# Find continuous stochastic variable $X$ with PDF $f_X = \frac{1}{x^2}$

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) $$f_X(x) = \begin{cases} \frac{1}{x^2}, & \text{for x \geq1} \\ 0, & \text{otherwise} \end{cases}$$

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) $$F_X(x) = \begin{cases} 0, & \text{for x \leq0} \\ x, & \text{for x \in [0,1]} \\ 1, & \text{for x \geq 1} \end{cases}$$

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

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