Solutions

1.11

a.

$\mathcal{B} = \{\emptyset, S\}$ is a $\sigma$ -algebra:

(i)

$\emptyset \in \mathcal{B}$ .

(ii)

$\emptyset^{c}=S \in \mathcal{B}$ and $S^{c} = \emptyset \in \mathcal{B}$ .

(iii)

Suppose $A_{1}, A_{2}, \ldots \in \mathcal{B}$ . If $A_{i}=\emptyset$ for all $i$ , then $\bigcup A_{i} = \emptyset \in \mathcal{B}$ . If any $A_{i} \neq \emptyset$ then it is $S$ , and $\bigcup A_{i} = S \in \mathcal{B}$ .

b.

$\mathcal{B} =$   all subsets of $S$$= 2^{S}$ is a $\sigma$ -algebra:

(i)

$\emptyset$ is a subset of $S$ , so $\emptyset \in \mathcal{B}$ .

(ii)

If $A \subset S$ then $A^{c} \subset S$ , so $A^{c} \in \mathcal{B}$ .

(iii)

If $A_{1}, A_{2}, \ldots$ are subsets of $S$ , then so is $\bigcup A_{i}$ .

c.

Suppose $\mathcal{B}_{\alpha}$ is a $\sigma$ -algebra on $S$ for every $\alpha$ . Then $\mathcal{C} = \bigcap \mathcal{B}_{\alpha}$ is a $\sigma$ -algebra on $S$ .

(i)

Since each $B_{\alpha}$ is a $\sigma$ -algebra, $\emptyset \in \mathcal{B}_{\alpha}$ for all $\alpha$ , so $\emptyset \in \mathcal{C}$ .

(ii)

If $A \in \mathcal{C}$ , then $A \in \mathcal{B}$ for all $\alpha$ . So $A^{c} \in \mathcal{B}$ for all $\alpha$ , i.e. $A \in \mathcal{C}$ .

(iii)

If $A_{1}, A_{2}, \ldots \in \mathcal{B}_{\alpha}$ for all $\alpha$ . then $\bigcup A_{i} \in \mathcal{B}_{\alpha}$ for all $\alpha$ . So $\bigcup A_{i} \in \mathcal{C}$ .

So $\mathcal{C}$ is a $\sigma$ -algebra. This means ``smallest $\sigma$ -algebra satisfying ...'' is well defined.

1.12

a.

Suppose $A$ and $B$ are disjoint. Let $C_1=A, C_2=B$ , and $C_i = \emptyset$ for $i > 2$ . Then, since $P(\emptyset) = 0$ and the $C_i$ are pairwise disjoint,

$$P(A \cup B) = P\left(\bigcup_{i=1}^\infty C_i\right)$$

$$= \sum_{i=1}^\infty P(C_i) = P(A) + P(B)$$

b.

Let $A_{1}, A_{2}, \ldots$ be pairwise disjoint. For each $i$ , let

$$B_{i} = \bigcup_{j=i}^{\infty} A_{j}$$

Then for each $n$ ,

$$\bigcup_{i=1}^{\infty} A_{i} = A_{1} \cup \cdots \cup A_{n} \cup B_{n+1}$$

Since $A_{1}, A_{2}, \ldots, A_{n}, B_{n+1}$ are pairwise disjoint,

$$P\left(\bigcup_{i=1}^{\infty} A_{i}\right) = \sum_{i=1}^{n} P(A_{i}) + P(B_{n+1})$$

for every $n$ by finite additivity. But

$$B_{1} \supset B_{2} \supset \cdots$$

and $\bigcap B_{i} = \emptyset$ . So by continuity, $P(B_{n+1}) \downarrow 0$ . So

$$\sum_{i=1}^{n} P(A_{i}) \rightarrow P\left(\bigcup_{i=1}^{\infty} A_{i}\right)$$

and thus

$$\sum_{i=1}^{\infty} P(A_{i}) = P\left(\bigcup_{i=1}^{\infty} A_{i}\right)$$

1.14

A set of $n$ elements has $2^{n}$ subsets:

$$x = \sum_{k=0}^{n} \binom{n}{k}$$

Can show $x = 2^{n}$ by induction. Can show it algebraically using

$$(a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{k} b^{n-k}$$

taking $a=b=1$ .

Can also use a counting argument: Each subset corresponds to an ordered list of $n$ 0's and 1's.

$$x_{i} = 1 i$$

$$x_{i} = 0 i$$

For example, if $S=\{s_{1},s_{2},s_{3},s_{4}\}$ , then

$$(1, 0, 0, 1) \leftrightarrow \{s_{1},s_{4}\}$$

How many such lists are there? $2^{n}$ .

1.16

Number of different sets of initials.

a.

Two given plus sur: $\sqcup\sqcup\sqcup$ . $26^{3}$ .

b.

One or two given plus sur: $\sqcup\sqcup$ or $\sqcup\sqcup\sqcup$ . $26^{2} + 26^{3}$ .

c.

