# Let <mi class="MJX-tex-caligraphic" mathvariant="script">X be a sample space, T a test s

Let $\mathcal{X}$ be a sample space, $T$ a test statistic and $G$ be a finite group of transformations (with M elements) from $\mathcal{X}$ onto itself. Under the null-hypothesis the distribution of the random variable $X$ is invariant under the transformations in $G$. Let
$\stackrel{^}{p}=\frac{1}{M}\sum _{g\in G}{I}_{\left\{T\left(gX\right)\ge T\left(X\right)\right\}}.$
Show that $P\left(\stackrel{^}{p}\le u\right)\le u$ for $0\le u\le 1$ under the null hypothesis
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Astok3mpd
Clearly, the distribution of $\stackrel{^}{p}$ is supported by $S:=\left\{0,\frac{1}{M},\frac{2}{M},\dots ,1\right\}$. Since$\mathrm{P}\left(\stackrel{^}{p}\le u\right)=\mathrm{P}\left(\stackrel{^}{p}\le \frac{⌊Mu⌋}{M}\right),$
it is enough to prove the statement for $u\in S$.
To prove the claim, we will use kind of "double counting" argument.
Let $u=\frac{k}{M}\in S$. What does $\stackrel{^}{p}\left(x\right)\le u$ mean? It means that the value $T\left(x\right)$ is among the $k$ largest values in the orbit ${O}_{x}:=\left\{hx,h\in G\right\}$ of $x$. That said, for any $x\in \mathcal{X}$,
$|\left\{y\in {O}_{x}:\stackrel{^}{p}\left(y\right)\le u\right\}|\le k.$
(In the orbit, there are no more than $k$ values, which are among the $k$ largest. There can be less: e.g. in $\left\{1,1,1,2,2,2\right\}$ there are only $3$ values which are among the $4$ largest.) In other words,
$\sum _{h\in G}{\mathbf{1}}_{\stackrel{^}{p}\left(hx\right)\le u}\le k.$
Substitute $x=X$ and take the expectation:
$k\ge \mathrm{E}\left[\sum _{h\in G}{\mathbf{1}}_{\stackrel{^}{p}\left(hX\right)\le u}\right]=\sum _{h\in G}\mathrm{E}\left[{\mathbf{1}}_{\stackrel{^}{p}\left(hX\right)\le u}\right]\phantom{\rule{0ex}{0ex}}=\sum _{h\in G}\mathrm{P}\left(\stackrel{^}{p}\left(hX\right)\le u\right)=\sum _{h\in G}\mathrm{P}\left(\stackrel{^}{p}\left(X\right)\le u\right)=M\phantom{\rule{thinmathspace}{0ex}}\mathrm{P}\left(\stackrel{^}{p}\left(X\right)\le u\right)$
(in the penultimate equality we have used the invariance of distribution under $G$). Equivalently,
$\mathrm{P}\left(\stackrel{^}{p}\left(X\right)\le u\right)\le \frac{k}{M}=u,$
as required.
###### Not exactly what you’re looking for?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee