What is the meaning of admissibility within a class, that every decision rule in a class is admissible in that class? Suppose that I have that $X$ is a Poisson random variable with mean $\lambda$. Suppose a decision rule is to estimate $\lambda$ by using $\delta(Y) = aY$. Now, let $K$ be the class of all decision rules where $0< a<1$. I would like to see that using the squared loss function, every decision rule in $K$ is admissible in $K$. However, I am not sure what it is meant to be admissible in $K$. It seems that for example, If I take $a = \frac{1}{2}$, then $\delta(Y) = \frac{1}{2}Y$, but I can always find another $a$ within $0< a<1$ such that the loss is less. In other words, it seems that to satisfy this criteria, the only way this can happen is if the risk were equal among all rules within the class. Can anyone help me see what is going on? thanks!
 A: To truly see what's going on, graphics help.
Let's begin with the definitions.  I will present them both mathematically and in a symbolic computing language Mathematica (to show how concrete and practical they are).


*

*An estimator $t$ of a parameter $\theta$ is a procedure to convert any sample $X$ into a number $t(X)$.  In this situation you are comparing a set of estimators indexed by a number $a$.  Let us call them $t_a$.  You are told that $t_a(X) = aX$.  Code:
estimator[sample_, a_:] := a sample


*The loss $\Lambda$ penalizes discrepancies between the result of an estimator and the true parameter value.  Squared loss is the function $\Lambda(t, \theta) = (t-\theta)^2$. Code:
loss[estimate_, parameter_] := (estimate - parameter)^2


*The risk $r$ is the expected loss for any given parameter value $\theta$ when using a particular estimator $t$; that is, $r_t(\theta) = \mathbb{E}_\theta(\Lambda(t(X), \theta)).$  Code:
risk[parameter_, estimator_, loss_, family_] := Module[{x},
  Expectation[loss[estimator[x], parameter], x \[Distributed] family[parameter]]]

Risk is explicitly a function of the parameters.  Comparing functions is tricky, so let's reason it out. To compare estimators $t$ and $t^\prime$, consider a parameter value $\theta$.  If the risk of using $t$ is less than the risk of using $t^\prime$, then obviously you would want to use $t$: its expected loss is smaller.  Unfortunately you don't know the value of $\theta$--that's why you're estimating it!  Thus, you would want to survey the landscape by comparing $t$ to $t^\prime$ for all plausible values of $\theta$.  In the language of risk, you want to graph the functions $r_{t}(\theta)$ and $r_{t^\prime}(\theta)$.  In most cases, for some values of $\theta$ the risk of $t$ will be smaller and for other values of $\theta$ the risk of $t^\prime$ will be smaller.  However, if--given a procedure $t$ you are thinking of using--you can find another procedure $t^\prime$ whose risk at any $\theta$ is no worse than the risk of $t$ and sometimes is definitely better, then why use $t$ at all?  Such a $t$ is called inadmissible.  All other $t$ are admissible.
Graphically, the plot of risk versus parameter for an inadmissible $t$ will lie above (or on) the plot for some other $t^\prime$.  Thus, to assess admissibility you need to plot the risk functions of all the procedures and check whether any two of them cross.  After all, if it's always the case that the plot for $t$ and the plot for $t^\prime$ cross, no matter what $t$ or $t^\prime$ might be, then neither one is always superior to the other.
Let's plot the risks for a careful selection of the procedures $t_a$ in the question.  I chose $a \in \{1/1000, 1/3, 1/2, 6/7\}$ as being good representatives of the whole family.  Since the parameter is any positive number obviously I can't plot the entirety of each risk function, but by looking at a good range of parameter values I can at least get a sense of what's going on.  The range from $0$ to $3$ works well.  Code:
Plot[Evaluate@Table[risk[parameter, estimator[#, a] &, loss, PoissonDistribution], 
  {a, {1/1000, 1/3, 1/2, 6/7}}], {parameter, 0, 3}]


Now you can see that the risk functions all overlap.  For small $a$ they rise quickly and attain truly awfully high risks, but nevertheless for very small values of the parameter they still appear to have lower risk than other risk functions.  This is intuitive: a small value of $a$ is steering the estimate to low values and this will work well when the parameter is small.  A larger value of $a$ will work better for larger values of the parameter.
It would be nice to prove that.  To that end, you have to knuckle down and do the calculation.  Code:
risk[\[Lambda], estimator[#, a] &, loss, PoissonDistribution]


$(1-a)^2 \lambda^2 + a^2 \lambda$

All that remains is to show that any two distinct such functions cross at some positive value of the parameter $\lambda$.  That's a matter of elementary algebra of little interest here.  I will point out, though, that since all the risks are multiples of the parameter $\lambda$, you may factor it out, reducing the problem to comparing the graphs of $\lambda \to a^2 + (1-a)^2 \lambda$, which is a family of (mutually crossing) lines:

Each value of $a$ produces a line showing $1/\lambda$ times the risk function for $t_a$.  The colors between the lines graduate from red through blue as $a$ increases from $0$ to $1$.
A: If you consider the Poisson problem, $X\sim \mathcal{P}(\lambda)$, under quadratic loss, if $\delta_a(x)=ax$, then
$$\begin{align*}
R(\lambda,\delta_a)&=\mathbb{E}_\lambda[(aX-\lambda)^2]\qquad\qquad\qquad\qquad\\
&=a^2\text{var}_\lambda(X)+(\mathbb{E}_\lambda[aX]-\lambda)^2\\
&=a^2\lambda+(a\lambda-\lambda)^2\qquad\qquad\\
&=\lambda\{a^2+\lambda(a-1)^2\}
\end{align*}$$
which should help you compare the estimators $\delta_a$. Namely,
$$\begin{align*}R(\lambda,\delta_a)-R(\lambda,\delta_b)&=\lambda\{a^2+\lambda(a-1)^2-b²-\lambda(b-1)²\}\\&=\lambda\{[a²-b²]+\lambda[(a-1)^2-(b-1)²]\}\\&=\lambda\{(a-b)(a+b)+\lambda(a-b)(a+b-2)\}\\&=(a-b)\,\lambda\{(a+b)+\lambda(a+b-2)\}\end{align*}$$
