Proof of the PDF Projection Theorem

Let

$\displaystyle f_x({\bf x}) \stackrel{\mbox{\tiny$\Delta$}}{=}
{ p_x({\bf x}\ve...
...bf z}\vert H_0)} \; f_z({\bf z}) ;\; \mbox{ where } \;\;\; {\bf z}=T({\bf x}),
$

where $ f_z({\bf z})$ is any PDF defined on $ {\cal Z}$. We show now that $ f_x({\bf x})$ is a PDF on $ {\cal X}$, thus

$\displaystyle \int_{{\bf x}\in {\cal X}} \; f_x({\bf x}) \; {\rm d} {\bf x}= 1.
$

Proof:

$\displaystyle \begin{array}{rcl}
{\displaystyle \int_{{\bf x}\in {{\cal X}}}} ...
...\in {\cal Z}}} \; f_z({\bf z}) \; {\rm d} {\bf z} \\
& = & 1.
\end{array} $

The equivalence of the expected values ( $ {\rm E}_{x\vert H_0}$ vs. $ {\rm E}_{z\vert H_0}$ in lines 2 and 3) is an application of the change of variables theorem [75]. For example, let $ h({\bf z})$ be any function defined on $ {\cal Z}$. If $ {\bf z}=T({\bf x})$, then $ {\rm E}_{z}\{ h({\bf z})\} =
{\rm E}_{x}\{h(T({\bf x}))\}$. This can be seen when the expected values are written as the limiting form of the sample mean of a size-$ K$ sample set as $ K\rightarrow \infty$, i.e.

$\displaystyle {\rm E}_{z}\{ h({\bf z})\} = \lim_{k\rightarrow \infty}
\; \frac...
...\frac{1}{K} \; \sum_{k=1}^K\; h(T({\bf x}_k))
= {\rm E}_{x}\{h(T({\bf x}))\}.
$

We show now that $ f_x({\bf x})$ is a member of $ {\cal P}(T,f_z)$.

Proof: Let $ {\bf x}$ be drawn from the PDF $ f_x({\bf x})$ and let $ {\bf t}=T({\bf x})$. We now show that the PDF of $ {\bf t}$ is indeed $ f_z({\bf t})$. We prove this by showing that the moment generating function (MGF) of $ {\bf t}$ is equal to the MGF corresponding to $ f_z({\bf t})$. Let $ M_t({\bf y})$ be the joint moment generating function (MGF) of $ {\bf t}$. By definition,

$\displaystyle \begin{array}{rcl}
M_t({\bf y}) &=& {\rm E}_{t}\left\{ e^{{\bf y...
...\; e^{{\bf y}^\prime {\bf z}}\;
f_z({\bf z}) \; {\rm d} {\bf z},
\end{array} $

from which we may conclude that the PDF of $ {\bf t}$ is $ f_z({\bf t})$.

Another proof of the PPT is available [76].

Baggenstoss 2017-05-19