handouts/Advanced/Random Walks/parts/2 equivalence.tex

\section{The Equivalence}
In the last problem, we found that the equations for $V(x)$ were the same as the equations for $P(x)$ on the same graph.
It turns out that this is true in general: problems about voltage in circuits directly correspond to problems about probability
in graphs. We'll spend the next section proving this fact.

\definition{}
For the following problems, \textit{conductance} will be more convenient than resistance. \par
The definition of conductance is quite simple: 
$$
	C(a, b) = \frac{1}{R(a,b)}
$$
\note[Aside]{
	Resistance is usually measured in Ohms, denoted $\Omega$. \\
	A few good-natured physicists came up with the \say{mho} (denoted \reflectbox{\rotatebox[origin=c]{180}{$\Omega$}})
	as a unit of conductance, which is equivalent to an inverse Ohm.
	Unfortunately, NIST discourages the use of Mhos in favor of the equivalent (and less amusing) \say{Siemens.}
}





\problem{}
Let $x$ be a node in a graph. \par
Let $N_x$ be the set of $x$'s neighbors, $w(x, y)$ the weight of the edge between nodes $x$ and $y$, and $W_x$
the sum of the weights of all edges connected to $x$.

We saw earlier that the probability function $P$ satisfies the following sum:
$$
	P(x) = \sum_{b \in N_x} \biggl( P(b) \times \frac{w(x, b)}{W_x} \biggr)
$$

\note{This was never explicitly stated, but is noted in \ref{weightedgraph}.}

\vspace{4mm}

Use Ohm's and Kirchoff's laws to show that the voltage function $V$ satisfies a similar sum:
$$
	V(x) = \sum_{b \in N_x} \biggl( V(b) \times \frac{C(x, b)}{C_x} \biggr)
$$
where $C(x, b)$ is the conductance of edge $(x, b)$ and $C_x$ is the sum of the conductances of all edges connected to $x$.


\begin{solution}
	First, we know that 
	$$
		\sum_{b \in N_x} I(x, b) = 0
	$$
	for all nodes $x$. Now, substitute $I(x, b) = \frac{V(x) - V(b)}{R(x, y)}$ and pull out $V(x)$ terms to get
	$$
		V(x) \sum_{b \in N_x} \frac{1}{R(x, b)} - \sum_{b \in N_x} \frac{V(b)}{R(x, b)} = 0
	$$

	Rearranging and replacing $R(x, b)^{-1}$ with $C(x, b)$ and $\sum C(x, b)$ with $C_x$ gives us 
	$$
		V(x) = \sum_{b \in N_x} V(b) \frac{C(x, b)}{C_x}
	$$
\end{solution}


\vfill

\pagebreak

Thus, if $w(a, b) = C(a, b)$, $P$ and $V$ satisfy the same system of linear equations. To finish proving that
$P = V$, we now need to show that there can only be one solution to this system. We will do this in the next
two problems.



\problem{}<generaleq>
Let $q$ be a solution to the following equations, where $x \neq a, b$.
$$
	q(x) = \sum_{b \in N_x} \biggl( q(b) \times \frac{w(x, b)}{W_x} \biggr)
$$
Show that the maximum and minimum of $q$ are $q(a)$ and $q(b)$ (not necessarily in this order).

\begin{solution}
	The domain of $q$ is finite, so a maximum and minimum must exist.

	\vspace{2mm}

	Since $q(x)$ is a weighted average of all $q(b), ~b \in N_x$, there exist $y, z \in N_x$ satisfying
	$q(y) \leq q(x) \leq q(z)$. Therefore, none of these can be an extreme point.
	
	\vspace{2mm}

	$A$ and $B$ are the only vertices for which this may not be true, so they must be the minimum and maximum.
\end{solution}

\vfill



\problem{}
Let $p$ and $q$ be functions that solve our linear system \par
and satisfy $p(A) = q(A) = 1$ and $p(B) = q(B) = 0$. \par

\vspace{1mm}

Show that the function $p - q$ satisfies the equations in \ref{generaleq}, \par
and that $p(x) - q(x) = 0$ for every $x$. \note{Note that $p(x) - q(x) = 0 ~ \forall x \implies p = q$}

\begin{solution}
	The equations in \ref{generaleq} for $p$ and $q$ directly imply that
	$$
		[p - q](x) = \sum_{b \in N_x} \biggl( [p - q](b) \times \frac{w(x, b)}{W_x} \biggr)
	$$
	Which are the equations from \ref{generaleq} for $(p - q)$.
	
	\vspace{2mm}

	Hence, the minimum and maximum values of $p - q$ are $[p - q](a) = 1 - 1 = 0$
	and $[p - q](b) = 1 - 1 = 0$.
	
	\vspace{2mm}

	Therefore $p(x) - q(x) = 0$ for all $x$, so $p(x) = q(x)$.
\end{solution}

\vfill
\pagebreak