Question

A real-valued source has PDF $f(x)$ as shown. 
A 1-bit quantizer maps positive samples to value $\alpha$ and others to $\beta$. 
If \(\alpha\) and \(\beta\)minimize the MSE, then \((\alpha^\ - \beta)\) is ____________ 
(rounded off to two decimal places). 
 

Hint:

For optimal quantizers, outputs equal conditional means of each region.

Answer 1

Correct Answer: 1.15

PDF description from the figure:
- Linear from $x=-2$ to $x=0$, rising to maximum.  
- Constant from $x=0$ to $x=1$. 
Compute probabilities: Total area must be 1. 
Area of triangle ($-2$ to $0$): \[ \frac{1}{2}(2)(1) = 1. \] 
Area of rectangle ($0$ to $1$): \[ 1 \times 1 = 1. \] 
Normalize PDF: divide by 2. 
Thus: \[ P(x<0) = \frac{1}{2}, \quad P(x>0) = \frac{1}{2}. \] 
Optimum quantizer outputs are conditional means: \[ \beta^\ = E[x|x<0], \qquad \alpha^\ = E[x|x>0]. \] 
Compute \(\alpha:\)  \(\alpha^\ = E[x|0 Compute \beta:\)Shape is linear from \(-2\ to\ 0\): \[ f(x) \propto (x+2). 
Normalized on $[-2,0]$: \[ f(x) = \frac{x+2}{2}. \] 
\[ \beta^\ = \int_{-2}^{0} x \frac{x+2}{2} dx = \frac{1}{2}\int_{-2}^{0}(x^2+2x)dx. \] 
Compute integral: \[ \int(x^2+2x)dx = \frac{x^3}{3} + x^2. \] 
Evaluate from -2 to 0: At 0 → 0. 
At -2 → \[ \frac{-8}{3} + 4 = \frac{4}{3}. \] 
Thus: \[ \beta^\ = -\frac{1}{2}\cdot\frac{4}{3} = -\frac{2}{3} \approx -0.67. \] 
Difference: \[ \alpha^\ - \beta^\ = 0.5 - (-0.67) \approx 1.17. \] 
Thus: \[ \boxed{1.17} \quad (\text{acceptable range: } 1.15\text{–}1.18) \]