I'm doing a question that discusses a six-sided fair die being rolled and then discussing the numbers that appear on the top and on the side facing you. The rest of the question discusses the conditional and joint probability distributions. The last part of the question (appears to be unrelated to the previous parts unless I'm missing something) asks to guess a function $\phi : {1,2,3,4,5,6} \to [0,1]$ that maximises $Var(\phi(T))$ and then compute it (T is the variable describing the number on the top of the die, which takes a value from ${1,2,3,4,5,6}$). I'm not sure where to start with my guess (computing the variance will not be an issue for me). Is there an intuitive function that comes to mind when looking at this problem?
Thanks!
To maximize the variance of a variable over a uniform distribution reduces to maximing the distance of the values of the variable from their average. That means that $\phi (n)\in\{0,1\}$ or otherwise the variance would decrease. Placing $N$ out of the 6 variables on $1$ and the rest on $0$ results in a variance of
$$Var(\phi(T))=\frac{N}{6}\left(1-\frac{N}{6}\right)$$
which is maximized for $N=3$.
EDIT: Per the request, here's a more rigorous way to show that the procedure above yields a maximizer. Note that $\phi^2(n)\leq \phi(n)$. Use this to obtain an upper bound for the variance
$$Var(\phi(T))\leq \frac{1}{36}[6\sum \phi(n)-\left(\sum\phi(n)\right)^2]\leq \frac{1}{4}$$
The equality here can be achieved only when $\sum\phi(n)=3$ and $\phi(n) \in \{0,1\}$. It is now easy to show that $$\phi(1)=\phi(2)=\phi(3)=1,\phi(4)=\phi(5)=\phi(6)=0 $$
saturates the bound, which proves that $\max_\phi Var(\phi(T))=1/4$.