Question

A random variable $X$, distributed normally as $N(0,1)$, undergoes the transformation $Y = h(X)$, given in the figure. The form of the probability density function of $Y$ is (In the options given below, $a, b, c$ are non-zero constants and $g(y)$ is a piece-wise continuous function)

• A. $a\delta(y-1) + b\delta(y+1) + g(y)$
• B. $a\delta(y+1) + b\delta(y-1) + c\delta(y) + g(y)$
• C. $a\delta(y+2) + b\delta(y) + c\delta(y-2) + g(y)$
• D. $a\delta(y+2) + b\delta(y-2) + g(y)$

Hint:

When a nonlinear mapping has flat segments, the pdf of the output includes Dirac delta impulses at those constant output values. Linear parts contribute continuous density via the change-of-variable rule.

Answer 1

The Correct Option is B

Step 1: Analyze the mapping $Y = h(X)$.
From the figure: - For $x \leq -2$, $y = -1$ (constant).
- For $-2<x<-1$, $y$ varies linearly from $-1$ to $0$.
- For $-1 \leq x \leq 1$, $y = 0$ (constant).
- For $1<x<2$, $y$ varies linearly from $0$ to $1$.
- For $x \geq 2$, $y = 1$ (constant).
Step 2: Implications for pdf.
- Flat segments ($x \leq -2$, $x \geq 2$, $-1 \leq x \leq 1$) produce delta functions in $y$ at $y=-1$, $y=0$, and $y=1$.
- Sloped regions ($-2<x<-1$, $1<x<2$) map intervals of $x$ to intervals of $y$, giving continuous pdf contributions $g(y)$.
Step 3: Form of pdf.
Thus, the pdf of $Y$ is a sum of impulses at $y=-1,0,1$ with nonzero weights, plus a continuous part $g(y)$.
This matches option (B): \[ a\delta(y+1) + b\delta(y-1) + c\delta(y) + g(y). \] \[ \boxed{\text{(B)}} \]