Given that x~U[0,1] and y~U[0,1], derive the conditional CDF of W=x-b*(x-y)^2 where 0<b<1? Condition on x (i.e. treat x as a constant).

Question

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.

Answer 1

I am running into difficulties with this, given the two to one transformation in parts.

It is just due to the square root.

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

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