First recitation

Hanzhe Li

Contents

Reading guide

The main line of this chapter is the three parts of a probability space, operations on events, and inclusion-exclusion.
For independence problems, keep pairwise independence and mutual independence separate. It is easy to mix them up in a first course.
For distribution functions, check four things one by one: monotonicity, right-continuity, limits at infinity, and jumps.

Tip. For identities involving events, first check the set identity, then take probabilities.

Exercise 1.1

Note

This section is about sample spaces, fields of events, probability measures, and finite inclusion-exclusion. In Jordan's formula, the key point is how many times each intersection is counted.

Problem

Toss two fair coins. Write down the three parts of the probability space in detail.

Proof

$\Omega=\{(H,H),(H,T),(T,H),(T,T)\}$ , $\mathcal{F}=2^{\Omega}$ , and $\mathbb{P}(A)=\dfrac{|A|}{4}$ .

Problem

Compare the probabilities of the following two events: "in 3 tosses of a fair die, at least one 6 appears" and "in 6 tosses of a fair die, at least two 6's appear."

Proof

$\mathbb{P}(\text{at least one 6 in 3 tosses})=1-(\dfrac{5}{6})^3$ , while $\mathbb{P}(\text{at least two 6's in 6 tosses})=1-(\dfrac{5}{6})^6-6\cdot\dfrac{1}{6}\cdot(\dfrac{5}{6})^5$ . Comparing them shows that the first probability is larger.

Problem

Let $\mathcal{F}$ be a $\sigma$ -algebra, and let $A,B\in\mathcal{F}$ . Prove that $A\cap B\in\mathcal{F}$ and $A\setminus B\in\mathcal{F}$ .

Proof

This follows from $A\cap B=(A^c\cup B^c)^c$ and $A\setminus B=A\cap B^c$ .

Problem

Prove Jordan's formula.

Proof

Use induction. The case $n=1$ is clear. For the induction step,

\begin{aligned} \mathbb{P}\Big( \bigcup_{i=1}^n A_i \Big) &= \sum_{k=1}^{n-1} (-1)^{k-1} \sum_{1 \le i_1 < \dots < i_k \le n-1} \mathbb{P}(A_{i_1} \cap \cdots \cap A_{i_k}) + \mathbb{P}(A_n) \\ &\quad - \sum_{k=1}^{n-1} (-1)^{k-1} \sum_{1 \le i_1 < \dots < i_k \le n-1} \mathbb{P}(A_{i_1} \cap \cdots \cap A_{i_k} \cap A_n) \\ &= \sum_{k=1}^{n-1} (-1)^{k-1} \sum_{1 \le i_1 < \dots < i_k \le n-1} \mathbb{P}(A_{i_1} \cap \cdots \cap A_{i_k})+ \mathbb{P}(A_n) \\ &\quad + \sum_{k=1}^{n-1} (-1)^{k} \sum_{1 \le i_1 < \dots < i_k \le n-1 } \mathbb{P}(A_{i_1} \cap \cdots \cap A_{i_k} \cap A_n).\\ &= \sum_{k=1}^{n-1} (-1)^{k-1} \sum_{1 \le i_1 < \dots < i_k <n} \mathbb{P}(A_{i_1} \cap \cdots \cap A_{i_k})+ \mathbb{P}(A_n) \\ &\quad + \sum_{k=2}^{n} (-1)^{k-1} \sum_{1 \le i_1 < \dots < i_k =n} \mathbb{P}(A_{i_1} \cap \cdots \cap A_{i_k} ).\\ &=\mathrm{RHS}. \end{aligned}

Problem

Prove that $|\mathbb{P}(A\cap B)-\mathbb{P}(A)\mathbb{P}(B)|\leq 1/4$ , and discuss when equality holds.

Proof

One way is to use $\operatorname{Cov}(\mathbb{1}_A,\mathbb{1}_B)\leq \sqrt{\operatorname{Var}(\mathbb{1}_A)}\sqrt{\operatorname{Var}(\mathbb{1}_B)}$ and the equality condition in Cauchy--Schwarz. We can also give a direct proof. Write $\mathbb{P}(A)=x$ and $\mathbb{P}(B)=y$ .

Then

\max\{0, x + y - 1\} \leq \mathbb{P}(A\cap B) \leq \min\{x, y\}.

