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Given the uniform stochastic variable $U$ defined on the interval [0,1]. Using $U$, define a continuous stochastic variable $X$ with probability density function (PDF) $$f_X(x) = \begin{cases} \frac{1}{x^2}, & \text{for x $\geq$1} \\ 0, & \text{otherwise} \end{cases}$$

Now I now that the uniform distribution $U$ has PDF $$f_U(x) = \begin{cases} 1, & \text{for x $\in$ [0,1]} \\ 0, & \text{otherwise} \end{cases}$$ and cumulative distribution function (CDF) $$F_X(x) = \begin{cases} 0, & \text{for x $\leq$0} \\ x, & \text{for x $\in$ [0,1]} \\ 1, & \text{for x $\geq$ 1} \end{cases}$$

However, I have no idea on how to construct $X$ using $U$. Any idea?

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  • $\begingroup$ This is a very well known result in the first course of simulation. en.wikipedia.org/wiki/Inverse_transform_sampling $\endgroup$ – BGM May 30 '16 at 8:56
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    $\begingroup$ For every $x\geqslant1$, $P(X\geqslant x)=\frac1x=P(U\leqslant\frac1x)=P(\frac1U\geqslant x)$ hence... $\endgroup$ – Did May 30 '16 at 9:10
  • $\begingroup$ @Did How does $P(\frac{1}{U} > x)$ imply that $f_X(x) = \frac{1}{x^2}$ $\endgroup$ – Whizkid95 May 30 '16 at 11:33
  • $\begingroup$ @Whizkid95 ?? $P(1/U>x)$ implies nothing, $P(1/U>x)$ is a number. $\endgroup$ – Did May 30 '16 at 11:43
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    $\begingroup$ @Whizkid95 This shows that $P(X>x)=P(1/U>x)$ for every $x$, hence, as requested, that $X$ and $1/U$ have the same distribution. $\endgroup$ – Did May 30 '16 at 17:13

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