Manipulating a random variable that is uniformly distributed

Question

I had to answer the following question: $X$ has a uniform distribution between 0 and 1. What is the distribution of $X^3$?

Now the pdf for X is simply:

$f(x)= \left\{ \begin{array}{ll} 1, &\mbox{if } 0\leq x \leq1 \\ 0, & \mbox{otherwise} \end{array} \right.$

After substituting $x^3$ for $x$ in the cdf for $X$ and then differentiating it with respect to $x$, I got the following pdf for $X$:

$f(x)= \left\{ \begin{array}{ll} \frac 13x^{-\frac{2}{3}}, &\mbox{if } 0\leq x \leq1 \\ 0, & \mbox{otherwise} \end{array} \right.$

Is this correct?

Now, the last part of the question asks "What is the distribution of $X^3$? Is my answer sufficient or does this pdf correspond to a particular distribution? Because it doesn't ring a bell for me.

Thanks

You are correct. For the last part, I don't recognize it. I'm guess it is enougth. Explicitely the distribution is $F_{X^3}(x)=\sqrt[x]{x}1_{[0,1]}(x)+1_{(1\infty)}(x)$ — sinbadh, Feb 15 '16 at 11:01
The distribution is completely determined by its PDF which you calculated correctly except that $0\leq x\leq 1$ should be replaced by $0<x\leq1$. If "they" make troubles about the absence of a label then protest against that firmly. Also in my head bells are not ringing. — drhab, Feb 15 '16 at 11:07
Can you please explain why it is strict inequality after the 0? Thanks for your help, much appreciated — Vladimir Nabokov, Feb 15 '16 at 11:15

Em. · Answer 1 · 2016-02-15T11:25:53.693

4

It looks right, and it is a $\text{Beta}(1/3,1)$; $$f_X(x) = \frac{\Gamma(1/3+1)}{\Gamma(1/3)\Gamma(1)} x^{1/3-1}(1-x)^{1-1} = \frac{1}{3}x^{-2/3}$$ over $(0,1]$.

edited Feb 15 '16 at 11:25

answered Feb 15 '16 at 11:04

Em.

15,981

Over $(0,1]$ you mean. Actually I hoped for the OP that a name would be lacking :(. – drhab Feb 15 '16 at 11:13
@drhab Yes, you are right. I was sloppy. At least it is not an obscure distribution. :) – Em. Feb 15 '16 at 11:17
Also termed a PowerFunction(1/3) with domain of support on (0,1): $$f(x) = a x^{a - 1} \quad \text{ for } 0<x<1$$ and parameter $a>0$ – wolfies Feb 15 '16 at 11:19
@wolfies That's obscure :p – Em. Feb 15 '16 at 11:20
Isn't your fraction of Gamma functions upside down? I'm pretty sure $\beta (\frac 13,1)=3$. In that case, what is the distribution? – Vladimir Nabokov Feb 15 '16 at 11:24
@VladimirNabokov It is a Beta distribution, not to be confused with the Beta function. – Em. Feb 15 '16 at 11:26
Ah yes, sorry. But isn't true that the Beta distribution is only defined on (0,1)? – Vladimir Nabokov Feb 15 '16 at 11:27
Or does it not make a difference since it is a continuous distribution? – Vladimir Nabokov Feb 15 '16 at 11:28
@VladimirNabokov I do not believe that it should make a difference. – Em. Feb 15 '16 at 11:31

score 1 · Answer 2 · edited Feb 15 '16 at 11:15

Yes, that is the probability density function (pdf).$$\begin{align}f_{X^3}(z) = & ~ f_X(z^{1/3})~\lvert(z^{1/3})'\rvert \\[1ex] = & ~ \dfrac{ z^{-2/3} }{3} ~\raise{0.4ex}\chi_{z\in(0;1]}\end{align}$$

You can also integrate to obtain the Cumulative Distribution Function:

$$F_{X^3}(z) =z^{1/3}~\raise{0.5ex}\chi_{z\in[0;1)}+ \raise{0.5ex}\chi_{z\in[1;\infty)}$$

This isn't any particularly named distribution, except perhaps a member of the family of Beta distributions.

Manipulating a random variable that is uniformly distributed

2 Answers2