GATE 2018 EC Q.40: solution and answer

Official answer: -0.5Source: Official answer key, GATE 2018 EC (IIT Guwahati)

1. The noise density is $f_Z(z)=\frac{1}{2}$ for $-1<z<1$ and zero elsewhere. Given $X=-0.5$, $Y=X+Z$ is uniform on $(-1.5,0.5)$ with density $\frac{1}{2}$. Given $X=0.5$, $Y$ is uniform on $(-0.5,1.5)$ with density $\frac{1}{2}$.

2. The MAP rule decides $\hat{X}=0.5$ when $P(X=0.5)f(y\mid 0.5)\ge P(X=-0.5)f(y\mid -0.5)$. Otherwise it decides $\hat{X}=-0.5$.

3. For $y<-0.5$ we have $f(y\mid 0.5)=0$ and $f(y\mid -0.5)=\frac{1}{2}$. Only $X=-0.5$ can have produced $y$, so decide $\hat{X}=-0.5$.

4. For $y>0.5$ we have $f(y\mid -0.5)=0$ and $f(y\mid 0.5)=\frac{1}{2}$. Only $X=0.5$ can have produced $y$, so decide $\hat{X}=0.5$.

5. For $-0.5\le y\le 0.5$ both likelihoods equal $\frac{1}{2}$, so the priors decide. Compare $\frac{3}{4}\cdot \frac{1}{2}=\frac{3}{8}$ with $\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}$. Since $\frac{3}{8}>\frac{1}{8}$, decide $\hat{X}=0.5$.

6. So $\hat{X}=-0.5$ for $Y<-0.5$ and $\hat{X}=0.5$ for $Y\ge -0.5$. Matching this to the form $\hat{X}=-0.5$ for $Y<\alpha$ and $\hat{X}=0.5$ for $Y\ge \alpha$ gives $$\alpha =-0.5$$