PDF of a uniform distribution

Question

Let the random variable $X\sim\operatorname{unif}(0,1)$. Find the PDF of $Y=X^n$ where $n\in\Bbb N$.

Initially, I misread the problem, somehow mistaking the $n$ for a $2$. As a result, I ended up finding that the PDF for this was $$ \frac{1}{2 \sqrt{x}}$$So my question is: is it logical to adapt my botched attempt by simply replacing the $2$'s with $n$'s? That is to say that the PDF of $Y=X^n$ is $$\frac{1}{2 (x^{(1/2)-n})}$$

Thanks!

Answer 1

For general positive $n$, repeat what you did with $n=2$, presumably something like

• the density of $X$ is $f_1(x)=1$ for $0 \le x \le 1$
• so $P(X \le x)=x$ for $0 \le x \le 1$
• so $P(X^n \le x^n)=x$ for $0 \le x \le 1$
• so $P(X^n \le y)=y^{1/n}$ for $0 \le y \le 1$
• so the density of $X^n$ is $f_n(y)=\frac1ny^{\frac1n-1}$ for $0 \le y \le 1$

For $n=2$, the density of $X^2$ is $f_2(y)=\frac12y^{-\frac12}$ for $0 \le y \le 1$, as you have found