Step 1: Concept Explanation:
The problem requires finding the moment generating function (MGF) of \(Z = XY\), where \(X\) follows a Beta distribution and \(Y\) follows a Gamma distribution, and they are independent. Solving this relies on a specific probability theorem. Note that \(\gamma(1, u+v)\) is potentially ambiguous, representing either Gamma(shape, scale) or Gamma(shape, rate). The solution hinges on the correct interpretation of these parameters and a known theorem.
Step 2: Core Theorem:
The relevant theorem 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: \( Z \sim \text{Gamma}(u, \lambda) \).
The MGF of a Gamma distribution with shape \(k\) and rate \(\lambda\) is given by \( M_Z(t) = \left(\frac{\lambda}{\lambda-t}\right)^k = \left(1 - \frac{t}{\lambda}\right)^{-k} \).
Step 3: Step-by-step Solution:
We have:
- \( X \sim \beta_1(u, v) \), a Beta distribution with parameters \(u\) and \(v\).
- \( Y \sim \gamma(1, u+v) \). The notation is assumed to mean \( Y \sim \text{Gamma}(u+v, 1) \), where shape=\(u+v\) and rate=\(1\). This assumption is crucial for applying the theorem and obtaining a matching answer choice.
Assuming \( Y \sim \text{Gamma}(\text{shape}=u+v, \text{rate}=1) \), i.e., \(\lambda = 1\):
Since \( X \sim \text{Beta}(u, v) \) and \( Y \sim \text{Gamma}(u+v, 1) \) are independent, then \( Z = XY \sim \text{Gamma}(u, 1) \) according to the theorem.
Now, determine the MGF of \(Z\), which is Gamma distributed with shape \(k=u\) and rate \(\lambda=1\).
Using the Gamma MGF formula:
\[ M_Z(t) = \left(1 - \frac{t}{\lambda}\right)^{-k} \]
Substitute \( k=u \) and \( \lambda=1 \):
\[ M_Z(t) = \left(1 - \frac{t}{1}\right)^{-u} = (1-t)^{-u} \]
This result aligns with option (C). Other interpretations of the Gamma distribution's notation do not yield a standard solution.
Step 4: Conclusion:
The moment generating function of Z is \( (1-t)^{-u} \).