One, two, or three given plus sur: $\sqcup\sqcup$ or $\sqcup\sqcup\sqcup$ or $\sqcup\sqcup\sqcup\sqcup$ . $26^{2} + 26^{3} + 26^{4}$ .

1.18

$n$ balls at random into $n$ cells.

$S$ has $n^n$ elements--repetition is allowed.

Assume equally likely outcomes. Equivalently, assume balls are assigned independently.

Exactly one cell empty means:

• one empty cell
• one with two
• n-2 with 1

Choices:

• $n$ for the one empty cell
• $n-1$ for the cell with two balls
• $\binom{n}{2}$ for the balls to use for the two-ball cell.
• $(n-2)!$ arrangements for the other balls in their cells.

So the number of ways to get one empty is

$$n(n-1)\binom{n}{2}(n-2)! = \binom{n}{2}n!$$

The probability of this arrangement is

$$\frac{\binom{n}{2}n!}{n^{n}}$$

1.20

Assigning calls at random to each day can be viewed as making 12 ordered draws with replacement from an urn with 7 balls. The total number of possible draws is $7^{12}$ . To count the number of ways we can get a sample with each day appearing at least once we need to look at the possible patterns of repetitions. For example, one day can have six calls; then each of the other days must have exactly one. The number of draws consistent with this pattern can be found by breaking up the selection of the draw into stages:

• select the day to be repeated--7 choices
• select the 6 calls to occur on that day-- $\binom{12}{6}$ choices
• assign the remaining 6 calls to the remaining 6 days--$6!$ choices

So the total number of ways we can form this pattern is $7 \binom{12}{6} 6!$ . The collection of possible patterns and the number of ways they can occur is:

Pattern	Count
6111111	$7 \binom{12}{6} 6!$ =	4,656,960
5211111	$7 \binom{12}{5} 6 \binom{7}{2} 5!$ =	83,825,280
4221111	$7 \binom{12}{4} \binom{6}{2} \binom{8}{2} \binom{6}{2} 4!$ =	523,908,000
4311111	$7 \binom{12}{4} 6 \binom{8}{3} 5!$ =	139,708,800
3321111	$\binom{7}{2} \binom{12}{3} \binom{9}{3} 5 \binom{6}{2} 4!$ =	139,708,800
3222111	$7 \binom{12}{3} \binom{6}{3} \binom{9}{2}\binom{7}{2} \binom{5}{2} 3!$ =	1,397,088,000
2222211	$\binom{7}{5}\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2} 2!$ =	314,344,800
		3,162,075,840

So the probability of at least one call on each day is $\frac{3,162,075,840}{7^{12}} = 0.22845$ .

A simpler alternative is to use the union-intersection formula. Let $A_i$ be the event that there are no calls on day $i$ . The event that there is at least one call each day is then $A_1^c \cap \dots \cap A_7^c$ , and the complement of this event is $A_1 \cup \dots \cup A_7$ . For any particular $k$ days $i_1, \dots, i_k$

$$P(A_{i_1} \cap \dots \cap A_{i_k}) = \left(\frac{7 - k}{7}\right)^{12}$$

and there are $\binom{7}{k}$ ways to choose $k$ days, so

$$P(A_1 \cup \dots \cup A_7) = \sum_{k=1}^7 (-1)^{k+1}\binom{7}{k}\left(\frac{7 - k}{7}\right)^{12}$$

The last term is zero and can be dropped, so

$$P(A_1 \cup \dots \cup A_7) = \sum_{k=1}^6 (-1)^{k+1}\binom{7}{k}\left(\frac{7 - k}{7}\right)^{12}$$

The number of ways to assign calls to days with each day receiving at least one call is therefore

$$7^{12}(1 - P(A_1 \cup \dots \cup A_7)) = 7^{12} - \sum_{k=1}^6 (-1)^{k+1}\binom{7}{k}(7 - k)^{12} = 3,162,075,840$$

and the corresponding probability is

$$1 - P(A_1 \cup \dots \cup A_7) = 1 - \sum_{k=1}^6 (-1)^{k+1}\binom{7}{k}\left(\frac{7 - k}{7}\right)^{12} = 0.22845$$

1.21

$n$ pairs of shoes, $2r$ chosen at random. Number of possible selections is $\binom{2n}{2r}$ . Number of ordered selections that contain no pairs:

$$2n (2n - 2) (2n - 4) \cdots (2n - 2(2r-1)) = 2^{2r}\frac{n!}{(n-2r)!}$$

Number of unordered selections:

$$2^{2r}\frac{n!}{(n-2r)!(2r)!} = 2^{2r}\binom{n}{2r}$$

So the probability of no matching pair is

$$\frac{2^{2r}\binom{n}{2r}}{\binom{2n}{2r}}$$

Or: Choose pair for each shoe in $\binom{n}{2r}$ ways, then shoes in pairs in $2^{2r}$ ways.