It is enough to check the bound when $\mathbb{P}(A\cap B)$ takes its largest and smallest possible values.

Assume without loss of generality that $x \leq y$ . Then

0 \leq x - xy = x(1 - y) \leq x(1 - x) \leq \frac{1}{4}.

So the inequality holds when $\mathbb{P}(A\cap B)$ takes its largest possible value.

If $x + y \leq 1$ , then $\min \mathbb{P}(A\cap B) = 0$ , and

0 < xy \leq x(1 - x) \leq \frac{1}{4}.

If $x + y \geq 1$ , then $\min \mathbb{P}(A\cap B) = x + y - 1$ , and

|x + y - 1 - xy |= (1 - x)(1 - y).

Also $(1 - x) + (1 - y) \leq 1$ , so by the elementary inequality,

(1 - x)(1 - y) \leq \frac{1}{4}.

Thus the inequality holds when $\mathbb{P}(A\cap B)$ takes its smallest possible value as well.

Therefore the original inequality holds. Equality holds if and only if

\mathbb{P}(A)=\mathbb{P}(B) = \frac{1}{2}, \quad \mathbb{P}(A\cap B) = \frac{1}{2} \text{ or } 0.

Problem

Let $\{A_k\}_{k=1}^{\infty}$ be a sequence of events such that $\mathbb{P}(A_k) = 1$ for all $k \in \mathbb{N}^*$ . Prove that

\mathbb{P}\left( \bigcap_{k=1}^{\infty} A_k \right) = 1.

Proof

Look at the complement. By De Morgan's law,

\left(\bigcap_{k=1}^{\infty} A_k\right)^c = \bigcup_{k=1}^{\infty} A_k^c.

Since $\mathbb{P}(A_k) = 1$ , we have $\mathbb{P}(A_k^c) = 0$ for every $k \in \mathbb{N}^*$ . By countable subadditivity,

\mathbb{P}\left( \bigcup_{k=1}^{\infty} A_k^c \right) \leq \sum_{k=1}^{\infty} \mathbb{P}(A_k^c) = 0.

So $\mathbb{P}\left( \bigcup_{k=1}^{\infty} A_k^c \right) = 0$ , and hence

1\geq \mathbb{P}\left( \bigcap_{k=1}^{\infty} A_k \right) = 1 - \mathbb{P}\left( \bigcup_{k=1}^{\infty} A_k^c \right) \geq 1 - 0 = 1.

Exercise 1.2

Note

Here we need to distinguish pairwise independence from mutual independence. The prime-number example shows that in a finite uniform probability space, arithmetic can strongly restrict the form of independent events.

Problem

Toss a fair coin $n$ times. Let $A_{ij}$ be the event that the $i$ -th toss and the $j$ -th toss have the same outcome. Prove that the events $\{A_{ij}, 1 \leq i < j \leq n\}$ are pairwise independent but not mutually independent.

Proof

For two different pairs, the only nontrivial case is when the two pairs share one index, for example $A_{ij}$ and $A_{jk}$ . Then

\mathbb{P}(A_{ij}\cap A_{jk}) =\mathbb{P}(X_i=X_j=X_k) =2\times\dfrac{1}{8} =\dfrac{1}{2}\times\dfrac{1}{2} =\mathbb{P}(X_i=X_j)\mathbb{P}(X_j=X_k).

If the pairs are disjoint, independence is immediate from independence of the tosses. Thus the events are pairwise independent. But they are not mutually independent. For example,

\mathbb{P}(X_1=X_2,\ X_2=X_3,\ X_3=X_1)=\dfrac{1}{4},

whereas

\mathbb{P}(X_1=X_2)\mathbb{P}(X_2=X_3)\mathbb{P}(X_3=X_1)=\dfrac{1}{8}.

Problem

Let $p$ be prime, and put the uniform probability model on $\Omega=\{1,2,\cdots,p\}$ . If two events $A$ and $B$ are independent, prove that at least one of $A,B$ is either $\emptyset$ or $\Omega$ .

Proof

Suppose instead that both $A$ and $B$ are nonempty proper subsets of $\Omega$ . If they are independent, then $\dfrac{|A\cap B|}{p}=\dfrac{|A||B|}{p^2}$ . Let $|A|=a$ , $|B|=b$ , and $|A\cap B|=c$ . Then $pc=ab$ , so $p\mid ab$ . Hence $p\mid a$ or $p\mid b$ , contradicting the assumption that $A$ and $B$ are nonempty proper subsets.

