Problem 1 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X\) and \(Y\) be independent random variables with a discrete uniform distribution, i.e., with probability mass functions $$ p_{X}(k)=p_{Y}(k)=\frac{1}{N}, \quad \text { for } k=1, \ldots, N . $$ Use the addition rule for discrete random variables on page 152 to determine the probability mass function of \(Z=X+Y\) for the following two cases. a. Suppose \(N=6\), so that \(X\) and \(Y\) represent two throws with a die. Show that $$ p_{Z}(k)=\mathrm{P}(X+Y=k)= \begin{cases}\frac{k-1}{36} & \text { for } k=2, \ldots, 6 \\ \frac{13-k}{36} & \text { for } k=7, \ldots, 12\end{cases} $$ You may check this with Quick exercise 11.1. b. Determine the expression for \(p_{Z}(k)\) for general \(N\).

Step-by-Step Solution

Answer

The probability mass function of \(Z = X + Y\) for \(N=6\) is as given, verified by enumeration, and for general \(N\) is symmetric about \(N+1\).

Step 1: Understand the Problem

Two independent random variables, \(X\) and \(Y\), each represent a discrete uniform distribution. We are asked to find the probability mass function (p.m.f) of \(Z = X + Y\), for specific and general values of \(N\).

Step 2: Define the Probabilities

For each \(X\) and \(Y\), we have: \[ p_X(k) = p_Y(k) = \frac{1}{N}, \quad k = 1, \ldots, N. \] Since \(X\) and \(Y\) are independent, the joint probability is the product: \(p_{X,Y}(k_1,k_2) = p_X(k_1) \cdot p_Y(k_2) = \frac{1}{N^2}.\)

Step 3: Calculate p.m.f. for N=6

We need to calculate \(p_{Z}(k) = P(X + Y = k)\). To cover all possible sums that can occur when adding \(X\) and \(Y\), we identify that values range from 2 (if both \(X\) and \(Y\) are 1) to 12 (if both are 6). Then: - For \(k = 2, \ldots, 6\), \(p_{Z}(k) = \frac{(k-1)}{36}\) because there are \(k-1\) combinations to obtain \(k\).- For \(k = 7, \ldots, 12\), \(p_{Z}(k) = \frac{13-k}{36}\) as the number of combinations is \(13-k\). This forms a symmetric distribution around \(k=7\).

Step 4: Generalize the Expression for Any N

For a general \(N\), we calculate the number of ways to sum to a particular \(k\), which changes based on \(k\): - From \(k = 2\) to \(k = N\), there are \(k-1\) combinations.- From \(k = N+1\) to \(k = 2N\), symmetry implies there are \(2N-k+1\) combinations.- Thus, the general form: \[ p_{Z}(k) = \begin{cases} \frac{k-1}{N^2} & \text{for } k = 2, \ldots, N \ \frac{2N-k+1}{N^2} & \text{for } k = N+1, \ldots, 2N \end{cases}. \]

Step 5: Verify Solution with Example

For Quick Exercise 11.1, if needed, plug in small values of \(N\) and verify the counts through enumeration or simulation to confirm the derived probabilities.

Key Concepts

Independent Random Variables

In probability, independent random variables are variables that do not influence each other's outcomes. The occurrence or outcome of one event does not affect the likelihood of another. For instance, if we roll a fair die twice, the result of the first roll has no impact on the second roll. Each roll represents an independent random variable.
For independent random variables, the joint probability of two events occurring is simply the product of their individual probabilities. Suppose variables \( X \) and \( Y \) are independent, each with a probability distribution. The joint probability can be expressed as \( p_{X,Y}(a, b) = p_X(a) \times p_Y(b) \). This concept is essential when dealing with discrete uniform distributions like in the die-rolling scenario, where each result is equally likely and independent of others.

Probability Mass Function

A probability mass function (p.m.f.) is a function that gives the probability of a discrete random variable being exactly equal to some value. It's a crucial concept in understanding how probabilities are distributed across different outcomes for discrete variables.
For our dice example, where \( X \) and \( Y \) each roll a number between 1 and \( N \), the p.m.f. for each is \( p_X(k) = \frac{1}{N} \), meaning each outcome is equally likely with probability \( \frac{1}{N} \).
When considering the sum \( Z = X + Y \), we analyze the sums that these dice rolls can result in. The resulting p.m.f. \( p_Z(k) = P(X + Y = k) \) provides the probability for each potential sum. Calculating this for various \(k\) involves summing over the probabilities of all pairs \((x, y)\) that result in the particular sum \(k\).

Addition Rule

The addition rule in probability allows us to determine the probability of the sum of two independent random variables. This rule applies when calculating the p.m.f. of \( Z = X + Y \).
The expectation is to find the distribution of \( Z \) by considering each possible sum. Using dice as an example, for each potential outcome \( k \), we need to count the combinations of \( X \) and \( Y \) that yield \( k \). For sums ranging from 2 to \( N \), the pattern follows \( k - 1 \) combinations. Conversely, for sums from \( N+1 \) to \( 2N \), we look at \( 2N-k+1 \) combinations. This systematic approach ensures every possible value is accounted for, leading to a complete p.m.f.
The rule highlights the symmetry in the distribution of outcomes, particularly for uniform discrete variables. Recognizing this symmetry simplifies the determination of probabilities, aiding in analyses and predictions for such systems.