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GATE 2015 EC – Question 62

Communications · Digital Communications · 2 marks · Multiple choice

A source emits bit 0 with probability $\frac{1}{3}$ and bit 1 with probability $\frac{2}{3}$. The emitted bits are communicated to the receiver. The receiver decides for either 0 or 1 based on the received value $R$. It is given that the conditional density functions of $R$ are as

$$f_{R|0}(r) = \begin{cases} \frac{1}{4}, & -3 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases} \text{ and } f_{R|1}(r) = \begin{cases} \frac{1}{6}, & -1 \leq x \leq 5 \\ 0, & \text{otherwise.} \end{cases}$$

The minimum decision error probability is

  1. 0
  2. 1/12
  3. 1/9
  4. 1/6

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Correct answer: (D) 1/6

Explanation

Outside the overlap $[-1, 1]$ only one bit is possible, so there is no error. In the overlap, compare the weighted densities: $\frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$ for bit 0 and $\frac{2}{3} \times \frac{1}{6} = \frac{1}{9}$ for bit 1. Bit 1 is larger, so the receiver decides 1 in the overlap. An error occurs when bit 0 was sent and $R$ lies in the overlap, which has probability $\frac{1}{3} \times \frac{2}{4} = \frac{1}{6}$.