Question

Let random variables \(X_1, X_2, \dots, X_k\) follows a multinomial distribution with parameter n and \(p_i\) where i= 1, 2,...,k. what is the value of Cov(\(X_i, X_j\))?
1. \(np_ip_j\)
2. \(-np_ip_j\)
3. \(p_ip_j\)
4. \(-p_ip_j\)

• A. \(np_ip_j\)
• B. \(-np_ip_j\)
• C. \(p_ip_j\)
• D. \(-p_ip_j\)

Hint:

Since a trial can only yield one outcome, an increase in \( X_i \) must decrease the chances for \( X_j \).
This inverse relationship means the covariance between different categories in a multinomial distribution must always be negative, immediately eliminating positive options.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question asks for a standard properties of the Multinomial Distribution, specifically the covariance between two different outcome categories \( X_i \) and \( X_j \) (\( i \neq j \)).

Step 2: Detailed Explanation:
A multinomial distribution model generalizes the binomial distribution.
Suppose we run a trial with \( n \) independent repetitions. Each trial can result in one of \( k \) possible mutually exclusive outcomes, with corresponding probabilities \( p_1, p_2, \dots, p_k \) such that:
\[ \sum_{i=1}^k p_i = 1 \] Let \( X_i \) represent the number of times outcome \( i \) occurs across the \( n \) trials.
Each individual variable \( X_i \) follows a marginal binomial distribution:
\[ X_i \sim \text{Binomial}(n, p_i) \implies \text{Var}(X_i) = n p_i (1 - p_i) \] Because the categories are mutually exclusive, an increase in the count of one category (\( X_i \)) reduces the remaining trials available for the other categories, creating an inverse relationship between any two classes \( X_i \) and \( X_j \) (where \( i \neq j \)).
Mathematically, the covariance between these two variables is negative and is given by the formula:
\[ \text{Cov}(X_i, X_j) = -n p_i p_j \quad \text{for } i \neq j \]

Step 3: Final Answer:
The value of \( \text{Cov}(X_i, X_j) \) is \( -n p_i p_j \).
Therefore, the correct choice is Option (B).