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Chapter 10: Problem 10

Let \(n\) be a prime number greater than \(2 ;\) and let \(X . Y\) be independent and uniformly distributed on \(\\{0,1, \ldots, n-1\\}\). (That is. \(P(X=i)=P(Y=\) i) \(=\frac{1}{n}\), for \(i=0.1 \ldots n-1 .\) For each \(r, 0 \leq r \leq n-1 .\) define \(Z_{r}=\) \(X+r Y(\bmod n)\). a) Show that the r.v,'s \(\left\\{Z_{r}: 0 \leq r \leq n-1\right\\}\) are pairwise independent. b) Is the same result true if \(n\) is no longer assumed to be prime? [Ans: No.]

Short Answer

Expert verified
a) Yes, when \(n\) is prime, \(Z_r\)'s are pairwise independent. b) No, not true if \(n\) is composite.

Step by step solution

01

Understanding the Variables

We are given two independent random variables, \(X\) and \(Y\), each uniformly distributed over the set \(\{0, 1, \ldots, n-1\}\). This implies that for any integer \(i\), \(P(X = i) = P(Y = i) = \frac{1}{n}\). The variable \(Z_r\) is defined as \(Z_r = X + rY \mod n\), where \(r\) is another integer such that \(0 \leq r \leq n-1\). Our goal is to show these \(Z_r\) are pairwise independent when \(n\) is prime.
02

Define Pairwise Independence

Two random variables \(Z_{r_1}\) and \(Z_{r_2}\) (where \(r_1 eq r_2\)) are pairwise independent if for any values \(a\) and \(b\), \(P(Z_{r_1} = a, Z_{r_2} = b) = P(Z_{r_1} = a)P(Z_{r_2} = b)\). We will verify this condition for the given random variables \(Z_r\).
03

Calculate Joint Probability

Consider \(P(Z_{r_1} = a, Z_{r_2} = b)\). Since \(Z_{r_1} = X + r_1Y \mod n\) and \(Z_{r_2} = X + r_2Y \mod n\), this probability is equivalent to finding the number of solutions \((x, y)\) such that \(x + r_1y \equiv a \mod n\) and \(x + r_2y \equiv b \mod n\), where \(x, y\) are uniformly distributed over \(\{0, 1, \ldots, n-1\}\).
04

Simplify Using Modulus Arithmetic

Using modular arithmetic, note that \(r_1 eq r_2\) implies \(r_1 - r_2 \) is coprime to \(n\), since \(n\) is prime. Thus, there always exists a unique solution \(y\) modulo \(n\) to \((r_1 - r_2)y \equiv a - b \mod n\). The choice of \(y\) uniquely determines \(x\), hence there are \(n\) possible combinations, one for each \(y\).
05

Calculate Individual Probabilities

Since \(X\) and \(Y\) are uniformly distributed, \(P(Z_r = a) = \frac{1}{n}\) for any \(a\). Therefore, both \(P(Z_{r_1} = a) = \frac{1}{n}\) and \(P(Z_{r_2} = b) = \frac{1}{n}\).
06

Verify Pairwise Independence

Since we can find exactly \(n\) valid solutions \((x, y)\) for the joint distribution and each occurs with probability \(\frac{1}{n^2}\) due to independence, \(P(Z_{r_1} = a, Z_{r_2} = b) = \frac{1}{n^2} = P(Z_{r_1} = a)P(Z_{r_2} = b)\). Thus, \(Z_{r_1}\) and \(Z_{r_2}\) are pairwise independent.
07

