Problem 4: Suppose the random variable $ X $ in Problem 3 is sent through a quantizer $ g(x) $ to produce the random variable $ Y = g(X) $, where\[g(x) = \begin{cases} -1, & x < -0.5, \\0, & -0.5 \leq x < 0.5, \\1, & x \geq 0.5.\end{cases}\](a) Find a mathematical expression for $ F_Y(y) $, the CDF of $ Y $.(b) Find a mathematical expression for $ f_Y(y) $, the PDF of $ Y $.(c) Find the value of $ E[Y] $.

a) For the given F(x) function, we have here:Fx(-0.5) = 0.2*(1 - 0.5)2 = 0.05Fx(0.5) = 0.5*(1 + 0.5)…

Answered: Problem 4: Suppose the random variable…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Problem 4: Suppose the random variable $ X $ in Problem 3 is sent through a quantizer $ g(x) $ to produce the random variable $ Y = g(X) $, where

\[
g(x) =
\begin{cases}
-1, & x < -0.5, \\
0, & -0.5 \leq x < 0.5, \\
1, & x \geq 0.5.
\end{cases}
\]

(a) Find a mathematical expression for $ F_Y(y) $, the CDF of $ Y $.

(b) Find a mathematical expression for $ f_Y(y) $, the PDF of $ Y $.

(c) Find the value of $ E[Y] $.

Expert Solution

Step 1

a) For the given F(x) function, we have here:
F_x(-0.5) = 0.2*(1 - 0.5)² = 0.05
F_x(0.5) = 0.5*(1 + 0.5) = 0.75

Therefore, the CDF for Y here is given as:

$F_y(y) = \left\{\begin{matrix} 0 & y < -1 \\ 0.05 & -1 \leq y < 0\\ 0.75 - 0.05 = 0.7 & 0 \leq y < 1 \\ 1 - 0.75 = 0.25 & y\geq 1 \end{matrix}\right.$

This is the required CDF here.

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