Problem 18 - A Computational Introduction to Number Theory and Algebra

Question

Suppose \(X\) and \(Y\) are random variables, each uniformly distributed over \(\mathbb{Z}_{2},\) but not necessarily independent. Show that the distribution of \((X, Y)\) is the same as the distribution of \((X+1, Y+1)\).

Step-by-Step Solution

Answer

Question: Show that the distribution of (X, Y) is the same as the distribution of (X+1, Y+1), where X and Y are random variables uniformly distributed over ℤ₂. Answer: The joint probability distributions of (X, Y) and (X+1, Y+1) are the same, which implies that their distributions are the same.

Step 1: Find the possible values for \(X\) and \(Y\)

Since \(X\) and \(Y\) are uniformly distributed over \(\mathbb{Z}_{2}\), they can take two possible integer values, either 0 or 1.

Step 2: Find the joint probability distribution of \((X, Y)\)

Let \(A = \{0,1\}\). Then \(p_{X,Y}(x, y) = P(X = x, Y = y)\) for all \((x, y) \in A \times A\). Since there are 4 possible pairs of values \((x,y)\), the joint probability distribution of \((X, Y)\) can be represented as: $$p_{X,Y}(0, 0), p_{X,Y}(0, 1), p_{X,Y}(1, 0), p_{X,Y}(1, 1)$$

Step 3: Find the joint probability distribution of \((X+1, Y+1)\)

The joint probability distribution of \((X+1, Y+1)\) is represented as: $$p_{X,Y}(x+1, y+1) = P(X+1 = x+1, Y+1 = y+1)$$ for all \((x, y) \in A \times A\). Adding 1 to the respective values of \(x\) and \(y\), we have: $$p_{X,Y}(1, 1), p_{X,Y}(1, 2), p_{X,Y}(2, 1), p_{X,Y}(2, 2)$$. However, since the values of \(X+1\) and \(Y+1\) are also in \(\mathbb{Z}_{2},\) any values of 2 should be replaced by 0. Therefore, we have: $$p_{X,Y}(1, 1), p_{X,Y}(1, 0), p_{X,Y}(0, 1), p_{X,Y}(0, 0)$$.

Step 4: Compare the joint probability distributions

Comparing the joint probability distributions of \((X, Y)\) and \((X+1, Y+1)\), we can see that: $$p_{X,Y}(0, 0) = p_{X,Y}(0, 0), p_{X,Y}(0, 1) = p_{X,Y}(1, 0), p_{X,Y}(1, 0) = p_{X,Y}(0, 1), p_{X,Y}(1, 1) = p_{X,Y}(1, 1).$$ All the joint probability distributions have the same value for both \((X, Y)\) and \((X+1, Y+1)\). Therefore, we can conclude that the distribution of \((X, Y)\) is the same as the distribution of \((X+1, Y+1)\).

Key Concepts

Random Variables

In probability theory, a random variable is essentially a way to assign numerical values to different outcomes of a random process. Instead of unpredictable outcomes, we now have a numeric value we can analyze.
Take for example the roll of a die. While each result tells us a face-value number, in probability, we would describe this with a random variable capturing the possibility of getting any face from 1 to 6.

In our earlier problem, we see the use of two random variables, \(X\) and \(Y\). Each of these random variables can take on values from a specific set, in this case, \(\mathbb{Z}_{2}\). This set consists of only two numbers: 0 and 1.

• \(X\): Represents a random process outcome which can be either 0 or 1.
• \(Y\): Similar to \(X\), can also be 0 or 1.

When dealing with multiple random variables, it's crucial to understand whether they operate independently or not. But both variables are defined over the same set and individually hold an equal chance of being either number.

Uniform Distribution

Uniform distribution is a fundamental concept in probability, where each outcome within a certain range is equally likely. Rather than having skewed chances of different results, a uniform distribution ensures each possible value has the same probability.

In our problem, both \(X\) and \(Y\) follow a uniform distribution over \(\mathbb{Z}_{2}\).

• What this means is: Both random variables can be 0 or 1 with an equal probability of 0.5.
• Neither value is favored over another; there's a balanced chance.

The principle behind this is simple but powerful, allowing for a clean slate in probability calculations. If you were to pick one number at random from \(\mathbb{Z}_{2}\), it’s as if you're flipping a coin, each side representing one number.
This uniform approach simplifies calculations since the structure doesn't skew toward any particular outcome.

Joint Probability Distribution

When studying two random variables together, we enter the realm of joint probability distributions. Here, we examine the probability of different combinations of outcomes from two random variables.

In the initial exercise, we were asked to look at joint distributions of \((X, Y)\) and \((X+1, Y+1)\).

• A joint probability distribution details probabilities for every combination of \(X\) and \(Y\). Given they are over \(\mathbb{Z}_{2}\), there are four possible outcome pairs: \((0,0), (0,1), (1,0), (1,1)\).
• Each pair has a probability based on the chance occurrences of \(X\) and \(Y\).

In the solution, we observed that altering \(X\) and \(Y\) with the addition of 1 doesn't change the overall distribution of outcomes. This demonstrates an important concept: operations performed on uniformly distributed variables can preserve homogeneity under applicable conditions.