Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $f(y)$.

share|improve this question
2  
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
1  
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

1 Answer 1

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 $a/1=a$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.