# Find the distribution of $X$.

$U \sim \mathrm {Unif} (0,1)$. Let $\alpha > 0$. Then find the density function of $X=U^{-\frac 1 {\alpha}}$. I have found that if $F$ is the cumulative distribution function of the random variable $X$ then $$F(x) = \left\{ \begin{array}{ll} 0 & \quad \frac {1} {x^{\alpha}} < 0 \\ 1 - x^{-\alpha} & \quad 0 < \frac 1 {x^{\alpha}} < 1 \\ 1 & \quad \frac {1} {x^{\alpha}} > 1 \end{array} \right.$$

Thank you very much.

• But the problem is that $F$ is not defined at $0$. – Dbchatto67 Sep 7 '18 at 8:06
• @Dbchatto67 CDF are right-continuous. – Yuta Sep 7 '18 at 8:11
• @drhab The random variable is greater than $1$ so the expressions obtained by OP are wrong. – Kabo Murphy Sep 7 '18 at 8:17

The correct answer is this: $F(x)=0$ for $x<1$ and $F(x)=1-x^{-\alpha}$ for $x \geq 1$. The density is given by $f(x)=0$ for $x<1$ and $f(x)=\alpha x^{-\alpha -1}$ for $x >1$
• sir that means $X \sim \mathrm {Pareto}\ ({\alpha})$. Right? – Dbchatto67 Sep 7 '18 at 8:16
• You could equally say $F(x)=0$ for $x\le 1$ and $F(x)=1-x^{-\alpha}$ for $x \gt 1$ – Henry Sep 7 '18 at 8:17