Let $X\sim \textbf{U}[0,1]$, and $r(X) = 0.5X^2 + 0.3X$, then what is $\textbf{E}[X|r(X) \leq w]$?
This honestly seems just so simple, but I am not confident. Is it simply, \begin{align} \int_0^w r(X)dX \approx 0.17w^3+0.15w^2? \end{align} Or is is because $X$ is constrained by $w$, $$0 \leq X \leq 1 \Rightarrow 0 \leq r(X) \leq w \leq 0.8,$$ I am supposed to find the inverse of $r(X)$, \begin{align} r^{-1}(X) = -0.3 \pm 0.1\sqrt{200X+9}, \end{align} making the expectation, $\textbf{E}[X|X \leq r^{-1}(w)]$ which is, \begin{align} \textbf{E}[X|X \leq r^{-1}(w)] = \int_0^{-0.3 \pm 0.1\sqrt{200w+9}} xdx = \frac{-0.3 \pm 0.1\sqrt{200w+9}}{2}? \end{align} Where $0 \leq w \leq 0.8$.
Yeah, I am just stuck because I am not terribly certain how to interpret and operate on the condition of the expectation...
$$.5x^2+.3x-w\le0$$
First solve the equality
$$.5x^2+.3x-w=0\Rightarrow x=-.3\pm\sqrt{.09+2w}$$
Now you may recall from high school math that this can be factored into $(x-(-.3+\sqrt{.09+2w}))(x-(-.3-\sqrt{.09+2w}))\le0$ and will be positive when $x$ is less than both of those values or greater than both of those values and negative when it is between those two values. In fact $-.3-\sqrt{.9+2w}<0$ and $X$ is uniform on $[0,1]$ so we know $X$ is between $0$ and $-.3+\sqrt{.09+2w}$.
Given this we now find the conditional expectation $\mathbb E(X|X<-.3+\sqrt{.09+2w})$. This involves reweighing the density of $X$ to $$\frac {f(X)}{\Pr(X<-.3+\sqrt{.09+2w})},0<X<-.3+\sqrt{.09+2w}\\ =\frac 1 {-.3+\sqrt{.09+2w}}$$
Then the expected value is
$$\begin{split}\int_0^{-.3+\sqrt{.09+2w}}\frac x{-.3+\sqrt{.09+2w}}dx&=\frac{x^2}{2(-.3+\sqrt{.09+2w})}\bigg|_{x=0}^{-.3+\sqrt{.09+2w}}\\ &=\frac{-.3+\sqrt{.09+2w}}{2}\end{split}$$