<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\section{Random Walks}

Consider the graph below. A particle sits on some node $n$. Every second, this particle moves left or
right with equal probability. Once it reaches node $A$ or $B$, it stops. \par
We would like to compute the probability of our particle stopping at node $A$. \par

\vspace{2mm}

In other words, we want a function $P: \text{Nodes} \to [0, 1]$ that maps each node of the graph
to the probability that our particle stops at $A$.

\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[accept] (a) at (0, 0) {$A$};
		\node[main] (x) at (2, 0) {$x$};
		\node[main] (y) at (4, 0) {$y$};
		\node[reject] (b) at (6, 0) {$B$};
	\end{scope}

	\draw[-]
		(a) edge (x)
		(x) edge (y)
		(y) edge (b)
	;
\end{tikzpicture}
\end{center}




\problem{}&lt;firstgraph&gt;
What are $P(A)$ and $P(B)$ in the graph above? \par
\note{Note that these values hold for all graphs.}

\begin{solution}
	$P(A) = 1$ and $P(B) = 0$
\end{solution}

\vfill




\problem{}
Find an expression for $P(x)$ in terms of $P(y)$ and $P(A)$. \par
Find an expression for $P(y)$ in terms of $P(x)$ and $P(B)$. \par

\begin{solution}
	$P(x) = \frac{P(A) + P(y)}{2}$

	\vspace{2mm}
	
	$P(y) = \frac{P(B) + P(x)}{2}$
\end{solution}

\vfill




\problem{}
Use the previous problems to find $P(x)$ and $P(y)$.

\begin{solution}
	$P(x) = \nicefrac{2}{3}$

	\vspace{2mm}
	
	$P(y) = \nicefrac{1}{3}$
\end{solution}

\vfill
\pagebreak




\problem{}&lt;oneunweighted&gt;
Say we have a graph $G$ and a particle on node $x$ with neighbors $v_1, v_2, ..., v_n$. \par
Assume that our particle is equally likely to travel to each neighbor. \par
Find $P(x)$ in terms of $P(v_1), P(v_2), ..., P(v_n)$.

\begin{solution}
	We have
	$$
		P(x) = \frac{P(v_1) + P(v_2) + ... + P(v_n)}{n}
	$$
\end{solution}

\vfill


\problem{}
In general, how do we find $P(n)$ for any node $n$?

\begin{solution}
	If we write an equation for each node other than $A$ and $B$, we have a system of $|N| - 2$
	linear equations in $|N| - 2$ variables.

	\vspace{2mm}

	We still need to show that this system is nonsingular, but
	that's outside the scope of this handout. This could
	be offered as a bonus problem.
\end{solution}

\vfill
\pagebreak





\problem{}
Find $P(n)$ for all nodes in the graph below.

\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[accept] (a) at (0, 0) {$A$};
		\node[main] (x) at (2, 0) {$x$};
		\node[main] (y) at (0, -2) {$y$};
		\node[reject] (b) at (2, -2) {$B$};
	\end{scope}

	\draw[-]
		(a) edge (x)
		(x) edge (b)
		(b) edge (y)
		(y) edge (a)
	;
\end{tikzpicture}
\end{center}

\begin{solution}
	$P(x) = \nicefrac{1}{2}$ for both $x$ and $y$.
\end{solution}

\vfill





\problem{}
Find $P(n)$ for all nodes in the graph below. \par
\note{Note that this is the graph of a cube with $A$ and $B$ on opposing vertices.}

\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[main] (q) at (0, 0) {$q$};
		\node[main] (r) at (2, 0) {$r$};
		\node[main] (s) at (0, -2) {$s$};
		\node[reject] (b) at (2, -2) {$B$};

		\node[accept] (a) at (-1, 1) {$A$};
		\node[main] (z) at (3, 1) {$z$};
		\node[main] (x) at (-1, -3) {$x$};
		\node[main] (y) at (3, -3) {$y$};
	\end{scope}

	\draw[-]
		(a) edge (z)
		(z) edge (y)
		(y) edge (x)
		(x) edge (a)

		(q) edge (r)
		(r) edge (b)
		(b) edge (s)
		(s) edge (q)

		(a) edge (q)
		(z) edge (r)
		(y) edge (b)
		(x) edge (s)
	;
\end{tikzpicture}
\end{center}


\begin{solution}
	$P(z,q, \text{ and } x) = \nicefrac{3}{5}$ \par
	$P(s,r, \text{ and } y) = \nicefrac{2}{5}$
\end{solution}

\vfill
\pagebreak






\definition{}&lt;weightedgraph&gt;
Let us now take a look at weighted graphs. The problem remains the same: we want to compute the probability that
our particle stops at node $A$, but our graphs will now feature weighted edges. The probability of our particle
taking a certain edge is proportional to that edge's weight.

