Admissibility does not imply minimax The answer to minimax estimator explains why minimax does not imply admissibility. The relevant statement is from https://www.stat.berkeley.edu/~yuekai/201b/lec6.pdf which says,

minimaxity does not imply admissibility: a minimax estimator has the
  best worst-case performance, but its performance at other parameters
  may be suboptimal. However, any estimator that dominates a minimax
  estimator is also minimax. Thus unique minimax estimators are
  admissible

I'm curious about the converse:

admissibility also does not imply minimaxity: an estimator is
  admissible if it has (strictly) smaller risk than any other estimator
  at a single θ ∈ Θ, but its risk at any other θ may be arbitrarily bad.

Can somebody explain this? Wouldn't being admissible mean you have uniformly lower risk and so your decision rule has the best worst-case-risk thereby making you a minimax?
 A:  As shown in this picture taken from this set of slides by Slava Chernoi, minimax (blue,red?) and admissible (black, red?) estimators have different characteristics: minimaxity is a global property obtained all over the parameter space, since the maximum risk is involved, while admissibility is much more local since it is enough that no other estimator does better at a point $\theta_0$ or by continuity in a neighbourhood of a point $\theta_0$. Hence, they are not implying one another and there may be an infinite number of inadmissible minimax estimators (as in the $\mathrm{N}_p(\theta,I_p)$ case when $p>2$) and there is almost always an infinite number of admissible estimators that are not minimax.
If the model is regular enough to handle limiting arguments, like the limit of a sequence of minimax estimators is a minimax estimator and so on, there should exist admissible minimax estimators, not necessarily unique (as in the $\mathrm{N}_p(\theta,I_p)$ case when $p>2$).
Another result to add to those you mention is the following one:

If $\delta_0$ is admissible with constant risk, $\delta_0$ is the
  unique minimax estimator.

Here is a detailed comparison in the simplest possible case, taken from my book:

Consider a Bernoulli random variable, $X\sim {\mathcal B}e(\theta)$ 
  with realisation $X=x$ and $\theta\in\{0.1,0.5\}$. Four nonrandomized
  estimators are available, \begin{align*} \delta_1(x) &=  0.1,
 \\ \delta_2(x) &= 0.5, \\ \delta_3(x) & =  0.1\, \mathbb{I}_{x =
 0} + 0.5\, \mathbb{I}_{x = 1}, \\     \delta_4(x)  &= 0.5\,
 \mathbb{I} + 0.1\, \mathbb{I}_{x = 1}. \end{align*} We assume in
  addition that the penalty (loss) for a wrong answer is equal to $2$ when $\theta
 = 0.1$ and equal to $1$ when $\theta = 0.5$. The risk vectors  $(R(0.1,\delta),R(0.5,\delta))$ of the four estimators are then,
  respectively, $(0,1)$, $(2,0)$, $(0.2,0.5)$, and $(1.8,0.5)$. It is
  straightforward to see that the risk vector of any randomized
  estimator is a convex combination of these four vectors or,
  equivalently, that the risk set, $\mathcal{R}$, is the convex hull of the
  above four vectors, as represented by the parallelogram in the Figure.
In this case, the minimax estimator is obtained at the intersection of
  the diagonal of $\mathbb{R}^2$ with the lower boundary of
  $\mathcal{R}$. As shown by the Figure, this estimator $\delta^*$ is
  randomized and takes the value $\delta_3(x)$ with probability $\alpha
 = 0.87$ and $\delta_2(x)$ with probability $1-\alpha$. The weight $\alpha$ is actually derived from the equation $$
 0.2 \alpha + 2(1-\alpha)  =  0.5 \alpha. $$ This estimator $\delta^*$ is also a (randomized) Bayes estimator with respect to the prior $$
 \pi(\theta)  =  0.22\, \mathbb{I}_{0.1}(\theta) + 0.78\,
 \mathbb{I}_{0.5} (\theta); $$ the prior probability $\pi_1 = 0.22$
  corresponds to the slope between $(0.2,0.5)$ and $(2,0)$, i.e., $$
 {\pi_1\over 1-\pi_1}  =  {0.5 \over 2-0.2}. $$ Notice that every
  randomized estimator that is a combination of $\delta_2$ and of
  $\delta_3$ is an admissible Bayes estimator for this distribution, but that
  $\delta^*$ only is also a minimax estimator.


