Question

Let, \(X \sim \beta_1(u, v)\) and \(Y \sim \gamma(1, u+v)\); (\(u, v>0\)) be independent random variables. If, \(Z = XY\), then the moment generating function of Z is given by

• A. \( \left(1-\frac{t}{v}\right)^{-u} \)
• B. \( (1-t)^{-v} \)
• C. \( (1-t)^{-u} \)
• D. \( \left(1-\frac{t}{u}\right)^{-v} \)

Hint:

Recognizing special relationships between distributions is a key skill. The identity that a Beta-distributed variable "selects" the shape parameter for a new Gamma distribution from the sum of the shape parameters is powerful. If a question seems complex, check if it fits a known statistical theorem.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The question asks for the moment generating function (MGF) of a product of two independent random variables, one following a Beta distribution and the other a Gamma distribution. This requires knowledge of a specific identity in probability theory relating these distributions. The notation \(\gamma(1, u+v)\) is ambiguous and could mean Gamma(shape, scale) or Gamma(shape, rate). However, a known theorem works if we interpret the parameters correctly.

Step 2: Key Formula or Approach:
There is a fundamental theorem in probability that states: If \( X \sim \text{Beta}(u, v) \) and \( S \sim \text{Gamma}(u+v, \lambda) \) are independent, then their product \( Z = XS \) follows a Gamma distribution, specifically \( Z \sim \text{Gamma}(u, \lambda) \). The MGF of a Gamma distribution with shape parameter \(k\) and rate parameter \(\lambda\) is \( M_Z(t) = \left(\frac{\lambda}{\lambda-t}\right)^k = \left(1 - \frac{t}{\lambda}\right)^{-k} \).

Step 3: Detailed Explanation:
Let's analyze the given distributions: - \( X \sim \beta_1(u, v) \) is a Beta distribution with parameters \(u\) and \(v\). - \( Y \sim \gamma(1, u+v) \). The notation is ambiguous. For the theorem to apply, \(Y\) should be a Gamma distribution with shape \(u+v\). The provided notation is likely a typo for \( Y \sim \text{Gamma}(u+v, 1) \), i.e., shape=\(u+v\) and rate=\(1\). Let's proceed with this assumption, as it leads to one of the given answers. Assuming \( Y \sim \text{Gamma}(\text{shape}=u+v, \text{rate}=1) \). According to the theorem, with \( \lambda=1 \): If \( X \sim \text{Beta}(u, v) \) and \( Y \sim \text{Gamma}(u+v, 1) \) are independent, then \( Z = XY \sim \text{Gamma}(u, 1) \). Now, we find the MGF of \(Z\), which follows a Gamma distribution with shape \(k=u\) and rate \(\lambda=1\). Using the MGF formula for a Gamma distribution: \[ M_Z(t) = \left(1 - \frac{t}{\lambda}\right)^{-k} \] Substitute \( k=u \) and \( \lambda=1 \): \[ M_Z(t) = \left(1 - \frac{t}{1}\right)^{-u} = (1-t)^{-u} \] This result matches option (C). Any other interpretation of the notation for the Gamma distribution does not lead to a simple, standard result.
Step 4: Final Answer:
The moment generating function of Z is \( (1-t)^{-u} \).