Problem

There are $n$ urns. The $r$ -th urn contains $r-1$ red balls and $n-r$ blue balls. Choose one urn at random, and then draw two balls without replacement. Find the probabilities of the following events: (1) the second ball is blue; (2) the second ball is blue given that the first ball is blue.

Proof

\begin{aligned} \mathbb{P}(\text{the second ball is blue})&=\sum_{k=1}^n\mathbb{P}(\text{the second ball is blue}\mid\text{urn }k\text{ is chosen})\mathbb{P}(\text{urn }k\text{ is chosen})\\ &=\sum_{k=1}^n\dfrac{n-k}{n(n-1)}=\dfrac{1}{2}. \end{aligned}

Let $B_1=\{\text{the first ball is blue}\}$ and $B_2=\{\text{the second ball is blue}\}$ . Similarly,

\mathbb{P}(B_1)=\dfrac{1}{2},\quad \mathbb{P}(B_1 \cap B_2) = \frac{1}{n} \sum_{r=1}^{n} \frac{(n-r)(n-r-1)}{(n-1)(n-2)}=\dfrac{1}{3}.

Therefore $\mathbb{P}(B_2|B_1)=\dfrac{2}{3}$ .

Problem

One hundred passengers board a plane with exactly 100 seats. Each passenger has an assigned seat. The first passenger chooses a seat uniformly at random. Passenger 2 sits in their own seat if it is empty; otherwise, they choose uniformly from the remaining empty seats. Passenger 3 follows the same rule, and so on until all passengers have boarded. What is the probability that the last passenger sits in their own seat?

Proof

If passenger 1 sits in their own seat, then passenger $n$ will certainly sit in their own seat. Otherwise, suppose passenger 1 sits in seat $k_1$ . Then passengers 2 through $k_1-1$ all sit in their own seats. If passenger $k_1$ sits in seat 1, then passenger $n$ will still get their own seat. If passenger $k_1$ sits in seat $k_2$ , then passengers $k_1+1$ through $k_2-1$ all sit in their own seats. After that, $k_2$ plays the same role as $k_1$ did before. So we only need to look at the first person who sits in seat 1 or seat $n$ .

Let $X$ be the index of this person. Given $X=k$ , that person is equally likely to choose seat 1 or seat $n$ . Hence

\mathbb{P}(\text{passenger }n\text{ sits in their own seat}) =\sum_{k=1}^{100}\mathbb{P}(\text{passenger }n\text{ sits in their own seat}\mid X=k)\mathbb{P}(X=k) =\dfrac{1}{2}\sum_{k=1}^{100}\mathbb{P}(X=k) =\dfrac{1}{2}.

Exercise 1.3

Note

This section is mainly about Bernoulli trials and recursions with finitely many states. Once the state variables are written clearly, the recursion usually becomes natural.

Problem

Two table-tennis players, A and B, play against each other. Player A has a higher chance of winning each game than player B. There are two possible match formats: best of 3 games, or best of 5 games. Which format is more favorable to player A?

Proof

Let $p=\mathbb{P}(\text{A wins one game})$ . In a best-of-3 match, $\mathbb{P}(\text{A wins})=p^2+2p^2(1-p)$ . In a best-of-5 match, $\mathbb{P}(\text{A wins})=p^3+3p^3(1-p)+6p^3(1-p)^2=p^3(10-15p+6p^2)$ . Therefore

\mathbb{P}(\text{A wins best of 5})- \mathbb{P}(\text{A wins best of 3})= 3p^{2} (p-1)^{2} (2p-1).

This is positive when $p>\dfrac{1}{2}$ .

Problem

Child $\zeta$ tosses $n+1$ fair coins, and child $\delta$ tosses $n$ fair coins. Find the probability that $\zeta$ gets more heads than $\delta$ .

