Question

Let \( X_1 \) and \( X_2 \) be i.i.d. continuous random variables with the probability density function 

Using Chebyshev's inequality, the lower bound of \( P \left( |X_1 + X_2 - 1| \leq \frac{1}{2} \right) \) is 
 

• A. \( \frac{5}{6} \)
• B. \( \frac{4}{5} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{1}{3} \)

Hint:

Chebyshev’s inequality gives a bound on probabilities, useful when you don't know the exact distribution of the random variable.

Answer 1

The Correct Option is C

Step 1: Applying Chebyshev's inequality.
Chebyshev's inequality states that: \[ P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2} \] Here, we need to find \( P \left( |X_1 + X_2 - 1| \leq \frac{1}{2} \right) \). This involves calculating the mean and variance of \( X_1 + X_2 \) and applying Chebyshev’s inequality.
Step 2: Calculating the variance.
The mean and variance of \( X_1 + X_2 \) are calculated based on the individual distribution of \( X_1 \) and \( X_2 \). The resulting probability is \( \frac{3}{5} \).