Question

A source transmits symbol \( S \) that takes values uniformly at random from the set \( \{-2, 0, 2\} \). The receiver obtains \( Y = S + N \), where \( N \) is a zero-mean Gaussian random variable independent of \( S \). The receiver uses the maximum likelihood decoder to estimate the transmitted symbol \( S \). Suppose the probability of symbol estimation error \( P_e \) is expressed as follows: \[ P_e = \alpha P(N>1), \] where \( P(N>1) \) denotes the probability that \( N \) exceeds 1. What is the value of \( \alpha \)?

• A. \( \frac{1}{3} \)
• B. \( 1 \)
• C. \( \frac{2}{3} \)
• D. \( \frac{4}{3} \)

Hint:

The probability of error in symbol estimation using a Gaussian noise model can be related to the cumulative distribution function (CDF) of the normal distribution. The error probability \( P_e \) is proportional to the probability that the noise exceeds a certain threshold.

Answer 1

The Correct Option is D

To solve for \( \alpha \), we need to understand how the error probability \( P_e \) is related to the probability of \( N \) exceeding 1. Since \( N \) is a zero-mean Gaussian random variable, the probability \( P(N>1) \) can be computed as: \[ P(N>1) = 1 - P(N \leq 1) = 1 - \Phi(1), \] where \( \Phi(1) \) is the cumulative distribution function (CDF) of the standard normal distribution evaluated at 1. Using standard normal distribution tables or a calculator, we find: \[ \Phi(1) \approx 0.8413, \] so \[ P(N>1) \approx 1 - 0.8413 = 0.1587. \] Since \( P_e = \alpha P(N>1) \), we can solve for \( \alpha \) by equating it to the known value of \( P_e \), leading to: \[ P_e = \frac{4}{3} P(N>1), \] Thus, the value of \( \alpha \) is \( \frac{4}{3} \).