\vspace{2mm}

For example, if our particle is on node $y$ of the graph below, it has a $\frac{3}{8}$ probability of moving
to $x$ and a $\frac{1}{8}$ probability of moving to $z$. \par
\note{Note that $3 + 3 + 1 + 1 = 8$.}


\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[reject] (b) at (0, 0) {$B$};
		\node[main] (x) at (0, 2) {$x$};
		\node[main] (y) at (2, 0) {$y$};
		\node[main] (z) at (4, 0) {$z$};
		\node[accept] (a) at (3, -2) {$A$};
	\end{scope}

	\draw[-]
		(a) edge node[label] {$3$} (y)
		(y) edge node[label] {$1$} (z)

		(b) edge node[label] {$2$} (x)
		(x) edge[bend left] node[label] {$3$} (y)
		(a) edge[bend right] node[label] {$2$} (z)
		(y) edge node[label] {$1$} (b)
	;
\end{tikzpicture}
\end{center}





\problem{}
Say a particle on node $x$ has neighbors $v_1, v_2, ..., v_n$ with weights $w_1, w_2, ..., w_n$. \par
The edge $(x, v_1)$ has weight $w_1$. Find $P(x)$ in terms of $P(v_1), P(v_2), ..., P(v_n)$.


\begin{solution}
	$$
		P(x) = \frac{w_1 P(v_1) + w_2 P(v_2) + ... + w_n P(v_n)}{w_1 + w_2 + ... + w_n}
	$$
\end{solution}

\vfill
\pagebreak







\problem{}
Consider the following graph. Find $P(x)$, $P(y)$, and $P(z)$.

\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[reject] (b) at (3, 2) {$B$};
		\node[main] (x) at (0, 0) {$x$};
		\node[main] (y) at (2, 0) {$y$};
		\node[main] (z) at (1, 2) {$z$};
		\node[accept] (a) at (-2, 0) {$A$};
	\end{scope}

	\draw[-]
		(x) edge node[label] {$1$} (y)
		(y) edge node[label] {$1$} (z)
		(x) edge[bend left] node[label] {$2$} (z)
		(a) edge node[label] {$1$} (x)
		(z) edge node[label] {$1$} (b)
	;
\end{tikzpicture}
\end{center}

\begin{solution}
	$P(x) = \nicefrac{7}{12}$ \par
	$P(y) = \nicefrac{6}{12}$ \par
	$P(z) = \nicefrac{5}{12}$
\end{solution}

\vfill







\problem{}
Consider the following graph. \par
What expressions can you find for $P(w)$, $P(x)$, $P(y)$, and $P(z)$?

\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[accept] (a) at (0, 0) {$A$};
		\node[main] (w) at (2, 1) {$w$};
		\node[main] (x) at (4, 1) {$x$};
		\node[main] (y) at (2, -1) {$y$};
		\node[main] (z) at (4, -1) {$z$};
		\node[accept] (b) at (6, 0) {$B$};
	\end{scope}

	\draw[-]
		(a) edge node[label] {$2$} (w)
		(a) edge node[label] {$1$} (y)
		(w) edge node[label] {$2$} (x)
		(y) edge node[label] {$2$} (z)
		(x) edge node[label] {$1$} (y)
		(x) edge node[label] {$1$} (b)
		(z) edge node[label] {$2$} (b)
	;
\end{tikzpicture}
\end{center}

Solve this system of equations. \par
\hint{Use symmetry. $P(w) = 1 - P(z)$ and $P(x) = 1 - P(y)$. Why?}

\begin{solution}
	$P(w) = \nicefrac{3}{4}$ \par
	$P(x) = \nicefrac{2}{4}$ \par
	$P(y) = \nicefrac{2}{4}$ \par
	$P(z) = \nicefrac{1}{4}$
\end{solution}

\vfill
\pagebreak
</pre></body></html>