Question

If X and Y are independent Poisson variates, What is the distribution of \(\frac{X}{X+Y}\)?
1. Binomial distribution
2. Poisson distribution
3. Negative binomial distribution
4. Multinomial distribution

• A. Binomial distribution
• B. Poisson distribution
• C. Negative binomial distribution
• D. Multinomial distribution

Hint:

A classic theorem in mathematical statistics states:
"The conditional distribution of independent Poisson variables given their sum is always a Binomial distribution."

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question tests the conditional distribution properties of independent Poisson random variables.

Step 2: Detailed Explanation:
Let \( X \) and \( Y \) be two independent Poisson random variables with rate parameters \( \lambda_1 \) and \( \lambda_2 \) respectively:
\[ X \sim \text{Poisson}(\lambda_1) \quad \text{and} \quad Y \sim \text{Poisson}(\lambda_2) \] The sum of two independent Poisson variables also follows a Poisson distribution:
\[ W = X + Y \sim \text{Poisson}(\lambda_1 + \lambda_2) \] Let us find the conditional probability of \( X = x \) given that their sum is a fixed integer \( n \) (representing the distribution of the fraction \( \frac{X}{X+Y} \)):
\[ P(X = x \mid X + Y = n) = \frac{P(X = x \cap X + Y = n)}{P(X + Y = n)} \] Since \( X \) and \( Y \) are independent, we can rewrite the numerator:
\[ P(X = x \mid X + Y = n) = \frac{P(X = x) \cdot P(Y = n - x)}{P(X + Y = n)} \] Substitute the Poisson probability mass functions into the expression:
\[ P(X = x \mid X + Y = n) = \frac{\left( \frac{e^{-\lambda_1} \lambda_1^x}{x!} \right) \cdot \left( \frac{e^{-\lambda_2} \lambda_2^{n-x}}{(n-x)!} \right)}{\frac{e^{-(\lambda_1 + \lambda_2)} (\lambda_1 + \lambda_2)^n}{n!}} \] Rearranging terms:
\[ P(X = x \mid X + Y = n) = \frac{n!}{x! (n-x)!} \cdot \frac{\lambda_1^x \lambda_2^{n-x}}{(\lambda_1 + \lambda_2)^n} \] \[ P(X = x \mid X + Y = n) = \binom{n}{x} \left( \frac{\lambda_1}{\lambda_1 + \lambda_2} \right)^x \left( \frac{\lambda_2}{\lambda_1 + \lambda_2} \right)^{n-x} \] This is the probability mass function of a Binomial distribution with parameters:
\[ n = X + Y \quad \text{and} \quad p = \frac{\lambda_1}{\lambda_1 + \lambda_2} \] Therefore, the conditional distribution representing the fraction \( \frac{X}{X+Y} \) is the Binomial distribution.

Step 3: Final Answer:
The distribution of the variable is the Binomial distribution.
Therefore, the correct choice is Option (A).