Question

Let \( X \) be a continuous random variable with probability density function \[ f(x) = \frac{1}{\sigma x \sqrt{2\pi}} \exp\left( -\frac{1}{2} \left( \frac{\log x - \mu}{\sigma} \right)^2 \right), \quad x>0, \] where \( \mu \in \mathbb{R}, \sigma>0 \). If \( \log_e \left( \frac{E(X^2)}{(E(X))^2} \right) = 4 \), then \( {Var}(\log_e X) \) equals:

• A. 2
• B. 4
• C. 16
• D. 64

Hint:

For log-normal distributions, the variance of the logarithm of the variable is just \( \sigma^2 \), the variance of the underlying normal distribution.

Answer 1

The Correct Option is B

Step 1: Recognize the distribution.
The given probability density function suggests that \( X \) follows a log-normal distribution. For a log-normal random variable \( X \), if \( \log_e X \sim N(\mu, \sigma^2) \), then:
\( E(X) = \exp(\mu + \frac{\sigma^2}{2}) \),
\( E(X^2) = \exp(2\mu + 2\sigma^2) \),
\( {Var}(X) = \exp(2\mu + 2\sigma^2) - \exp(2\mu + \sigma^2) \).
Step 2: Analyze the given condition.
We are given \( \log_e \left( \frac{E(X^2)}{(E(X))^2} \right) = 4 \). Let's first compute \( \frac{E(X^2)}{(E(X))^2} \): \[ \frac{E(X^2)}{(E(X))^2} = \frac{\exp(2\mu + 2\sigma^2)}{\left( \exp(\mu + \frac{\sigma^2}{2}) \right)^2} = \frac{\exp(2\mu + 2\sigma^2)}{\exp(2\mu + \sigma^2)} = \exp(\sigma^2). \] Thus, \( \log_e (\exp(\sigma^2)) = \sigma^2 \). Given that this equals 4, we have \( \sigma^2 = 4 \).
Step 3: Find \( {Var}(\log_e X) \).
Since \( \log_e X \sim N(\mu, \sigma^2) \), we know that the variance of \( \log_e X \) is simply \( \sigma^2 \). Therefore, \( {Var}(\log_e X) = 4 \).
Thus, the answer is \( \boxed{4} \).