Question

Let \( X \) be a continuous random variable with the probability density function \[ f(x) = \frac{e^x}{(1 + e^x)^2}, \quad -\infty<x<\infty \] Then \( E(X) \) and \( P(X>1) \), respectively, are

• A. 1 and \( (1 + e)^{-1} \)
• B. 0 and \( 2(1 + e)^{-2} \)
• C. 2 and \( (2 + 2e)^{-1} \)
• D. 0 and \( (1 + e)^{-1} \)

Hint:

For continuous random variables, the expected value is calculated as the integral of \( x \) multiplied by the PDF over the range of the variable.

Answer 1

The Correct Option is D

Step 1: Calculate \( E(X) \).
The expected value of \( X \), \( E(X) \), is given by: \[ E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx \] For the given probability density function (PDF), we calculate \( E(X) \) using integration. After performing the integration, we find that \( E(X) = 0 \). Step 2: Calculate \( P(X>1) \).
To find \( P(X>1) \), we use the PDF and integrate: \[ P(X>1) = \int_{1}^{\infty} \frac{e^x}{(1 + e^x)^2} \, dx \] After calculating this integral, we get \( P(X>1) = (1 + e)^{-1} \). Step 3: Conclusion.
Thus, the correct answer is (D) 0 and \( (1 + e)^{-1 \)}.