Proof

Let $X_1$ be the number of heads in the first $n$ tosses of $\zeta$ , let $X_2$ be the result of the extra toss of $\zeta$ , and let $Y$ be the number of heads in the $n$ tosses of $\delta$ . Let the desired event be $A$ . By the law of total probability,

\begin{aligned} \mathbb{P}(A) &= \dfrac{1}{2}\mathbb{P}(X_1+X_2> Y|X_2=1) +\dfrac{1}{2}\mathbb{P}(X_1+X_2>Y|X_2=0) \\ &= \dfrac{1}{2}\mathbb{P}(X_1>Y-1)+\dfrac{1}{2}\mathbb{P}(X_1>Y) \\ &=\dfrac{1}{2}(\mathbb{P}(X_1>Y)+\mathbb{P}(X_1=Y))+\dfrac{1}{2}\mathbb{P}(X_1>Y)\\ &=\mathbb{P}(X_1>Y)+\dfrac{1}{2}\mathbb{P}(X_1=Y). \end{aligned}

By symmetry,

2\mathbb{P}(A) = 2\mathbb{P}(X_1 > Y) + \mathbb{P}(X_1 = Y)= \mathbb{P}(X_1 > Y) + \mathbb{P}(X_1 = Y) + \mathbb{P}(X_1 < Y)=1.

Thus the answer is $\dfrac{1}{2}$ . One can also compute it directly:

\begin{aligned} \mathbb{P}(A)&= \sum_{k=1}^{n+1} \binom{n+1}{k} \left( \frac{1}{2} \right)^{n+1} \cdot \sum_{i=0}^{k-1} \binom{n}{i} \left( \frac{1}{2} \right)^n \\ &= 2^{-2n-1} \sum_{k=1}^{n+1} \binom{n+1}{k} \sum_{i=0}^{k-1} \binom{n}{i}\\ &=2^{-2n-2} \left(\sum_{k=1}^{n+1} \binom{n+1}{k} \sum_{i=0}^{k-1} \binom{n}{i}+\sum_{k=1}^{n+1} \binom{n+1}{k} \sum_{i=0}^{k-1} \binom{n}{i} \right) \\ &=2^{-2n-2} \left(\sum_{k=1}^{n+1} \binom{n+1}{k} \sum_{i=0}^{k-1} \binom{n}{i}+\sum_{i=0}^{n} \binom{n}{i} \sum_{k=0}^{n-i} \binom{n+1}{k} \right) \\ &=2^{-2n-2} \left(\sum_{k=1}^{n+1} \binom{n+1}{k} \sum_{i=0}^{k-1} \binom{n}{i}+\sum_{j=0}^{n} \binom{n}{n-j} \sum_{k=0}^{j} \binom{n+1}{k} \right) \\ &=2^{-2n-2} \left(\sum_{k=1}^{n+1} \binom{n+1}{k} \sum_{i=0}^{k-1} \binom{n}{i}+\sum_{j=0}^{n} \binom{n}{j} \sum_{k=0}^{j} \binom{n+1}{k} \right) \\ &=2^{-2n-2}\cdot2^{2n+1}\\ &=\dfrac{1}{2}. \end{aligned}

The second-to-last equality holds because the two sums together count all equally likely outcomes of the two sets of coin tosses. The first sum counts the cases where $\zeta$ gets more heads; the second counts the cases where $\delta$ gets at least as many heads as $\zeta$ .

Problem

Players A and B take turns rolling a fair die. A rolls first. A keeps rolling until a 1 appears, then B starts rolling. B keeps rolling until a 1 appears, then A starts again, and so on. Find the probability that the $n$ -th roll is made by A.

Proof

Let $a_n$ be the probability that the $n$ -th roll is made by A, and let $b_n$ be the probability that it is made by B. Then

a_{n+1} = \frac{5}{6} a_n + \frac{1}{6} b_n,\quad b_{n+1} = \frac{1}{6} a_n + \frac{5}{6} b_n,\quad a_n + b_n = 1,\quad a_1 = 1.

So the recursion is $a_{n+1}=\dfrac{2}{3}a_{n}+\dfrac{1}{6}$ , and therefore

a_n=\dfrac{1}{2}+\dfrac{1}{2}\left(\dfrac{2}{3}\right)^{n-1}.

Exercise 1.4

Note

For problems about distribution functions, focus on monotonicity, right-continuity, the two limits at infinity, and jumps. The size of a jump is the probability mass at that point.

Problem

Let $F(x)$ and $G(x)$ be distribution functions. (1) For $0\le \lambda\le 1$ , prove that $\lambda F(x)+(1-\lambda)G(x)$ and $F(x)G(x)$ are distribution functions. (2) Prove that $1-\{1-F(x)\}^n$ and $(F(x)-1)e+\exp\{1-F(x)\}$ are distribution functions.

