Step 1: Concept Overview:
The problem requires finding the expected value, \(E(X)\), of a discrete random variable \(X\), given its probability mass function (PMF) which includes a constant \(C\). The solution involves directly applying the definition of expected value. Solving for \(C\) is unnecessary as the answer choices are in terms of \(C\).
Step 2: Core Formula:
The expected value for a discrete random variable is:
\[ E(X) = \sum_{x} x . P(X=x) \]
Here, \( P(X=x) = f(x) = \frac{C\theta^x}{x} \).
Step 3: Step-by-Step Solution:
Apply the expected value formula to the given PMF, summing over all possible \(x\) values, from \(x=1\) to infinity.
\[ E(X) = \sum_{x=1}^{\infty} x . f(x) \]
\[ E(X) = \sum_{x=1}^{\infty} x . \left(\frac{C\theta^x}{x}\right) \]
Simplify by canceling \(x\):
\[ E(X) = \sum_{x=1}^{\infty} C\theta^x \]
Factor out the constant \(C\):
\[ E(X) = C \sum_{x=1}^{\infty} \theta^x \]
The summation is a geometric series with initial term \(a = \theta\) and ratio \(r = \theta\). Given \(0<\theta<1\), the series converges. The sum is given by \( S = \frac{a}{1-r} \).
\[ \sum_{x=1}^{\infty} \theta^x = \frac{\theta}{1-\theta} \]
Substitute this back into the \(E(X)\) equation:
\[ E(X) = C \left(\frac{\theta}{1-\theta}\right) = \frac{C\theta}{1-\theta} \]
Step 4: Solution:
The expected value \(E(X)\) is \( \frac{C\theta}{1-\theta} \).