概率论与数理统计 (Spring 2023)/Problem Set 1

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Assumption throughout Problem Set 1

Without further notice, we are working on probability space [math]\displaystyle{ (\Omega,\mathcal{F},\mathbf{Pr}) }[/math].

Problem 1 (Principle of Inclusion and Exclusion)

Let [math]\displaystyle{ n\ge 1 }[/math] be a positive integer and [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] be [math]\displaystyle{ n }[/math] events.

• [Union bound] Prove [math]\displaystyle{ \mathbf{Pr}\left( \bigcup_{i=1}^{n} A_i \right) \le \sum_{i=1}^{n} \mathbf{Pr}\left(A_i\right) }[/math] using the definition of probability space.
• [Principle of Inclusion and Exclusion (PIE)] Prove that [math]\displaystyle{ \mathbf{Pr}\left( \bigcup_{i=1}^n A_i\right) = \sum_{\emptyset \neq S \subseteq [n]} (-1)^{|S|-1} \mathbf{Pr}\left( \bigcap_{i \in S} A_i \right) }[/math], where [math]\displaystyle{ [n]=\{1,2,\ldots,n\} }[/math].
• [Proper [math]\displaystyle{ q }[/math]-colorings of a cycle] Given an arbitrary integer [math]\displaystyle{ q\ge 2 }[/math], find the number of [math]\displaystyle{ n }[/math]-tuples [math]\displaystyle{ p = (p_1,p_2,\ldots,p_n)\in[q]^n }[/math] such that [math]\displaystyle{ p_{n} \neq p_1 }[/math] and [math]\displaystyle{ p_i \neq p_{i+1} }[/math] for all [math]\displaystyle{ 1 \le i \le n-1 }[/math]. You are required to apply PIE to solve this problem.
• [Bonferroni's inequality and Kounias' inequality] Prove that [math]\displaystyle{ \sum_{i=1}^n \mathbf{Pr}(A_i) - \sum_{1 \le i\lt j \le n} \mathbf{Pr}(A_i \cap A_j)\le \mathbf{Pr}\left(\bigcup_{i=1}^n A_i\right) \le \sum_{i=1}^n \mathbf{Pr} \left( A_i\right) - \sum_{i=2}^n \mathbf{Pr}(A_1 \cap A_i). }[/math] (Hint: This is sometimes called Kounias' inequality which is weaker than the Bonferroni's inequality. You can try using Venn diagram to understand these inequalities.)

Problem 2 (Symmetric 1D random walk)

A gambler plays a fair gambling game: At each round, he flips a fair coin, earns [math]\displaystyle{ 1 }[/math] point if it's HEADs, and loses [math]\displaystyle{ 1 }[/math] point if otherwise.

• [Symmetric 1D random walk (I)] Let [math]\displaystyle{ A_i }[/math] be the event that the gambler earns [math]\displaystyle{ 0 }[/math] points after playing [math]\displaystyle{ i }[/math] rounds of the game. Compute [math]\displaystyle{ \sum_{i=1}^{+\infty} \mathbf{Pr}(A_i) }[/math].

• [Symmetric 1D random walk (II)] Suppose that the game ends upon that the gambler loses all his [math]\displaystyle{ m }[/math] points. Let [math]\displaystyle{ B_i }[/math] be the event that the game ends within [math]\displaystyle{ i }[/math] rounds. Compute [math]\displaystyle{ \sum_{i=1}^{+\infty} \mathbf{Pr}(\overline{B_i}) }[/math].

• [Symmetric 1D random walk (III)] Suppose that the game ends upon that the gambler loses all his [math]\displaystyle{ m }[/math] points or earns another [math]\displaystyle{ m }[/math] points (i.e., the number of points he possesses reaches [math]\displaystyle{ 2m }[/math]). Let [math]\displaystyle{ C }[/math] be the event that the game ends within [math]\displaystyle{ n }[/math] rounds. Compute [math]\displaystyle{ \mathbf{Pr}(C) }[/math]. (Hint: A variant of the problem is sometimes called the ballot problem, and there is a famous trick known as André's reflection principle.)

Problem 3 (Probability space)

• [Nonexistence of probability space] Prove that it is impossible to define a uniform probability law on natural numbers [math]\displaystyle{ \mathbb{N} }[/math]. More precisely, prove that there does not exist a probability space [math]\displaystyle{ (\mathbb{N},2^{\mathbb{N}},\mathbf{Pr}) }[/math] such that [math]\displaystyle{ \mathbf{Pr}(\{i\}) = \mathbf{Pr}(\{j\}) }[/math] for all [math]\displaystyle{ i, j \in \mathbb{N} }[/math]. Please explain why the same argument fails to prove that there is no uniform probability law on the real interval [math]\displaystyle{ [0,1] }[/math], that is, there is no such probability space [math]\displaystyle{ ([0,1],\mathcal{F},\mathbf{Pr}) }[/math] that for any interval [math]\displaystyle{ (l,r] \subseteq [0,1] }[/math], it holds that [math]\displaystyle{ (l,r] \in \mathcal{F} }[/math] and [math]\displaystyle{ \mathbf{Pr}( (l,r] ) = r-l }[/math]. (Actually, such probability measure does exist and is called the Lebesgue measure on [math]\displaystyle{ [0,1] }[/math]).