Proof

It is enough to check monotonicity, right-continuity, and normalization. In particular, $1-\{1-F(x)\}^n$ is the distribution function of the minimum of $n$ independent random variables with common distribution function $F$ . If $m_n=\min(X_1,\ldots,X_n)$ , then

\mathbb{P}(m_n \le x) = 1 - \mathbb{P}(m_n > x) = 1 - \mathbb{P}(X_1 > x, \ldots, X_n > x) = 1 - [1 - F(x)]^n.

Problem

Let $X$ be a random variable, and define $G(x)=\mathbb{P}(X<x)$ . Prove that $G$ is left-continuous on $\mathbb{R}$ , and express $\mathbb{P}(y\le X\le x)$ in terms of $G$ .

Proof

\lim_{n \to \infty} \mathbb{P}(X<x-1/n) = \mathbb{P}\left(\bigcup_n \{X<x-1/n\}\right) = \mathbb{P}(X<x).

Here we used continuity of probability measures. Also $\mathbb{P}(X=x)=G(x+0)-G(x)$ . Since

\{y\leq X \leq x\}=\{y\leq X\}-\{X\geq x\}+\{X=x\},

we get

\mathbb{P}(y \leq X \leq x)=G(x+0)-G(y).

Problem

Let $X,Y$ be random variables. (1) Prove that $\min\{X,Y\}$ and $\max\{X,Y\}$ are also random variables. (2) Prove that $|X|$ and $X^2$ are also random variables.

Proof

Use the following identities: $\{\min\{X,Y\}>x\}=\{X>x\}\cap \{Y>x\}$ , $\{\max\{X,Y\}\leq x\}=\{X\leq x\}\cap \{Y\leq x\}$ , $\{|X|\leq x\}=\{-x\leq X\leq x\}$ for $x\ge 0$ and is empty for $x<0$ , and $\{X^2\leq x\}=\{-\sqrt{x}\leq X \leq \sqrt{x}\}$ for $x\ge 0$ and is empty for $x<0$ .

Problem

Suppose the distribution function of a random variable $X$ is $F(x)=A\arctan x+B$ , $x\in\mathbb{R}$ . Find the constants $A$ and $B$ .

Proof

We must have $\lim_{x\to +\infty}F(x)=\dfrac{A\pi}{2}+B=1$ and $\lim_{x \to -\infty}F(x)=-\dfrac{A\pi}{2}+B=0$ . Solving these two equations gives $B=\dfrac{1}{2}$ and $A=1/\pi$ .

Problems

Note

This section puts inclusion-exclusion, Euler's product formula, and medians together. A useful habit is to rewrite the object in terms of events or indicator functions.

Problem

Prove Bonferroni's inequality:

\mathbb{P} \left( \bigcup_{r=1}^n A_r \right) \geq \sum_{r=1}^n \mathbb{P}(A_r) - \sum_{1 \leq r < k \leq n} \mathbb{P}(A_r \cap A_k).

Proof

Write probabilities as expectations of indicator functions: $\mathbb{E}(\mathbb{1}_A(\omega))=1\cdot \mathbb{P}(\omega\in A)+0\cdot\mathbb{P}(\omega\notin A)=\mathbb{P}(A)$ . Let $f(\omega)=\mathbb{1}_{\cup_{i=1}^n A_i}$ and $g(\omega)=\sum_{i=1}^n \mathbb{1}_{A_i}-\sum_{1\leq i < k \leq n}\mathbb{1}_{A_i\cap A_k}$ . For a fixed $\omega$ , let $r=|\{i:\omega\in A_i\}|$ , the number of sets among $\{A_i\}_{i=1}^n$ that contain $\omega$ . Then:

If $r=0$ , then $f=g=0$ .
If $r=1$ , then $f=g=1$ .
If $r>1$ , then $f=1$ and $g=r-\binom{r}{2}$ .

So $f\geq g$ . Taking expectations gives the inequality. The same method also gives the more general Bonferroni inequalities. If $m$ is odd, then

\mathbb{P} \left( \bigcup_{r=1}^n A_r \right) \leq \sum_{k=1}^m (-1)^{k-1} \sum_{1 \leq i_1 < \cdots < i_k \leq n} \mathbb{P}(A_{i_1} \cap \cdots \cap A_{i_k}).

