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: $$ P_e = \alpha P(N1), $$ where $ P(N1) $ denotes the probability that $ N $ exceeds 1. What is the value of $ \alpha $?

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 $?

cdquestions Admin · Accepted Answer

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} $.

Answer

$ \frac{1}{3} $

Answer

Answer

$ \frac{2}{3} $

Answer

$ \frac{4}{3} $

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The Correct Option is D

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