Find the PDF $f_Y(y)$ for a Random Variable $Y=X^n$ for Negative and Non-Integer Values of n

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?

Also, here is the Original Post. Thank you to @Ubez and @Srivatsan for posting and helping on this question respectively.

• Take $n=-1$. You will find the CDF is $1-\frac 1 t$ for $t \geq 1$ and 0 for $t<1$ – Kavi Rama Murthy Dec 7 '17 at 7:26
• Could you show an example of how you found that @KaviRamaMurthy – strwars Dec 7 '17 at 14:44
• Can you try what they did in the other post? There will not change much.. – Shashi Dec 7 '17 at 16:10
• Wouldn't it come out the same, positive or negative, if you go by the last post though? I'm not sure how it changes @Shashi – strwars Dec 7 '17 at 16:20
• No, it is almost the same though. I have posted a solution – Shashi Dec 7 '17 at 16:35

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 $f_Y(y)$.