Question

Let \( (X, Y) \) have joint probability mass function \[ p(x, y) = \begin{cases} \frac{e^{-2}}{x! (y - x)!} & \text{if } x = 0, 1, 2, \dots, y; \, y = 0, 1, 2, \dots \\ 0 & \text{otherwise}. \end{cases} \] Then which of the following statements is/are true?

• A. \( E(X | Y = 4) = 2 \)
• B. The moment generating function of \( Y \) is \( e^2 (e^{v-1}) \) for all \( v \in \mathbb{R} \)
• C. \( E(X) = 2 \)
• D. The joint moment generating function of \( (X, Y) \) is \( e^{-2} (1 + e^u) e^v \) for all \( (u, v) \in \mathbb{R}^2 \)

Hint:

- The MGF for a Poisson-like distribution can be derived based on its moment properties.
- For conditional expectations, the distribution of \( X \) given \( Y = y \) can often be modeled as binomial or other distributions depending on the problem structure.

Answer 1

The Correct Option is A

1) Understanding the joint probability mass function:
The given joint probability mass function (PMF) is for a distribution where \( X \) and \( Y \) are dependent. This distribution resembles a truncated Poisson distribution. We will now analyze each of the options based on this setup.
2) Analysis of the options:
(A) Correct: \( E(X | Y = 4) = 2 \).
To calculate \( E(X | Y = 4) \), we use the fact that given \( Y = 4 \), the distribution of \( X \) is binomial with parameters \( n = 4 \) and \( p = \frac{1}{2} \). The expected value of a binomial distribution is \( n . p = 4 . \frac{1}{2} = 2 \).
(B) Correct: The moment generating function of \( Y \) is \( e^2 (e^{v-1}) \) for all \( v \in \mathbb{R} \).
The moment generating function (MGF) of \( Y \) can be computed from the given PMF. The form of the MGF for a Poisson-like distribution leads to the result \( e^2 (e^{v-1}) \).
(C) Incorrect: \( E(X) = 2 \).
The unconditional expectation \( E(X) \) is not simply 2. The distribution of \( X \) depends on \( Y \), and without conditioning on \( Y \), we cannot directly conclude that \( E(X) = 2 \).
(D) Correct: The joint moment generating function of \( (X, Y) \) is \( e^{-2} (1 + e^u) e^v \) for all \( (u, v) \in \mathbb{R}^2 \).
The joint MGF can be derived from the given PMF, and the resulting formula matches the one provided in this option.
The correct answers are (A), (B), and (D).