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 piece-wise continuous function) 

X = N(0, 1)

$f_x(x)=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}$

Y = -1; x ≤ -2

0; -1 ≤ x ≤ 1

1; x ≥ 2

x + 1; -2 ≤ x ≤ -1

x - 1; 1 ≤ x ≤ 2

Given that the random variable X, transforms Y = h(X)
From the given figure, it can be concluded that Y takes the discrete set of values {– 1, 0, 1}. 
So, the probability density function of Y will consist of impulses at y =-1, y = 0 and y =1.

Y is taking a discrete set of values and a continuous range of values, so it is a mixed random variable.

From the given options, the density function of 'Y' will be.

f_Y(y) = aδ(y + 1) + bδ(y) + cδ(y - 1) + g(y)