# Find the CDF and PDF of $W$ where $W =.7 - b(.7 - y)^2, 0<b<1$, with $Y \sim U[0,1]$

A random variable $$Y \sim U[0,1]$$. Let $$W = \frac7{10} - b\cdot(\frac7{10} - y)^2, 0.
Completely specify the CDF and PDF of $$W$$.
Also show that the PDF of W integrates to $$1$$.

So I have worked through most of this problem. Importantly, the transformation is $$2:1$$ on the interval y $$\in (\frac25,1]$$ and $$1:1$$ on the interval y $$\in [0, \frac25)$$. Thus I think the CDF is:

$$\begin{cases} \frac3{10}+\sqrt{\frac{7/10 - w}b } & \text{for}~~ w \in (\frac7{10} - \frac{49b}{100} ,\, \frac7{10} - \frac{9b}{100}) \\ 2\sqrt{ \frac{7/10 - w}b } & \text{for}~~ w \in (\frac7{10} - \frac{9b}{100} ,\, \frac7{10}) \\ \end{cases}$$

Can anyone confirm this? If I know this is right, I can figure out the rest on my own. When integrating the PDF, I am running into some strange ($${}+{}$$ or $${}-{}$$) situations that integrate to $$1$$ only if I select a certain sign, so I am having trouble verifying the CDF is correct by integrating the PDF over its support.

\begin{align}F_W(w) &=\mathsf P(0.7 -b(0.7-Y)^2\leqslant w) \\[1ex]&=\mathsf P(Y\leq 0.7-\sqrt{\tfrac{0.7-w}b})+\mathsf P (Y\geq 0.7+\sqrt{\tfrac{0.7-w}b}) \\[1ex]&=(0.7-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0\leq (0.7-\sqrt{\tfrac{0.7-w}b})\leq 1}+(0.3-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0\leq (0.7+\sqrt{\tfrac{0.7-w}b})\leq 1}+\mathbf 1_{0.7
To check, I have $$F_W([0.7-0.49b]^-)=0$$ and $$F_W([0.7]^+)=1$$ as we require.