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GATE 2025 EE – Question 36

Engineering Mathematics · Probability and Statistics: Random variables · 2 marks · Multiple choice

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

$$P_X(x)=\begin{cases}1,&x\in(0,1]\\0,&\text{otherwise}\end{cases}$$

Which one of the following options is correct?

  1. $P_Y(y)=\begin{cases}\dfrac1{2\sqrt y},&y\in(0,1]\\0,&\text{otherwise}\end{cases}$
  2. $P_Y(y)=\begin{cases}1,&y\in(0,1]\\0,&\text{otherwise}\end{cases}$
  3. $P_Y(y)=\begin{cases}1.5\sqrt y,&y\in(0,1]\\0,&\text{otherwise}\end{cases}$
  4. $P_Y(y)=\begin{cases}2y,&y\in(0,1]\\0,&\text{otherwise}\end{cases}$

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Show answer and explanation

Correct answer: (A) $P_Y(y)=\begin{cases}\dfrac1{2\sqrt y},&y\in(0,1]\\0,&\text{otherwise}\end{cases}$

Explanation

For $Y=X^2$ with $X\in(0,1]$ the map is monotonic, so $P_Y(y)=P_X(\sqrt y)\left|\dfrac{dx}{dy}\right|=1\cdot\dfrac1{2\sqrt y}$ for $y\in(0,1]$. (Check: $\int_0^1\dfrac{dy}{2\sqrt y}=1$.)