basis for the induction. We assume that $C_{n-1}=\bigcup_{i=1}^{n-1} B_i$ and prove that $C_n=\bigcup_{i=1}^n B_i$.
\begin{aligned} C_n & =C_{n-1} \bigcup A_n=C_{n-1} \bigcup A_n C_{n-1}^c \\ & =C_{n-1} \bigcup B_n=\bigcup_{i-1}^n B_i . \end{aligned}
In the second equality, we used (A.1), letting $C_{n-1}$ play the role of $A_1$ and $A_n$ play the role of $A_2$. From this same application of (A.1), we also see that $C_{n-1}$ and $B_n=A_n-C_{n-1}$ are disjoint. Since $C_{n-1}=\bigcup_{i=1}^{n-1} B_i$, this also shows that $B_n$ is disjoint from $B_1, \ldots, B_{n-1}$.
c) Show that
$$\operatorname{Pr}\left\{\bigcup_{n=1}^{\infty} A_n\right\}=\operatorname{Pr}\left\{\bigcup_{n=1}^{\infty} B_n\right\}=\sum_{n=1}^{\infty} \operatorname{Pr}\left\{B_n\right\}$$

Exercise 1.4: Consider a sample space of 8 equiprobable sample points and let $A_1, A_2, A_3$ be three events each of probability $1 / 2$ such that $\operatorname{Pr}\left\{A_1 A_2 A_3\right\}=\operatorname{Pr}\left\{A_1\right\} \operatorname{Pr}\left\{A_2\right\} \operatorname{Pr}\left\{A_3\right\}$.
a) Create an example where $\operatorname{Pr}\left\{A_1 A_2\right\}=\operatorname{Pr}\left\{A_1 A_3\right\}=\frac{1}{4}$ but $\operatorname{Pr}\left\{A_2 A_3\right\}=\frac{1}{8}$. Hint: Make a table with a row for each sample point and a column for each of the above 3 events and try different ways of assigning sample points to events (the answer is not unique).

Solution: Note that $\operatorname{Pr}\left\{A_2 A_3\right\}=1 / 8 \neq \operatorname{Pr}\left\{A_2\right\} \operatorname{Pr}\left\{A_3\right\}$, so $A_2$ and $A_3$ are dependent. We also note that $\operatorname{Pr}\left\{A_1^c A_2^c A_3^c\right\}=0 \neq \operatorname{Pr}\left\{A_1^c\right\} \operatorname{Pr}\left\{A_2^c\right\} \operatorname{Pr}\left\{A_3^c\right\}$, further reinforcing the conclusion that $A_1, A_2, A_3$ are not statistically independent. Although the definition in (1.14) of statistical independence of more than 2 events looks strange, it is clear from this example that (1.15) is insufficient in the sense that it only specifies part of the above table.

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