Hypothesis testing, power diagram Consider the statistical model $(\mathcal X,\mathcal A,\mathbb P)$.
Let $ϕ$ be a statistic from $(\mathcal X^n, \mathcal A^n)$ to $([0,1],\mathcal B[0,1])$.
Consider the test $H_0$ versus $H_1$ with the decision rule: conditional on $X=x$ the decision will be to reject $H_0$ with probability $ϕ(x)$. 
For example, if $x$ is observed and $ϕ(x)=1/2$ , we flip a coin to decide between the rejection or not of $H_0$. 
In a Neyman Pearson test, we have that $ϕ(x)$ is either $1$, $0$ or $γ$ under some conditions on the likelihoods and a well defined $γ$
$E_0[\phi] = \int \phi(x) dP_0(x)$ is the error that occurs when $H_0$ is rejected even though it's true.
Furthermore we have $E_1[\phi] = \int \phi(x) dP_1(x)$.
The type 1 error is $E_0[\phi]$.
The type 2 error is $1-E_1[\phi]$.
The set
$\mathcal M = \{(E_0,E_1) | (E_0,E_1)=(E_0[\phi],E_1[\phi])$ for some $\phi \}$
is called the power diagram for $(P_0,P_1)$.
My teacher said that this power diagram looks like this: 

where the "upper boundary" represents the Neyman Pearson tests.
He then listed some properties of the power diagram (which we can obviously see on his graph), but I dont understand how he found these ones: 
$(1)$ $\mathcal M$ is symmetric with respect to $y= x$,
$(2)$ the "lower bound" of $\mathcal M$ is convex,
$(3)$ $\mathcal M$ is compact.
How can you find these from the definition of $\mathcal M$? 
And I have another question: Why in a Neyman Pearson test $\phi$ of level $\alpha$ (meaning that $E_0[\phi]=\alpha$) do we necessarily have that $E_1[\phi]>\alpha$ for $P_0 \neq P_1$?
Concerning the diagram: I think that compactness and symmetry are obtained by considering $1-\phi$ and $\alpha \phi$ for $\alpha \in [0,1]$, which are still statistics from $(\mathcal X^n, \mathcal A^n)$ to $([0,1],\mathcal B[0,1])$
It only remains to show the convexity of the lower boundary/concavity of the upper boundary
 A: Consider a simple null $H_0:\theta=\theta_0$ against a simple alternative $H_1:\theta=\theta_1$, assuming that the supports of the underlying distributions under $H_0$ and $H_1$ are not disjoint. 
Define the set $\mathcal M=\{(\alpha_0,\alpha_1): \phi\text{ is a test with }E_{\theta_0}\phi=E_0\phi=\alpha,E_{\theta_1}\phi=E_1\phi=1-\alpha_1\}$.
So $\phi$ is any test with probability of type I error $\alpha_0$ and probability of type II error $\alpha_1$.
Couple of straightforward observations:


*

*$\mathcal M$ is contained in the unit square, i.e. $\mathcal M\subseteq [0,1]\times [0,1]$.

*Suppose $(a,b)\in\mathcal M$. Then $1-\phi$ is a test with $E_0(1-\phi)=1-a$ and $E_1(1-\phi)=1-(1-b)=b$. This means $(1-a,1-b)\in \mathcal M$ whenever $(a,b)\in \mathcal M$.
Now take any two tests $\phi'$ and $\phi''$ with $E_0\phi'=a',E_1\phi'=1-b'$ and $E_0\phi''=a'',E_1\phi''=1-b''$.
Then for any $\lambda\in(0,1)$, the convex combination $\phi=\lambda \phi'+(1-\lambda)\phi''$ is also a test.
Clearly $E_0\phi=\lambda a'+(1-\lambda)a''$ and $E_1\phi=\lambda(1-b')+(1-\lambda)(1-b'')=1-(\lambda b'+(1-\lambda)b'')$.
Therefore, $\lambda(a',b')+(1-\lambda)(a'',b'')\in \mathcal M$ for any $\lambda\in(0,1)$ whenever $(a',b')$ and $(a'',b'')$ are members of $\mathcal M$. This shows $\mathcal M$ is a convex set.
For some more details regarding the properties of $\mathcal M$ you can check out Lehmann-Romano's Testing Statistical Hypotheses (pages 62-63) where the picture of the region is used in a corollary of NP lemma:


