GATE 2018 EC – Question 50

Communications · Digital Communications · 2 marks · Numerical answer

A random variable $X$ takes values $-0.5$ and $0.5$ with probabilities $\frac14$ and $\frac34$, respectively. The noisy observation of $X$ is $Y=X+Z$, where $Z$ has uniform probability density over the interval $(-1,1)$. $X$ and $Z$ are independent. If the MAP rule based detector outputs $\hat X$ as

$$\hat X=\begin{cases}-0.5,&Y<\alpha\\0.5,&Y\ge\alpha,\end{cases}$$

then the value of $\alpha$ (accurate to two decimal places) is ________.

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: -0.51 to -0.49

Explanation

For $X=-0.5$, $Y$ is uniform on $(-1.5,0.5)$ and for $X=0.5$ it is uniform on $(-0.5,1.5)$, each with density $\frac12$. For $Y<-0.5$ only $X=-0.5$ is possible, and for $Y>0.5$ only $X=0.5$. In the overlap $(-0.5,0.5)$ the likelihoods are equal, so the prior $\frac34>\frac14$ makes the MAP decision $0.5$. Hence the threshold is $\alpha=-0.5$.