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Let's say there exists f(x) = 3(x-1)^2 0 < x < 1, 0 otherwise Then the CDF would surely be F(x) = 0 if x < 0, 3x^3 -9x^2 + 9x for 0 < x < 1 and 1 otherwise.

How would I go about finding the inverse CDF of this RV? I can't find any possible way to invert the CDF.

guest122313
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    Your cdf is incorrect. It should be $3 x - 3 x^2 + x^3$. Then lookup "roots of cubic equation". – JimB Nov 26 '21 at 04:02
  • Oh that was honestly just a typo... I'm a bit confused as to how the roots will help me here though in terms of finding the inverse. – guest122313 Nov 26 '21 at 04:51
  • @guest122313 You see, the meaning of an inverse of CDF F^-1(p) is simply finding the x such that for that x, p=F(x). So the way of solving the equation is to solve the polynomial equation 0=F(x)-p, which is possible – Preston Lui Nov 26 '21 at 05:31
  • Is there an easy method to solving a cubic where the constant is unknown? I've never encountered this problem before, and I'm not entirely sure where to start looking – guest122313 Nov 26 '21 at 06:07

1 Answers1

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If $f(x) = 3(x-1)^2, x \in (0,1),\hspace{1mm} 0\hspace{1mm} \text{otherwise}$, then

$p = F(x) = \left\{\begin{array}{lr}0 & x\leq 0 \\ (x-1)^3 + 1 & x \in (0,1) \\ 1 & x \geq 1\end{array}\right.$

Then you can invert by solving for $x$:

$x = F^{-1}(p) = \sqrt[3]{p-1}+1, p \in [0,1]$

gibson25
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