• [Smallest [math]\displaystyle{ \sigma }[/math]-field] For any subset [math]\displaystyle{ S \subseteq 2^\Omega }[/math], prove that the smallest [math]\displaystyle{ \sigma }[/math]-field containing [math]\displaystyle{ S }[/math] is given by [math]\displaystyle{ \bigcap_{\substack{S \subseteq \mathcal{F} \subseteq 2^\Omega\\ \mathcal{F} \text{ is a } \sigma\text{-field }} } \mathcal{F} }[/math]. (Hint: You should show that it is indeed a [math]\displaystyle{ \sigma }[/math]-field and also it is the smallest one containing [math]\displaystyle{ S }[/math].)

• [Limit of events] Let [math]\displaystyle{ (A_i)_{i \in \mathbb{N}} }[/math] be a sequence of events. Define [math]\displaystyle{ \limsup A_n = \bigcap_{n=1}^{+\infty} \bigcup_{k=n}^{+\infty} A_k }[/math] and [math]\displaystyle{ \liminf A_n = \bigcup_{n=1}^{+\infty} \bigcap_{k=n}^{+\infty} A_k }[/math]. Prove that

  • [math]\displaystyle{ \liminf A_n, \limsup A_n \in \mathcal{F} }[/math] and [math]\displaystyle{ \liminf A_n \subseteq \limsup A_n }[/math];
  • [math]\displaystyle{ \liminf A_n = \{\omega \in \Omega \mid \omega \in A_n \text{ for all but finitely many values of } n\} }[/math];
  • [math]\displaystyle{ \limsup A_n = \{\omega \in \Omega \mid \omega \in A_n \text{ for infinite many values of } n\} }[/math];
  • If [math]\displaystyle{ A = \liminf A_n = \limsup A_n }[/math], then [math]\displaystyle{ \mathbf{Pr}(A_n) \rightarrow \mathbf{Pr}(A) }[/math] as [math]\displaystyle{ n\to\infty }[/math];
  • Furthermore, if the limit [math]\displaystyle{ A = \liminf A_n = \limsup A_n }[/math] exists and for every [math]\displaystyle{ n }[/math], the two events [math]\displaystyle{ \bigcup_{k=n}^{+\infty} A_k }[/math] and [math]\displaystyle{ \bigcap_{k=n}^{+\infty} A_k }[/math] are independent, then [math]\displaystyle{ \mathbf{Pr}(A) }[/math] is either [math]\displaystyle{ 0 }[/math] or [math]\displaystyle{ 1 }[/math].

Problem 4 (Conditional probability)

• [Conditional version of the total probability] Let [math]\displaystyle{ C_1,\cdots, C_n }[/math] be disjoint events that form a partition of the sample space. Let [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] be events such that [math]\displaystyle{ \textbf{Pr}(B\cap C_i)\gt 0 }[/math] for all [math]\displaystyle{ i }[/math]. Show that [math]\displaystyle{ \textbf{Pr}(A|B) = \sum_{i = 1}^n \textbf{Pr}(C_i|B)\cdot \textbf{Pr}(A|B\cap C_i) }[/math]

• [Balls in urns (I)] There are [math]\displaystyle{ n }[/math] urns of which the [math]\displaystyle{ r }[/math]-th contains [math]\displaystyle{ r-1 }[/math] white balls and [math]\displaystyle{ n-r }[/math] black balls. You pick an urn uniformly at random (here, "uniformly" means that each urn has equal probability of being chosen) and remove two balls from that urn, uniformly at random without replacement (which means that each of the [math]\displaystyle{ {n-1\choose 2} }[/math] pairs of balls are chosen to be removed with equal probability). Find the following probabilities:

  1. the second ball is black;
  2. the second ball is black, given that the first is black.

• [Balls in urns (II)] Suppose that an urn contains [math]\displaystyle{ w }[/math] white balls and [math]\displaystyle{ b }[/math] black balls. The balls are drawn from the urn one by one, each time uniformly and independently at random, without replacement (which means we do not put the chosen ball back after each drawing). Find the probabilities of the events:

  1. the first white ball drawn is the [math]\displaystyle{ (k+1) }[/math]th ball;
  2. the last ball drawn is white.

