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Suppose $X$ is an uniform random variable on the interval (0,1). Find the distribution and and density functions of $Y=X^n$ for $n\ge2$.

Since $n\ge2$ I say that $-\infty < y < \infty$. Where I am having trouble is solving $P(X^n\le y)$. My first reaction is to take the $nth$ root of both sides, but that really does not seem right. My other thought is to take the $\ln$ of each side. If the latter is correct would it be $P(X\le\ln(y))$, in which case you would have $1-F(\ln(y))$ for the distribution function?k After this point I now to then differentiate to get $f(y)$.

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Hint: Begin by not saying that $-\infty < y < \infty$. Since $0 < X < 1$, $X^n > 0$ for all $n \geq 2$ and so, as long as you restrict yourself to $y > 0$, you can take $n$-th roots without any problems in the expression $P\{X^n \leq y\}$ to get an expression for $F_Y(y) = P\{Y \leq y\}$ in terms of $F_X$, the CDF of $X$. Hopefully, you can figure out that $F_Y(y) = 0$ for $y \leq 0$ without having to work too much at it. – Dilip Sarwate Oct 1 '12 at 18:38
The logarithm possibility leads to complications. For $y$ in the interesting part, we have $X^n \le y$ iff $n\ln X \le \ln y$. So to use the strategy, one would want to know about the distribution of $\ln X$. Certainly not impossible, but for sure less nice than the distribution of $X$. – André Nicolas Oct 1 '12 at 18:57
up vote 1 down vote accepted

No. The power operation is not commutative ($x^y=y^x$ only for $2,4$ or $x=y$), so its right inverse and left inverse operations (taking $y$-th root / taking logarithm of base $x$) are not exchangeable..

Since, $X$ lives in $(0,1)$, $P(X>0)=1$ and for $y<0$ you will get $P(X^n\le y)=0$, similarly for $y\ge 1$, $\ P(X^n\le y) = 1$, and taking the $n$th root is the right way to do (you can do it because $X$ and $y$ are assumed positive): $$P(X^n\le y) =P(X\le\sqrt[n]y) = \sqrt[n]y $$ because the probability of being in $(0,a]$ for a $U(0,1)$ distribution is just $a/1=a$.

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I got $\sqrt[n]y$ for the distribution function and then took the derivative of that for the density function and got $\frac{\lambda}{n}e^{-\lambda (n-1)/n}$ – Sprock Oct 1 '12 at 23:25
Sprock: You took the derivative of $\sqrt[n]{y}$ and got... and got what exactly? – Did Oct 3 '12 at 6:38
@did-The density function. – Sprock Oct 16 '12 at 17:01

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