GATE 2025 EE Q.36: solution and answer

GATE 2025 · EEQ.36 · 2 marks · MCQ

Let $X$ and $Y$ be continuous random variables with probability density functions $P_{X} (x)$ and $P_{Y} (y)$ , respectively. Further, let $Y = X^{2}$ and

P_{X} (x) = {1, 0, x \in (0, 1] otherwise .

Which one of the following options is correct?

(A)
$P_{Y} (y) = ⎩ ⎨ ⎧ \frac{1}{2 y}, 0, y \in (0, 1] otherwise$
(B)
$P_{Y} (y) = {1, 0, y \in (0, 1] otherwise$
(C)
$P_{Y} (y) = {1.5 y, 0, y \in (0, 1] otherwise$
(D)
$P_{Y} (y) = {2 y, 0, y \in (0, 1] otherwise$

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Answer and solution

ANSWER

$P_{Y} (y) = \frac{1}{2 y}$ for $y \in (0, 1]$ and $0$ otherwise, which is option (A).

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

Range of $Y$ . $X$ takes values only in $(0, 1]$ , and $y = x^{2}$ is strictly increasing there, mapping $(0, 1]$ onto $(0, 1]$ . So $Y \in (0, 1]$ : $F_{Y} (y) = 0$ for $y \leq 0$ and $F_{Y} (y) = 1$ for $y > 1$ , which gives $P_{Y} (y) = 0$ outside $(0, 1]$ .
CDF method. For $y \in (0, 1]$ , $F_{Y} (y) = P (X^{2} \leq y) = P (- y \leq X \leq y) = P (X \leq y)$ , because $X > 0$ .
Use the uniform density. Since $y \in (0, 1]$ , $F_{Y} (y) = \int_{0}^{y} 1 d x = y$ .
Differentiate. $P_{Y} (y) = \frac{d}{d y} y = \frac{1}{2 y}$ for $y \in (0, 1]$ .
Check with the change-of-variable formula. For a monotone map, $P_{Y} (y) = P_{X} (x) \frac{d x}{d y}$ . Here $x = y$ and $\frac{d x}{d y} = \frac{1}{2 y}$ , so $P_{Y} (y) = 1 \cdot \frac{1}{2 y}$ . Same result.
Why normalisation can't decide it. Every option integrates to 1 over $(0, 1]$ : $\int_{0}^{1} \frac{d y}{2 y} = 1$ , $\int_{0}^{1} 1 d y = 1$ , $\int_{0}^{1} 1.5 y d y = 1.5 \cdot \frac{2}{3} = 1$ and $\int_{0}^{1} 2 y d y = 1$ . So checking that an option is a valid density doesn't pick the answer. The density derived above does, and it is option (A).

KEY IDEA

To find the density of a function of a random variable, work out the CDF first and then differentiate: $F_{Y} (y) = P (X^{2} \leq y) = P (X \leq y) = y$ on $(0, 1]$ , so $P_{Y} (y) = \frac{1}{2 y}$ . For a monotone map, the shortcut $P_{Y} (y) = P_{X} (x) \frac{d x}{d y}$ gives the same answer.

WATCH OUT FOR

Assuming $Y$ is uniform because $X$ is uniform (option B). Squaring squeezes values near 0 closer together, so $Y$ piles up near 0 and its density grows without bound as $y \to 0^{+}$ . A second trap is checking only that an option integrates to 1. All four options do, so that test can't decide the answer. A third is multiplying by $\frac{d y}{d x}$ where the formula needs $\frac{d x}{d y}$ .

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