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Given that $x\sim U[0,1]$ and $y\sim U[0,1]$, derive the conditional CDF of $W=x-b\cdot(x-y)^2$ where $0<b<1$? Condition on $x$ (i.e. treat $x$ as a constant). I am running into difficulties with this, given the two to one transformation in parts.

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    $\begingroup$ Conditional on what exactly and what is the meaning of the trailing phrase “where 0”? $\endgroup$ – Nap D. Lover May 1 at 1:48
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I am running into difficulties with this, given the two to one transformation in parts.

It is just due to the square root.

$$\begin{align}\mathsf P(W\leq w\mid x=s)&=\mathsf P(x-b(x-y)^2\leq w\mid x=s)\\[2ex]&=\mathsf P\left(s-b(s-y)^2\leq w\right)&\text{independence of }x,y\\[2ex]&=\mathsf P\left(\tfrac{s-w}b\leq (s-y)^2\right)&\text{since }0<b<1\\[1ex]&\ddots\end{align}$$


NB: Recall that: $\{c\leq Z^2\}=\{Z\leq -\surd c\}\cup\{ \surd c\leq Z\}$

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  • $\begingroup$ It's not the square root that is bothering me. There are two questions I have: 1) What you have written is true, but depending on the value of x=s, there should be part of the CDF that is one to one, and part that is many to one (as you demonstrated). 2) What is the correct support of the cdf? $\endgroup$ – user432299 May 1 at 3:58
  • $\begingroup$ EDIT to above, it should read "part of the transformation is 1-1 and part that is many to one.." $\endgroup$ – user432299 May 1 at 4:06

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