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\begin{document}
\title{Notes on the true distribution}
\author{John Kerl}
\date{\today}

\maketitle

\emph{The notes in this section are temporary scratchwork, needing to be folded
into the main discussion.}

Example with $N=2$:
\begin{align*}
	\pi_1 &= (1)(2) &
	\pi_2 &= (1  2) \\
	%
	r_1(\pi_1) &= 2 &
	r_1(\pi_2) &= 0 \\
	%
	r_2(\pi_1) &= 0 &
	r_2(\pi_2) &= 1 \\
	%
	H(\vecX,\pi_1) &= 2 \alpha_1 &
	H(\vecX,\pi_2) &= 
		\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2 + \alpha_2 \\
	%
	P_\vecX(\pi_1) &= \frac{
			e^{-2 \alpha_1}
		}{
			e^{-2 \alpha_1} +
			e^{-\alpha_2} e^{-\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2}
		} &
	P_\vecX(\pi_2) &= \frac{
			e^{-\alpha_2} e^{-\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2}
		}{
			e^{-2 \alpha_1} +
			e^{-\alpha_2} e^{-\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2}
		} \\
\end{align*}
Then
\begin{align*}
	Y(\vecX) &=
		e^{-2 \alpha_1} +
		e^{-\alpha_2} e^{-\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2} \\
	%
	Z(\Lambda,2) &= \int_{\Lambda^2} d\vecx_1 d\vecx_2
		(
		e^{-2 \alpha_1} +
		e^{-\alpha_2} e^{-\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2}
		) \\
	&= e^{-2 \alpha_1} L^d
		+ e^{-\alpha_2}
		\int_{\Lambda^2}
			d\vecx_1 d\vecx_2 \,
			e^{-\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2}
		\\
	%
	P(\vecX)\,d\vecX &= \frac{
			d\vecx_1 d\vecx_2
			(e^{-2 \alpha_1} +
			e^{-\alpha_2} e^{-\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2})
		} {
			\int_{\Lambda^2} d\vecx_1 d\vecx_2
				(e^{-2 \alpha_1} +
				e^{-\alpha_2} e^{-\frac{1}{2\beta} \|\vecx_1-\vecx_2\|^2})
		}\\
\end{align*}

Let $\vecx_1=(x_1,y_1,z_1)$
and $\vecx_2=(x_2,y_2,z_2)$.
Then (letting $L \to \infty$)
\begin{align*}
	&\int dx_1\, dy_1\, dz_1\, dx_2\, dy_2\, dz_2\,
		e^{-\frac{1}{2\beta} ((x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2)} \\
	&=
		\int dx_1\, dx_2\, e^{-\frac{1}{2\beta} (x_1-x_2)^2}
		\int dy_1\, dy_2\, e^{-\frac{1}{2\beta} (y_1-y_2)^2}
		\int dz_1\, dz_2\, e^{-\frac{1}{2\beta} (z_1-z_2)^2} \\
	&=
		\left(\int dx_1\, dx_2\, e^{-\frac{1}{2\beta} (x_1-x_2)^2}\right)^3.
\end{align*}
What is this integral?  Integrating over $dx_2$ only gives
$\sqrt{2\pi\beta}$, so the integral of that over $dx_1$ is infinite.

\end{document}
