I've seen this question kind of posted here before, but only solved for 1 case and I had some questions about why it wouldn't work for others. Wanted to comment on that post directly, but I don't have enough points yet…
Let random variable $X \sim U(0,1)$. Assume a random variable $Y = X^n$, where n is a fixed number. Find the probability density function (pdf) for the random variable $Y$.
When this was originally posted, this was the answer given:
For these problems, it may be easier to manipulate the cumulative distribution function (CDF) first.
Notice that $Y$ is also supported on (0,1) assuming that n is a positive integer. Let's calculate its CDF. Fix y ∈ (0,1). Then
$$ F_Y(y) = Pr[Y \le y] = Pr[0 \le X \le y^\frac{1}{n}] = F_x(y^\frac{1}{n}) = y^\frac{1}{n}$$
Since $F_X(z) = z$ for $0 \le z \le 1$.
My question is why this wouldn't work for negative and non-integer values of n as well? Also, how would I go about solving for this CDF if n were a negative value?
Thank you for any help you can provide!
Also, here is the Original Post. Thank you to @Ubez and @Srivatsan for posting and helping on this question respectively.
Best Answer
For positive noninteger values you can still use what was in the other post. Let $n\in (-\infty,0)$. Set $k=-n\in (0,\infty)$ (the reason I do this is because I like to work with positive exponents). We have $Y=X^n$ and we want to find $f_Y$. Let $y\in(0,\infty)$. \begin{align} F_Y(y)&=P(Y\leq y) = P(X^n \leq y) = P\left( \frac{1}{X^k}\leq y \right)\\ &= P\left( \frac{1}{y}\leq X^k \right) = P\left( \frac{1}{y^{1/k}}\leq X \right) \end{align} If $y<1$ we have: $F_Y(y)=0$ (why?). For $y\geq 1$ we have: \begin{align} F_Y(y) = 1-P\left( \frac{1}{y^{1/k}}> X \right)=1-\frac{1}{y^{1/k}} = 1-y^{1/n} \end{align} So: \begin{align} F_Y(y) = \mathbf{1}_{[1,\infty)} (1-y^{1/n}) \end{align} Differentiate to get $f_Y(y)$.