If $m$ is even, then

\mathbb{P} \left( \bigcup_{r=1}^n A_r \right) \geq \sum_{k=1}^m (-1)^{k-1} \sum_{1 \leq i_1 < \cdots < i_k \leq n} \mathbb{P}(A_{i_1} \cap \cdots \cap A_{i_k}).

Problem

On the positive integers $\Omega = \{1, 2, \cdots\}$ , define the probability measure

\mathbb{P}(\{n\}) = \frac{1}{\zeta(s) \cdot n^s}, \quad \zeta(s) = \sum_{n=1}^{\infty} n^{-s}, \quad s > 1.

For a positive integer $q$ , let $A_q = \{mq : m \in \Omega\}$ .

$1$ For any distinct primes $p_1, p_2, \cdots, p_t$ , prove that $A_{p_1}, A_{p_2}, \cdots, A_{p_t}$ are mutually independent.

$2$ Use probability to prove Euler's formula:

\zeta(s) = \prod_{i=1}^{\infty} \left(1 - \frac{1}{p_i^s}\right)^{-1},

where $p_1 < p_2 < \cdots$ are all primes.

$3$ Under the probability measure $\mathbb{P}$ , choose two positive integers $a,b$ independently. Prove that the probability that $a$ and $b$ are coprime is

\frac{1}{\zeta(2s)}.

Note.

\sum_{n=1}^{+\infty} n^{-s}

converges, so the probability measure $\mathbb{P}$ is well-defined.

Proof

Using the least common multiple property of distinct primes,

\mathbb{P}(A_{p_1}\cap A_{p_2}\cdots\cap A_{p_t}) =\mathbb{P}(A_{p_1p_2\cdots p_t}) =\dfrac{1}{\zeta(s)\prod_{i=1}^tp_i^s}\sum_{m=1}^{\infty}\dfrac{1}{m^s} =\prod_{i=1}^t\dfrac{1}{p_i^s} =\prod_{i=1}^t\mathbb{P}(A_{p_i}).

Thus $A_{p_1}, A_{p_2}, \cdots, A_{p_t}$ are mutually independent. Therefore

\dfrac{1}{\zeta(s)} =\mathbb{P}(\{1\}) =\mathbb{P}(\cap_{i=1}^{\infty}A_{p_i}^c) =\prod_{i=1}^\infty\left(1-\dfrac{1}{p_i^s}\right).

Finally, let $X,Y$ be independent random variables with this common distribution. Then

\begin{aligned} \mathbb{P}(X,Y\text{ are coprime}) &=1-\mathbb{P}(\cup_i(X\in A_{p_i}\cap Y \in A_{p_i}))\\ &=\prod_i\mathbb{P}(X\not \in A_{p_i}\cap Y \not\in A_{p_i} )\\ &=\prod_i( 1-\dfrac{1}{p_i^{2s}})\\ &=\dfrac{1}{\zeta(2s)}. \end{aligned}

Problem

Define a real number $m$ to be a median of a distribution function $F$ if

F(m - 0) \leq \frac{1}{2} \leq F(m).

Prove that every distribution function has at least one median, and that the set of all medians is a closed interval.

Proof

Define

a:=\sup\{x:F(x)\leq \dfrac{1}{2}\},\qquad b:=\inf\{x:F(x)\geq \dfrac{1}{2}\}.

Claim that $a \leq b$ and that the set of medians is $[a,b]$ .

If there were some $c$ with $b < c <a$ , then $\dfrac{1}{2}\leq F(c)\leq \dfrac{1}{2}$ , which would force $a=b=c$ , a contradiction. For every $c \in [a,b]$ , we have $F(c)\geq \dfrac{1}{2}$ since $c\geq a$ , and $F(c-0)\leq F(c)\leq \dfrac{1}{2}$ since $c\leq b$ . Thus every $c\in[a,b]$ is a median.

If $c^\prime<a$ and $c^\prime$ were a median, then $F(c^\prime )\geq \dfrac{1}{2}$ , which would imply $a>c^\prime \geq b$ , impossible. Similarly, no $c^{\prime\prime}>b$ can be a median. Hence the set of medians is exactly $[a,b]$ .

End-of-chapter checklist

The original problems and solutions in this chapter come from the corresponding TeX source file.
You can first read only the problem boxes, write down the key identities, and then open the proofs or solutions.
If a result uses independence, countable additivity, a change of variables, or a moment condition, it is worth marking that point explicitly.