Proof of the PDF Projection Theorem

Let

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

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

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

Proof:

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

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 [95]. 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.

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

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,

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

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

Another proof of the PPT is available [96].