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let $U$ be a random variable uniformly distributed between $0$ and $1$.

A. For any strictly increasing function $f:\Bbb R \rightarrow [0,1]$, find the CDF of $X =f^{-1}(U)$

B. Given any random variable $X$ with PDF $p(x)$, show that $X$ can be generated by $X =f^{-1}(U)$,where

$$f(x)=\int^x_{-\infty}p(v) \, dv$$

What does $X =f^{-1}(U)$ mean? inverse? or just like $5^{-1}=\frac{1}{5}$?

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  • $\begingroup$ Neither. $\ \ \ $ $\endgroup$ Feb 3, 2017 at 4:11

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The number $\dfrac 1 5$ is the multiplicative inverse of $5$.

The function $f^{-1}$ is the compositional inverse of $f$.

$5^8$ means $5\times5\times5\times5\times5\times5\times5\times5.$ The operation of multiplication of numbers is iterated through $8$ instances of the number $5$.

$f^8(x)$ means $f(f(f(f(f(f(f(f(x)))))))).$ The operation of composition of numbers is iterated through $8$ instances of the function $f$.

Multiplying by $5^{-1}$ means multiplying $-1$ times by $5$, which amounts to dividing by $5.$

Finding the value of the function $f^{-1}$ at a certain input means applying $-1$ times, the function $f$, i.e. finding the image of that input under the inverse function. (For example, if $f(31)=97$ then $f^{-1}(97) = 31.$)

So, yes, it means inverse, but it's a compositional inverse, not a multiplicative inverse.

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  • $\begingroup$ Thx,However,after knowing this,i still don't know how to solve it $\endgroup$
    – Shine Sun
    Feb 3, 2017 at 4:22
  • $\begingroup$ Since $f$ is strictly increasing, if you have $X\le x$ if and only if $U\le f(x).$ $\endgroup$ Feb 3, 2017 at 4:37

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