Question

Let \( X \) and \( Y \) be discrete random variables with joint probability mass function \[ p_{X, Y}(m, n) = \frac{\lambda^n e^{-\lambda} 2^n m! (n - m)!}{n!}, \quad m = 0, \dots, n, \quad n = 0, 1, 2, \dots, \] where \( \lambda \) is a fixed positive real number. Then which one of the following options is correct?

• A. The marginal distribution of \( X \) is Poisson with mean \( \lambda \)
• B. The marginal distribution of \( Y \) is Poisson with mean \( 2\lambda \)
• C. The conditional distribution of \( X \) given \( Y = 3 \) is \( {Bin}(3, \frac{1}{2}) \)
• D. \( E(Y | X = 2) = \frac{\lambda}{2} \)

Hint:

In joint distributions, check whether the conditional distributions are binomial or Poisson, and use that to identify relationships between the variables.

Answer 1

The Correct Option is C

Step 1: Understand the joint distribution.
The given joint probability mass function is: \[ p_{X, Y}(m, n) = \frac{\lambda^n e^{-\lambda} 2^n m! (n - m)!}{n!}, \quad m = 0, \dots, n, \quad n = 0, 1, 2, \dots. \] This represents a compound distribution where \( X \) is conditionally distributed as a binomial random variable given \( Y = n \), with parameters \( n \) and \( p = \frac{1}{2} \). The marginal distribution of \( Y \) is Poisson with mean \( 2\lambda \), and the conditional distribution of \( X \) given \( Y = n \) is \( {Binomial}(n, \frac{1}{2}) \).
Step 2: Analyze the options.
Option (A): The marginal distribution of \( X \) is not Poisson with mean \( \lambda \); it's a compound distribution.
Option (B): The marginal distribution of \( Y \) is indeed Poisson with mean \( 2\lambda \), but this is not the correct option for the question.
Option (C): Given \( Y = 3 \), the conditional distribution of \( X \) is \( {Binomial}(3, \frac{1}{2}) \). This is the correct option.
Option (D): The expected value \( E(Y | X = 2) \) does not equal \( \frac{\lambda}{2} \), so this option is incorrect.
Thus, the correct answer is \( \boxed{(C)} \).