Suppose you have shown that the variance of a given biochemical measurement depends on the time points you sample, namely:
$$\mbox{Var} (k) = \frac {\sigma^2} {\displaystyle\sum_{i=1}^N (t_i - \bar t)^2}$$
where $t_i$ are the time points and $\bar t$ is their mean, $\sigma$ is a positive constant, and $t_i \ge 0, \forall i \in [1,N]$. Due to biochemical constraints, you need to impose $t_i \le 45$, too. Due to cost constraints, you need to impose $N \le 4$. Within the above constraints, how many, and which time points should we sample to ensure that $\mbox{Var}(k)$ is minimal?
One solution that results from numerical approaches is $N = 4, t = (0,0,45,45)$. E.g., in R:
g <- function(p) {
4/(3*p[4]^2-2*p[3]*p[4]-2*p[2]*p[4]-2*p[1]*p[4]+3*p[3]^2-2*p[2]*p[3]-2*p[1]*p[3]+3*p[2]^2-2*p[1]*p[2]+3*p[1]^2)
}
o2 <- optim(par = c(1,5,25,40), fn = g, gr = NULL, method = "L-BFGS-B", lower = rep(0, 4), upper = rep(45, 4), control = list(factr = 1))
print(o2)
#$par
#[1] 0 0 45 45
#$value
#[1] 0.0004938272
#$counts
#function gradient
# 7 7
#$convergence
#[1] 0
#$message
#[1] "CONVERGENCE: NORM OF PROJECTED GRADIENT <= PGTOL"
Do you think any formal/theoretical argument could be made to show that this is a general result, namely that the above $\mbox{Var}(k)$ is absolutely minimal when $N = 4$ and the sampled times are $(0,0,T,T)$, where $0 \le t_i \le T, \forall i \in [1,N], T > 0, N \le 4$?
Or if you could please point me to posts / literature where this topic (box-constrained nonlinear optimisation) is discussed, it would be great.
For fixed $N$, you are solving $$\max_{t} \left\{ {\sum_{i=1}^N (t_i - \bar t)^2} : 0 \leq t_i \leq T \right\}.$$ Since $(t_i - \bar t)^2$ is convex in $t$, the maximum occurs at an extreme point of the feasible region. So to find the optimum, you try $N=1$, $N=2$, $N=3$ and $N=4$, and for each $N$ you try the $2^N$ extreme points.
For $N=1$ the objective is $0$. For $N=2$ the objective for $(0,0)$ and $(T,T)$ is $0$, for $(0,T)$ and $(T,0)$ it is $0.5T^2$. For $N=3$ you can attain $(2/3) T^2$ while for $N=4$ you can get $T^2$.