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Suppose that $X$ has the exponential distribution. Find the density for $X^3$.

Really not sure where to go with this problem, my notes from class weren't sufficient and after poking around online couldn't find anything to get me going. Any hints/help are really appreciated.

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    $\begingroup$ One can use what textbooks often call the method of transformations, or else find the cdf and differentiate. $\endgroup$ – André Nicolas May 9 '16 at 22:08
  • $\begingroup$ Would that other method look something like this? $\endgroup$ – 355durch113 May 9 '16 at 23:13
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As is the case here the transformation $z = g(x) = x^3$ is one-to-one over the support and has the inverse transformation

$$x = g^{-1}(z) = h(z) = \sqrt[3]{z}$$

then the pdf of $Z$ is given by

$$f_Z(z) = f_X(x)\left|\frac{dx}{dz}\right| = f_X[h(z)]\left|\frac{dh(z)}{dz}\right| = f_X[\sqrt[3]{z}]\left|\frac{d\sqrt[3]{z}}{dz}\right| $$

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Really not sure where to go with this problem

Alternatively, you can compute the cdf of $X^3$ by $$P(X^3\leq x) = P(X\leq \sqrt[3]{3}),\qquad x\geq0.$$

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The simplest way to describe an exponential distribution says $\Pr(X>x) = e^{-x/\alpha}$ (where $\alpha$ is the expected value) or $\Pr(X>x)= e^{-\beta x}$ (where of course $\beta=1/\alpha$). Then you can say \begin{align} f_{X^3}(x) & = \frac d {dx} \Pr(X^3 \le x) = \frac d {dx} \Pr(X\le \sqrt[3] x) = \frac d {dx} \left(1 - e^{-\sqrt[3]x /\alpha}\right) \\[10pt] & = \cdots\cdots \qquad \text{for }x>0. \end{align}

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