For fun, I am working through Probability Via Expectation by Peter Whittle. I am currently stuck on one of his exercises.
The statement: A steel billet is trimmed to a length $x$ and then rolled. After rolling, its length becomes $y = \beta x + \epsilon$, where $\epsilon$ is a r.v. expressing the variability of the rolling. It is then trimmed again to a final length $z$. If $y$ is greater than $z$ then there is a loss proportional to the excess: $a(y - z)$. If $y$ is less than $z$, then the billet must be remelted, and there is a flat loss $b$. The original trim-length $x$ must now be chosen so as to minimize the expected loss. Show that if $\epsilon$ has the probability density $f(\epsilon)$, then the equation determining the optimal value of $x$ is
$$b f(z - \beta x) = a \int_{z - \beta x} ^ \infty f(\epsilon) d\epsilon$$
Attempt: Since $y \geq z \iff \beta x + \epsilon \geq z \iff \epsilon \geq z - \beta x$, we can write the loss function as
$$g_x(\epsilon) = \begin{cases} a(y-z) \quad & \epsilon \geq z - \beta x \\ b \quad & \epsilon < z - \beta x \end{cases}$$
We want to minimize the expected loss, so we want to minimize $$ \begin{align}E[g_x(\epsilon)] =& a(y - z)P(\epsilon \geq z - \beta x) + b P(\epsilon < z - \beta x) \\ =& a(\beta x + \epsilon - z)P(\epsilon \geq z - \beta x) + b P(\epsilon < z - \beta x) \end{align}$$
From here, I took the derivative with respect to $x$ and set equal to $0$, but this does not result in the correct equation. Instead, I get
$$a P(\epsilon \geq z - \beta x ) - a(\beta x + \epsilon - z) f(z - \beta x) - b f(z - \beta x ) = 0$$
which is a result of using the fundamental theorem of calculus and the product rule. I feel like I'm missing something obvious but I can't figure it out. I'm using $\frac{\partial}{\partial x} P(X \leq g(x) ) = f(g(x))g'(x)$, which I think is right, but I'm a little rusty.
One problem lies in your calculation of the expected loss. Since you don't know what the value of $\ y\ $ is going to before the billet is rolled, this has to be integrated over when the expected loss is calculated and so cannot appear in the expression for the expected loss. The expected loss is, in fact, \begin{eqnarray} E\left(g_x\left(\epsilon\right)\right) &=& \int_{-\infty}^{\infty} g_x\left(\epsilon\right)f\left(\epsilon\right)d\epsilon \\ &=& \int_{-\infty}^{z-\beta x}bf\left(\epsilon\right)d\epsilon +\int_{z-\beta x}^\infty a\left(\beta x +\epsilon - z\right)f\left(\epsilon\right)d\epsilon\ . \end{eqnarray} Therefore: \begin{eqnarray} \frac{dE\left(g_x\left(\epsilon\right)\right)}{dx} &=& -\beta b f\left(z-\beta x\right) + a\beta\int_{z-\beta x}^\infty f\left(\epsilon\right)d\epsilon\ , \end{eqnarray} and setting this equal to $\ 0\ $ gives the result which the exercise asks you to show.