I've been self-studying my way through A First Course in Probability by Ross, and just finished working through an example dealing with optimization of a function of a continuous random variable. The expected value of this function looks like this:
\begin{align} E[C_t(X)]&=\int_{0}^{\infty} C_t(x)f(x)dx \\ &=ct\int_{0}^{t}f(x)dx-c\int_{0}^{t}xf(x)dx+k\int_{t}^{\infty}xf(x)dx-kt\int_t^{\infty}f(x)dx \\ \end{align}
We take the derivative with respect to $t$, equate the result to $0$, and solve (aside: it turns out that a minima is achieved at $t^*$, such that $F(t^*)=\frac{k}{k+c}$, where $F(t)$ is the c.d.f. of $X$).
Fair enough. My question is, how can we know beforehand that this ugly thing has a minimum? Furthermore, how can we know that this minimum is global at $t^*$? In more simple optimization problems, you might be able to look at the form of the function and reason it out. Here, it isn't obvious (to myself at least).
Suggestions on resurces related to optimization theory and general problems are also welcome. Thanks!
I'll denote $g(t) = E[C_t(X)]$.
Taking the first derivative of the expression you wrote and simplifying we get $$g'(t) = (k+c)F(t) - k$$
and equating that with $0$ we get $t^*$ as you said. So $t^*$ is a stationary point of $g$.
Taking the second derivative we get $$g'(t)=(k+c)F'(t) = (k+c)f(t) \geq 0, \forall t$$ so the function $g(t)$ is convex (assuming $k+c\geq 0$) and thus the stationary point $t^*$ is the global minimum of $g$.
If a function is convex (and differentiable) on a bounded interval we know that it has a global minimum, which must be at the stationary point, if one exist, or on one of the edges of the interval $g$ is defined on.
If $g$ is defined on an unbounded interval, the global minimum can only be at a stationary point, as we have no 'edges'. However, a convex function on an unbounded interval can also have no minimum if it has no stationary points (e.g. $e^x$).
So in your example, because the function $g $ is convex and we found a stationary point, we knew that it has a global minimum at that point.