GATE 2023 EC Q.36: solution and answer

GATE 2023 · ECQ.36 · 2 marks · MCQ

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)

GATE 2023 EC question 36: Graph of y = h(x), horizontal axis X with ticks at -2, -1, 1, 2 and vertical axis y = h(x) with ticks at 1 and -1. — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
$a δ (y - 1) + b δ (y + 1) + g (y)$
(B)
$a δ (y + 1) + b δ (y) + cδ (y - 1) + g (y)$
(C)
$a δ (y + 2) + b δ (y) + cδ (y - 2) + g (y)$
(D)
$a δ (y + 2) + b δ (y - 2) + g (y)$

Answer and solution

ANSWER

(B) $a δ (y + 1) + b δ (y) + cδ (y - 1) + g (y)$ . Each of the three flat pieces of $h$ (at output levels $y = - 1, 0, 1$ ) takes an interval of $x$ that has non-zero probability, so each produces an impulse. The two sloped pieces give the continuous part $g (y)$ .

Official answer: BSource: Official answer key, GATE 2023 EC (IIT Kanpur)

Read the figure as a piecewise function: $h (x) = - 1$ for $x \leq - 2$ , $h (x) = x + 1$ for $- 2 \leq x \leq - 1$ , $h (x) = 0$ for $- 1 \leq x \leq 1$ , $h (x) = x - 1$ for $1 \leq x \leq 2$ , and $h (x) = 1$ for $x \geq 2$ . So $Y$ only takes values in $[- 1, 1]$ .
Rule: wherever $h$ is flat over an interval of $x$ with non-zero probability, all of that probability lands on one value $y_{0}$ . In the PDF of $Y$ this appears as an impulse $P (interval) δ (y - y_{0})$ .
The flat piece $x \leq - 2$ maps to $y = - 1$ : $P (Y = - 1) = P (X \leq - 2) = Φ (- 2) \approx 0.0228 \neq = 0$ . This gives the term $a δ (y + 1)$ .
The flat piece $- 1 \leq x \leq 1$ (the part of the graph lying on the $x$ -axis) maps to $y = 0$ : $P (Y = 0) = P (- 1 \leq X \leq 1) = 2Φ (1) - 1 \approx 0.6827 \neq = 0$ . This gives the term $b δ (y)$ .
The flat piece $x \geq 2$ maps to $y = 1$ : $P (Y = 1) = P (X \geq 2) = 1 - Φ (2) \approx 0.0228 \neq = 0$ . This gives the term $c δ (y - 1)$ .
The two sloped pieces are one-to-one with slope $1$ , so they give an ordinary (continuous) density. For $- 1 < y < 0$ : $x = y - 1$ and $\frac{d x}{d y} = 1$ , so $f_{Y} (y) = f_{X} (y - 1)$ . For $0 < y < 1$ : $x = y + 1$ , so $f_{Y} (y) = f_{X} (y + 1)$ . Together these make the piecewise-continuous part $g (y)$ , which is zero outside $(- 1, 1)$ .
Collecting the terms: $f_{Y} (y) = a δ (y + 1) + b δ (y) + c δ (y - 1) + g (y),$ with $a = Φ (- 2)$ , $b = 2Φ (1) - 1$ , $c = 1 - Φ (2)$ , all non-zero. Check: $a + b + c + \int g (y) d y \approx 0.0228 + 0.6827 + 0.0228 + 2 (0.1359) = 1$ . This is option (B).

KEY IDEA

When a transformation $Y = h (X)$ is constant at $y_{0}$ over an interval of $x$ that has non-zero probability, $Y$ has a point mass at $y_{0}$ , and its PDF picks up an impulse $P (interval) δ (y - y_{0})$ . Where $h$ is strictly monotonic, the density is continuous: $f_{Y} (y) = f_{X} (x) /∣ h^{'} (x) ∣$ . So you get one delta for each flat level of the graph, and they sit at the output values ( $y$ ), not at the $x$ breakpoints.

WATCH OUT FOR

There are two traps. First, placing the impulses at the $x$ -axis breakpoints $\pm 2$ instead of at the output levels $y = \pm 1$ , which leads to (C) or (D). Second, overlooking that the segment lying along the $x$ -axis for $- 1 \leq x \leq 1$ is also a flat piece. It carries about 68% of the probability, so it produces the impulse $b δ (y)$ at $y = 0$ . Missing it leads to (A).

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 33 Probability & statistics questions, with notes

GATE 2023 EC Q.36: solution and answer

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