GATE 2022 EC Q.64: solution and answer

Official answer: 1.15 to 1.18Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

1. The total area under the pdf must be 1. The triangle from $x=-2$ to $0$ has area $\frac{1}{2}\cdot 2\cdot h=h$, and the rectangle from $0$ to $1$ has area $1\cdot h=h$. So $2h=1$ and $h=\frac{1}{2}$.

2. The pdf is $f(x)=\frac{x+2}{4}$ for $-2\le x\le 0$ (it rises linearly from $0$ to $\frac{1}{2}$), and $f(x)=\frac{1}{2}$ for $0<x\le 1$. It is zero elsewhere.

3. The quantizer has two cells: $X>0$ (output $\alpha$) and $X\le 0$ (output $\beta$). For a fixed cell, the mean square error is smallest when the output level is the centroid (conditional mean) of that cell. So ${\alpha }^{\ast }=E[X\mid X>0]$ and ${\beta }^{\ast }=E[X\mid X\le 0]$.

4. Positive cell: $P(X>0)=\frac{1}{2}\cdot 1=\frac{1}{2}$. Here the pdf is constant, so $X$ is uniform on $(0,1]$ and ${\alpha }^{\ast }=\frac{1}{2}$.

5. Non-positive cell: $P(X\le 0)=\frac{1}{2}$ (the triangle's area). The first moment over this cell is $${\int }_{-2}^{0}x\frac{x+2}{4}dx=\frac{1}{4}{\left[\frac{x^3}{3}+x^2\right]}_{-2}^0=\frac{1}{4}\left(0-\left(-\frac{8}{3}+4\right)\right)=-\frac{1}{3}.$$

6. So ${\beta }^{\ast }=\frac{-1/3}{1/2}=-\frac{2}{3}$.

7. ${\alpha }^{\ast }-{\beta }^{\ast }=\frac{1}{2}+\frac{2}{3}=\frac{7}{6}\approx 1.17$.