Considering Composite Number for n

If \(n\) is composite, the above reasoning fails because \(r_1 - r_2\) might share a common factor with \(n\), making \( (r_1 - r_2)y \equiv a - b \mod n\) potentially have more than one solution \(y\), impacting the independence condition. Thus, \(Z_r\)'s are not necessarily pairwise independent in the composite case.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Prime Numbers
A prime number is a fundamental concept in mathematics and probability theory. It is defined as a natural number greater than 1 that has no positive divisors other than 1 and itself. For example, numbers like 2, 3, 5, 7, and 11 are considered prime numbers. Prime numbers are important because they serve as the building blocks for other numbers, through the process called prime factorization.
  • Properties of Prime Numbers: They have exactly two distinct positive divisors: 1 and the number itself.
  • Use in Cryptography: Their properties make them an essential tool in cryptography, such as in RSA encryption.
Prime numbers are also inherently connected to concepts like pairwise independence in probability theory, because their unique characteristics ensure specific probabilities hold true, as seen in the exercise.
Uniform Distribution
Uniform distribution is a crucial concept in probability theory. A random variable is said to be uniformly distributed if each of its possible values has the same likelihood of occurring. In our exercise, both variables, X and Y, are uniformly distributed over the set \{0, 1, ..., n-1\}. This means:
  • Each value is equally likely, with probability calculated as \( \frac{1}{n} \).
  • Uniform distribution is often considered the most straightforward form of distribution due to its equal probability assignment.
Uniform distribution helps simplify calculations and is particularly significant in theoretical aspects of probability, such as when demonstrating pairwise independence.
Pairwise Independence
In probability theory, understanding independence is essential. Two events are independent if the occurrence of one does not affect the probability of the other occurring. Pairwise independence refers to this concept within a set, where any two of the events are independent of each other.
  • Pairwise vs. Mutual Independence: It's crucial to note that pairwise independence does not imply mutual independence, where all events in a set are independent.
  • Verification: For two variables \(Z_{r_1}\) and \(Z_{r_2}\), being pairwise independent means \(P(Z_{r_1} = a, Z_{r_2} = b) = P(Z_{r_1} = a)P(Z_{r_2} = b)\).
In the exercise, the fact that n is a prime number helps to ensure pairwise independence of \(Z_r\) variables, due to the modular arithmetic properties of prime numbers.
Modular Arithmetic
Modular arithmetic is often referred to as "clock arithmetic" because it deals with integers and a specific modulus, similar to how hours reset after 12 on a clock. It is written using the notation \(a \equiv b \mod n\), meaning \(a\) and \(b\) give the same remainder when divided by \(n\).
  • Basic Operation: If \(a \equiv b \mod n\), then \(a\) and \(b\) have the same remainder when divided by \(n\).
  • Use with Prime Numbers: When \(n\) is prime, modular arithmetic has unique solutions, which simplifies solving equations like \((r_1 - r_2)y \equiv a - b \mod n\), ensuring only one solution for \(y\).
In the context of the exercise, modular arithmetic is used to define the random variables \(Z_r\) and is crucial in demonstrating their pairwise independence.

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Most popular questions from this chapter

Let \(\mathbf{R}^{2}=\mathbf{R} \times \mathbf{R}\), and let \(\mathcal{B}^{2}\) be the Borel sets of \(\mathbf{R}^{2}\), while \(\mathcal{B}\) denotes the Borel sets of \(\mathbf{R}\). Show that \(\mathcal{B}^{2}=\mathcal{B} \otimes \mathcal{B}\).

Toss a coin with \(P\) (Heads) \(=p\) repeatedly. Let \(A_{k}\) be the event that \(k\) or more consecutive heads occurs amongst the tosses numbered \(2^{k}, 2^{k}+\) \(1, \ldots, 2^{k+1}-1 .\) Show that \(P\left(A_{k}\right.\) i.o. \()=1\) if \(p \geq \frac{1}{2}\) and \(P\left(A_{k}\right.\) i.o. \()=0\) if \(p<\frac{1}{2} .\)

Let \(X, Y\) be independent and suppose \(P(X+Y=\alpha)=1\), where \(\alpha\) is a constant. Show that both \(X\) and \(Y\) are constant random variables.

Construct an example to show that \(E\\{X Y\\}=E\\{X\\} E\\{Y\\}\) does not imply in general that \(X\) and \(Y\) are independent r.v.'s (we assume \(X, Y\) and \(X Y\) are all in \(L^{1}\) ).

Let \(X, Y\) be independent random variables taking values in \(\mathbf{N}\) with $$ P(X=i)=P(Y=i)=\frac{1}{2^{i}} \quad(i=1,2, \ldots) $$ Find the following probabilities: a) \(P(\min (X, Y) \leq i) \quad\left[\right.\) Ans.: \(\left.1-\frac{1}{4^{4}}\right]\) b) \(P(X=Y)\) [Ans.: \(\frac{1}{3}\) ] c) \(P(Y>X) \quad\left[\right.\) Ans.: \(\left.\sum_{i \geq 0}^{3} \frac{1}{2^{2}\left(2^{i}-1\right)}\right]\) d) \(P(X\) divides \(Y)\) [Ans,: \(\frac{1}{3}\) ] e) \(P(X \geq k Y)\) for a given positive integer \(k\) [Ans,: \(\frac{1}{2^{1+k}-1}\) ]
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