Question

If, \(f(X) = \frac{C\theta^x}{x}\); \(x = 1,2, \dots\); \(0<\theta<1\), then E(X) is equal to

• A. \( C\theta \)
• B. \( \frac{C\theta}{(1-\theta)} \)
• C. \( \frac{C}{(1-\theta)} \)
• D. C

Hint:

When a PMF is given, first check if you need to solve for the normalizing constant (C). If the options for E(X) or other moments include the constant, you can often proceed with the calculation directly without finding its specific value. Also, be quick to recognize standard mathematical series like the geometric series.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question asks for the expected value, E(X), of a discrete random variable X given its probability mass function (PMF). The PMF includes an unknown constant C. The key is to apply the definition of expected value directly. We do not need to solve for C because the options are given in terms of C.

Step 2: Key Formula or Approach:
The expected value of a discrete random variable X is defined as: \[ E(X) = \sum_{x} x . P(X=x) \] In this case, \( P(X=x) = f(x) = \frac{C\theta^x}{x} \).

Step 3: Detailed Explanation:
We apply the formula for E(X) to the given PMF. The sum is over all possible values of x, which are \(x=1, 2, 3, \dots\). \[ E(X) = \sum_{x=1}^{\infty} x . f(x) \] \[ E(X) = \sum_{x=1}^{\infty} x . \left(\frac{C\theta^x}{x}\right) \] The 'x' in the numerator and denominator cancels out: \[ E(X) = \sum_{x=1}^{\infty} C\theta^x \] We can factor out the constant C from the summation: \[ E(X) = C \sum_{x=1}^{\infty} \theta^x \] The summation is a geometric series with first term \(a = \theta\) and common ratio \(r = \theta\). Since \(0<\theta<1\), the series converges. The sum of this infinite geometric series is given by the formula \( S = \frac{a}{1-r} \). \[ \sum_{x=1}^{\infty} \theta^x = \frac{\theta}{1-\theta} \] Substituting this back into the expression for E(X): \[ E(X) = C \left(\frac{\theta}{1-\theta}\right) = \frac{C\theta}{1-\theta} \]
Step 4: Final Answer:
The expected value E(X) is equal to \( \frac{C\theta}{1-\theta} \).