Consider the binary operation ⊕on (-π, π] defined by a ⊕b=a+b-2 π⌈(a+b-π) /(2 π)⌉. In other word

Consider the binary operation ⊕on (-π, π] defined by a...

Consider the binary operation $\oplus$ on $(-\pi, \pi]$ defined by $$ a \oplus b=a+b-2 \pi\lceil(a+b-\pi) /(2 \pi)\rceil . $$ In other words, $a \oplus b$ is obtained by adding $a$ and $b$ and then adding or subtracting the appropriate multiple of $2 \pi$ so that the result lies in $(-\pi, \pi]$. Corresponding to this operation calculate $Q * R$, where $Q$ assigns probability $\frac{1}{4}$ to each member of $\left\{-\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi\right\}$ and $R$ assigns probability $\frac{1}{3}$ to each member of $\left\{-\frac{2 \pi}{3}, 0, \frac{2 \pi}{3}\right\}$.

Consider the binary operation $\oplus$ on $(-\pi, \pi]$ defined by
$$
a \oplus b=a+b-2 \pi\lceil(a+b-\pi) /(2 \pi)\rceil .
$$
In other words, $a \oplus b$ is obtained by adding $a$ and $b$ and then adding or subtracting the appropriate multiple of $2 \pi$ so that the result lies in $(-\pi, \pi]$. Corresponding to this operation calculate $Q * R$, where $Q$ assigns probability $\frac{1}{4}$ to each member of $\left\{-\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi\right\}$ and $R$ assigns probability $\frac{1}{3}$ to each member of $\left\{-\frac{2 \pi}{3}, 0, \frac{2 \pi}{3}\right\}$.

Key Concepts

Modular Arithmetic

Modular arithmetic deals with numbers where values wrap around upon reaching a certain value, called the modulus. In this context, the operation adjusts sums by adding or subtracting multiples of 2? so that the results fall within a specified interval. This process of 'wrapping around' is fundamental in situations where the domain is circular, such as angles, and ensures that every result is within a fixed range.

Circular Group

A circular group, or the group of angles, is formed by considering angles with addition modulo 2?. The binary operation defined in the problem is an example of such an operation, where the usual addition is modified by a modulus (2?) to keep the result within a bounded interval. This structure is central to the study of rotations and periodic phenomena in mathematics.

Convolution of Probability Measures

Convolution of probability measures on a group is a method of combining two probability distributions defined on the group into a new distribution. When the underlying operation of the group (in this case, a circular addition modulo 2?) is used to 'add' independent random variables, the resulting distribution is the convolution of the original measures. This concept is key in probability theory and harmonic analysis on groups.

Discrete Probability Distributions

Discrete probability distributions are measures that assign probabilities to a finite or countably infinite set of outcomes. In the provided problem, the measures Q and R assign specific probabilities to finite sets of angles. Understanding how these discrete distributions behave under operations like convolution is crucial for solving problems in probability on algebraic structures such as groups.

Transcript

00:01 Here we are given two random variables x and y, and we know they're independent, and they actually have the same distribution, the bernoulli distribution with parameter 1 over 2.

00:14 That means x is equal to 0 and 1 with probability 1 over 2.

00:20 The same for y, y is equal to 1 and 0 with probability 1 over 2.

00:28 Define a random variable u, which is defined as x plus y, and v is defined as absolute value of x minus y.

00:37 We want to find the, let's begin with the marginal distribution for u and v, and then combine those two distributions, and we'll do some discussion about their joint distribution.

00:58 By those two facts, we know that u can take value 0, 1, and 2.

01:11 0, 1, 2, there's the probability.

01:17 When u is equal to 0, that means both y and x takes a value of 0.

01:26 As they're independent, we know the probability of x being 0, y being 0 is equal to the product of x being 0 times the probability of y being 0, which is equal to 1 over 4.

01:46 The same thing for when u is equal to 2, that means both x and y take the value 1.

01:54 So, use the independence, we have 1 over 4.

01:58 What about 1? when u is equal to 1, we have two situations.

02:03 First, x is equal to 0 while y is equal to 1, or x is equal to 1, but y is equal to 0.

02:13 Use the independence, this happens with probability 1 .25, the same for the second thing.

02:22 Combining those two cases, we have 1 over 2.

02:27 Okay.

02:28 Now, let's do the same discussion for v.

02:32 V represents the difference between x and y, and notice from the values of x and y, we know v can take value 0 and 1.

02:44 Okay.

02:48 When v is equal to 0, that means x is equal to y.

02:51 So, we have two cases.