GATE 2025 EC Q.41: solution and answer

GATE 2025 · ECQ.41 · 2 marks · MCQ

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 $P_{e} = α P (N > 1)$ , where $P (N > 1)$ denotes the probability that $N$ exceeds 1. What is the value of $α$ ?

(A)
$1/3$
(B)
$1$
(C)
$2/3$
(D)
$4/3$

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Answer and solution

ANSWER

$α = 4/3$ , option (D).

Official answer: DSource: Official answer key, GATE 2025 EC (IIT Roorkee)

The three symbols are equally likely and the noise is zero-mean Gaussian with the same variance for every symbol. The ML decoder is therefore the minimum-distance decoder: it picks the symbol closest to $Y$ .
The decision thresholds sit midway between neighbouring symbols, at $Y = - 1$ and $Y = + 1$ . Decide $- 2$ if $Y < - 1$ , decide $0$ if $- 1 < Y < 1$ , and decide $2$ if $Y > 1$ .
Let $p = P (N > 1)$ . Since $N$ is zero-mean Gaussian, its density is symmetric, so $P (N < - 1) = p$ as well.
Given $S = - 2$ : $Y = - 2 + N$ . The decoder goes wrong only if $Y > - 1$ , that is $N > 1$ . So $P (error ∣ S = - 2) = p$ .
Given $S = 2$ : $Y = 2 + N$ . The decoder goes wrong only if $Y < 1$ , that is $N < - 1$ . So $P (error ∣ S = 2) = p$ .
Given $S = 0$ : $Y = N$ . The decoder goes wrong if $Y > 1$ or $Y < - 1$ , that is $∣ N ∣ > 1$ . So $P (error ∣ S = 0) = P (N > 1) + P (N < - 1) = 2 p$ .
Average over the three equally likely symbols: $P_{e} = \frac{1}{3} (p + 2 p + p) = \frac{4}{3} p = \frac{4}{3} P (N > 1) .$
Comparing with $P_{e} = α P (N > 1)$ gives $α = \frac{4}{3}$ , which is option (D).

KEY IDEA

With equal priors and equal-variance Gaussian noise, ML decoding is nearest-symbol decoding, so the thresholds lie midway between symbols. The two outer symbols have only one neighbouring threshold and the middle symbol has two. Average the error probabilities over the symbols.

WATCH OUT FOR

Giving the middle symbol only one tail, which gives $P_{e} = p$ and $α = 1$ . The symbol $0$ is wrong when the noise exceeds $1$ in either direction, so its error probability is $2 p$ . Another slip is to multiply every symbol's error by $2$ and get $2 p$ overall.

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GATE 2025 EC Q.41: solution and answer

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