Some have proposed that for a natural centrality measure, the most central a node can get is the cen

If $X\sim$ Poisson $(\lambda )$, prove that $P[X=k]$ is maximized at $k=\lambda$

How would I go about proving something like this so that the method would work to prove a maximization point of other random variables like say Binomial is max at $k=(n+1)\star p$?

Let $x\in {\mathbb{R}}^n$ and let $P$ be a $n\times n$ positive definite symmetrix matrix. It is known that the maximum of
$\begin{array}{ll}\text{maximize}& x^{T}Px\\ \text{subject to}& x^{T}x\le 1\end{array}$
is ${\lambda }_{\text{max}}(P)$, the largest eigenvalue of $P$. Now consider the following problem
$\begin{array}{ll}\text{maximize}& x^{T}Px\\ \text{subject to}& (x-a{)}^T(x-a)\le 1\end{array}$
where $a\in {\mathbb{R}}^n$ is known. What is the analytical solution?

Let $f(x,y):\mathbb{R}\times \mathbb{R}\mapsto \mathbb{R}$ be a continuous and differentiable function. When can we claim that the following holds true:
$\frac{d}{dx}\underset{y\in \mathbb{R}}{max}f(x,y)=\underset{y\in \mathbb{R}}{max}\frac{\mathrm{\partial }}{\mathrm{\partial }x}f(x,y)$
assuming that both $\underset{y\in \mathbb{R}}{max}f(x,y)$ and $\underset{y\in \mathbb{R}}{max}\frac{\mathrm{\partial }}{\mathrm{\partial }x}f(x,y)$ exists.

Let $X\in \{0,1{\}}^d$ be a Boolean vector and $Y,Z\in \{0,1\}$ are Boolean variables. Assume that there is a joint distribution $\mathcal{D}$ over $Y,Z$ and we'd like to find a joint distribution ${\mathcal{D}}^{\prime }$ over $X,Y,Z$ such that:

1. The marginal of ${\mathcal{D}}^{\prime }$ on $Y$,$Z$ equals $\mathcal{D}$.
2. $X$ are independent of $Z$ under ${\mathcal{D}}^{\prime }$, i.e., $I(X;Z)=0$.
3. $I(X;Y)$ is maximized,

where $I(\cdot ;\cdot )$ denotes the mutual information. For now I don't even know what is a nontrivial upper bound of $I(X;Y)$ given that $I(X;Z)=0$? Furthermore, is it possible we can know the optimal distribution ${\mathcal{D}}^{\prime }$ that achieves the upper bound?

My conjecture is that the upper bound of $I(X;Y)$ should have something to do with the correlation (coupling?) between $Y$ and $Z$, so ideally it should contain something related to that.

Let us consider a linear programming problem shown below:
$\begin{array}{r}\underset{x}{min}f(x)-c\end{array}$
where, $f(x)$ is a linear function of $x$. Now, can be following maximization problem be considered as an equivalent to the above minimization problem?
$\begin{array}{r}\underset{x}{max}\left(\frac{1}{f(x)-c}\right)\end{array}$

A huge conical tank to be made from a circular piece of sheet metal of radius 10m by cutting out a sector with vertex angle theta and welding the straight edges of the straight edges of the remaining piece together. Find theta so that the resulting cone has the largest possible volume.

Specifically, the question is asked in the context of wanting derivatives, multiple max/min equations, and hopefully more calc rather than trig or geo.

I have gotten as far as using 10m as the hypotenuse for a triangle formed by the height of the cone, radius of the base of the cone, and slant. I'm not sure where to go from there, because I can't determine how to find height and/or radius, without which I'm not sure I can continue.

Show that $f(x_1^{\ast },...x_n^{\ast })=max\{f(x_1,...,x_n):(x_1,...,x_n)\in \mathrm{\Omega }\}$ if and only if $-f(x_1^{\ast },...x_n^{\ast })=min\{-f(x_1,...,x_n):(x_1,...,x_n)\in \mathrm{\Omega }\}$
I am not exactly sure how to approach this problem -- it is very general, so I can't assume anything about the shape of $f$. It seems obvious that flipping the max problem with a negative turns it into a min problem. Thoughts?