• [Balls in urns (III)] There are [math]\displaystyle{ n }[/math] urns filled with black and white balls. Let [math]\displaystyle{ f_i }[/math] be the fraction of white balls in the [math]\displaystyle{ i }[/math]-th urn. In stage 1 an urn is chosen uniformly at random. In stage 2 a ball is drawn uniformly at random from the urn chosen in stage 1. Let [math]\displaystyle{ U_i }[/math] be the event that the [math]\displaystyle{ i }[/math]-th urn was chosen at stage 1. Let [math]\displaystyle{ W }[/math] be the event that a white ball is drawn at stage 2, and [math]\displaystyle{ B }[/math] be the event that a black ball is drawn at stage 2.
  1. Use Bayes's Law to express [math]\displaystyle{ \textbf{Pr}(U_i|W) }[/math] in terms of [math]\displaystyle{ f_1,\cdots, f_n }[/math].
  2. Let's say there are three urns and urn 1 has 30 white and 10 black balls, urn 2 has 20 white and 20 black balls, and urn 3 has 10 white balls and 30 black balls. Compute [math]\displaystyle{ \textbf{Pr}(U_1|B) }[/math], [math]\displaystyle{ \textbf{Pr}(U_2|B) }[/math] and [math]\displaystyle{ \textbf{Pr}(U_3|B) }[/math].

Problem 5 (Independence)

Let's consider a series of [math]\displaystyle{ n }[/math] outputs [math]\displaystyle{ (X_1, X_2, \cdots, X_n) \in \{0,1\}^n }[/math] of [math]\displaystyle{ n }[/math] independent Bernoulli trials, where each trial succeeds with the same probability [math]\displaystyle{ 0 \lt p \lt 1 }[/math].

• [Limited independence] Let [math]\displaystyle{ n = 2 }[/math]. Construct three events [math]\displaystyle{ A,B }[/math] and [math]\displaystyle{ C }[/math] out of the [math]\displaystyle{ n = 2 }[/math] Bernoulli trials such that [math]\displaystyle{ A, B }[/math] and [math]\displaystyle{ C }[/math] are pairwise independent but are not (mutually) independent. You need to prove that the constructed events [math]\displaystyle{ A, B }[/math] and [math]\displaystyle{ C }[/math] satisfy this.

• [Product distribution] Suppose someone has observed the output of the [math]\displaystyle{ n }[/math] trials, and she told you that precisely [math]\displaystyle{ k }[/math] out of [math]\displaystyle{ n }[/math] trials succeeded for some [math]\displaystyle{ 0\lt k\lt n }[/math]. Now you want to predict the output of the [math]\displaystyle{ (n+1) }[/math]-th trial while the parameter [math]\displaystyle{ p }[/math] of the Bernoulli trial is unknown. One way to estimate [math]\displaystyle{ p }[/math] is to find a [math]\displaystyle{ \hat{p} }[/math] which makes the known result most probable, namely you need to solve [math]\displaystyle{ \arg \max_{\hat{p}\in(0,1)} \mathbf{Pr}_{\hat{p}} [\text{the trial succeeded } k \text{ times}]. }[/math]

  1. Estimate [math]\displaystyle{ p }[/math] by solving the given objective function.
  2. If someone tells you exactly which [math]\displaystyle{ k }[/math] rounds in total [math]\displaystyle{ n }[/math] trials that are successful, would it help you to estimate [math]\displaystyle{ p }[/math] more accurately? Why?

Problem 6 (Probabilistic method)

A CNF formula over variables [math]\displaystyle{ x_1,\cdots, x_n }[/math] is a conjunction (AND) of clause, where each clause is an OR of literals. A literal is either a variable [math]\displaystyle{ x_1 }[/math] or its negation [math]\displaystyle{ \bar{x}_i }[/math]. A CNF formula is satisfiable if there is an assignment in [math]\displaystyle{ \{0,1\}^n }[/math] to the variables that makes the formula true. A [math]\displaystyle{ k }[/math]-CNF formula is a CNF formula where each clause of it has exactly [math]\displaystyle{ k }[/math] literals (without repetition).

• [Satisfiability (I)] Let [math]\displaystyle{ F }[/math] be a [math]\displaystyle{ k }[/math]-CNF with less than [math]\displaystyle{ 2^k }[/math] clauses. Use the probabilistic method to show that [math]\displaystyle{ F }[/math] must be satisfiable. You should clarify the probability space you used.
• [Satisfiability (II)] Give a constructive proof of same problem above. That is, prove that [math]\displaystyle{ F }[/math] is satisfiable by showing how to construct an assignment that does satisfy [math]\displaystyle{ F }[/math]. Your construction doesn't have to be efficient. Can you do it also by the probabilistic method?
• [Satisfiability (III)] Let [math]\displaystyle{ F }[/math] be a [math]\displaystyle{ k }[/math]-CNF with [math]\displaystyle{ m\geq 2^k }[/math] clauses. Use the probabilistic method to show that there exists an assignment satisfying at least [math]\displaystyle{ \lfloor m(1-1/2^k) \rfloor }[/math] clauses in [math]\displaystyle{ F }[/math]. (Hint: Consider overlaps of events in Venn diagram.) You should clarify the probability space you used.