Distribution of a random variable to the power of 3

Question

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

I'm not looking for an answer, just want to know how I should begin to answer the question.

Now the pdf is simply:

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

So do I simply substitute $x$ with $x^3$ and rearrange the inequalities?

Thanks

One way would be to compute the cdf for $X^3$ and differentiate to find the pdf. — hmakholm left over Monica, Feb 14 '16 at 15:14
How would I compute the CDF for $X^3$? By computing CDF for $X$ and then substituting $x^3$ for $x$? — Vladimir Nabokov, Feb 14 '16 at 15:16
@Vlad: Directly from the definition $F(t) = P(X^3\le t) = P(X\le \sqrt[3] t) = \cdots$? — hmakholm left over Monica, Feb 14 '16 at 15:17
@HenningMakholm: You mean $F(t)$? Normally variable names shouldn't matter but you didn't define $f$, in which case upper case is usually for CDF. — user21820, Feb 14 '16 at 15:20

score 2 · Answer 1 · answered Feb 14 '16 at 15:19

2

This is a basic problem in "transformations" Since the transformation is 1-1 ($x^3$ is one-to-one on $[0,1]$) the formula is $F_{g(X)}(x)=F_X(g^{-1}(x))$ where $F$ is the cumulative distribution (which in this case is just $F_X(x)=x$). So substitute $\sqrt[3]{x}$ into that. Then take the derivative to get the density function.

answered Feb 14 '16 at 15:19

Gregory Grant

14,874

So pdf of $X^3$ is $f(x)= \left{ \begin{array}{ll} \frac 1 3 x^-\frac{2}{3}, &\mbox{if } 0\leq x \leq1 \ 0, & \mbox{otherwise} \end{array} \right.$ The questions asks "What is the distribution of $X^3$?" - Is there a name for a distribution with this pdf? Because it doesn't seem to ring a bell for me... – Vladimir Nabokov Feb 15 '16 at 00:36

Distribution of a random variable to the power of 3

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