325 lines
6.4 KiB
TeX
325 lines
6.4 KiB
TeX
\section{Random Walks}
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Consider the graph below. A particle sits on some node $n$. Every second, this particle moves left or
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right with equal probability. Once it reaches node $A$ or $B$, it stops. \par
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We would like to compute the probability of our particle stopping at node $A$. \par
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\vspace{2mm}
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In other words, we want a function $P(n): N \to [0, 1]$ that returns the probability that our particle stops at $A$,
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where $N$ is the set of nodes in $G$.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[accept] (a) at (0, 0) {$A$};
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\node[main] (x) at (2, 0) {$x$};
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\node[main] (y) at (4, 0) {$y$};
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\node[reject] (b) at (6, 0) {$B$};
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\end{scope}
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\draw[-]
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(a) edge (x)
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(x) edge (y)
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(y) edge (b)
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;
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\end{tikzpicture}
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\end{center}
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\problem{}<firstgraph>
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What are $P(A)$ and $P(B)$ in the graph above? \par
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\note{Note that these values hold for all graphs.}
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\begin{solution}
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$P(A) = 1$ and $P(B) = 0$
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\end{solution}
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\vfill
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\problem{}
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Find an expression for $P(x)$ in terms of $P(y)$ and $P(A)$. \par
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Find an expression for $P(y)$ in terms of $P(x)$ and $P(B)$. \par
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\begin{solution}
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$P(x) = \frac{P(A) + P(y)}{2}$
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\vspace{2mm}
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$P(y) = \frac{P(B) + P(x)}{2}$
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\end{solution}
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\vfill
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\problem{}
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Use the previous problems to find $P(x)$ and $P(y)$.
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\begin{solution}
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$P(x) = \nicefrac{2}{3}$
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\vspace{2mm}
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$P(y) = \nicefrac{1}{3}$
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\end{solution}
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\vfill
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\pagebreak
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\problem{}<oneunweighted>
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Say we have a graph $G$ and a particle on node $x$ with neighbors $v_1, v_2, ..., v_n$. \par
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Assume that our particle is equally likely to travel to each neighbor. \par
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Find $P(x)$ in terms of $P(v_1), P(v_2), ..., P(v_n)$.
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\begin{solution}
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We have
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$$
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P(x) = \frac{P(v_1) + P(v_2) + ... + P(v_n)}{n}
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$$
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\end{solution}
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\vfill
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\problem{}
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How can we use \ref{oneunweighted} to find $P(n)$ for any $n$?
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\begin{solution}
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If we write an equation for each node other than $A$ and $B$, we have a system of $|N| - 2$
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linear equations in $|N| - 2$ variables.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Find $P(n)$ for all nodes in the graph below.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[accept] (a) at (0, 0) {$A$};
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\node[main] (x) at (2, 0) {$x$};
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\node[main] (y) at (0, -2) {$y$};
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\node[reject] (b) at (2, -2) {$B$};
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\end{scope}
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\draw[-]
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(a) edge (x)
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(x) edge (b)
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(b) edge (y)
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(y) edge (a)
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;
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\end{tikzpicture}
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\end{center}
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\begin{solution}
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$P(x) = \nicefrac{1}{2}$ for both $x$ and $y$.
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\end{solution}
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\vfill
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\problem{}
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Find $P(n)$ for all nodes in the graph below. \par
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\note{Note that this is the graph of a cube with $A$ and $B$ on opposing vertices.}
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (q) at (0, 0) {$q$};
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\node[main] (r) at (2, 0) {$r$};
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\node[main] (s) at (0, -2) {$s$};
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\node[reject] (b) at (2, -2) {$B$};
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\node[accept] (a) at (-1, 1) {$A$};
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\node[main] (z) at (3, 1) {$z$};
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\node[main] (x) at (-1, -3) {$x$};
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\node[main] (y) at (3, -3) {$y$};
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\end{scope}
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\draw[-]
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(a) edge (z)
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(z) edge (y)
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(y) edge (x)
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(x) edge (a)
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(q) edge (r)
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(r) edge (b)
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(b) edge (s)
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(s) edge (q)
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(a) edge (q)
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(z) edge (r)
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(y) edge (b)
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(x) edge (s)
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;
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\end{tikzpicture}
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\end{center}
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\begin{solution}
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$P(z,q, \text{ and } x) = \nicefrac{3}{5}$ \par
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$P(s,r, \text{ and } y) = \nicefrac{2}{5}$
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\end{solution}
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\vfill
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\pagebreak
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\definition{}<weightedgraph>
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Let us now take a look at weighted graphs. The problem remains the same: we want to compute the probability that
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our particle stops at node $A$, but our graphs will now feature weighted edges. The probability of our particle
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taking a certain edge is proportional to that edge's weight.
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\vspace{2mm}
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For example, if our particle is on node $y$ of the graph below, it has a $\frac{3}{8}$ probability of moving
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to $x$ and a $\frac{1}{8}$ probability of moving to $z$. \par
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\note{Note that $3 + 3 + 1 + 1 = 8$.}
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[reject] (b) at (0, 0) {$B$};
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\node[main] (x) at (0, 2) {$x$};
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\node[main] (y) at (2, 0) {$y$};
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\node[main] (z) at (4, 0) {$z$};
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\node[accept] (a) at (3, -2) {$A$};
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\end{scope}
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\draw[-]
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(a) edge node[label] {$3$} (y)
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(y) edge node[label] {$1$} (z)
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(b) edge node[label] {$2$} (x)
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(x) edge[bend left] node[label] {$3$} (y)
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(a) edge[bend right] node[label] {$2$} (z)
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(y) edge node[label] {$1$} (b)
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;
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\end{tikzpicture}
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\end{center}
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\problem{}
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Say a particle on node $x$ has neighbors $v_1, v_2, ..., v_n$ with weights $w_1, w_2, ..., w_n$. \par
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The edge $(x, v_1)$ has weight $w_1$. Find $P(x)$ in terms of $P(v_1), P(v_2), ..., P(v_n)$.
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\begin{solution}
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$$
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P(x) = \frac{w_1 P(v_1) + w_2 P(v_2) + ... + w_n P(v_n)}{w_1 + w_2 + ... + w_n}
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$$
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Consider the following graph. Find $P(x)$, $P(y)$, and $P(z)$.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[reject] (b) at (3, 2) {$B$};
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\node[main] (x) at (0, 0) {$x$};
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\node[main] (y) at (2, 0) {$y$};
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\node[main] (z) at (1, 2) {$z$};
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\node[accept] (a) at (-2, 0) {$A$};
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\end{scope}
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\draw[-]
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(x) edge node[label] {$1$} (y)
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(y) edge node[label] {$1$} (z)
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(x) edge[bend left] node[label] {$2$} (z)
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(a) edge node[label] {$1$} (x)
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(z) edge node[label] {$1$} (b)
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;
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\end{tikzpicture}
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\end{center}
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\begin{solution}
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$P(x) = \nicefrac{7}{12}$ \par
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$P(y) = \nicefrac{6}{12}$ \par
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$P(z) = \nicefrac{5}{12}$
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\end{solution}
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\vfill
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\problem{}
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Consider the following graph. \par
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What expressions can you find for $P(w)$, $P(x)$, $P(y)$, and $P(z)$?
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[accept] (a) at (0, 0) {$A$};
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\node[main] (w) at (2, 1) {$w$};
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\node[main] (x) at (4, 1) {$x$};
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\node[main] (y) at (2, -1) {$y$};
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\node[main] (z) at (4, -1) {$z$};
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\node[accept] (b) at (6, 0) {$B$};
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\end{scope}
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\draw[-]
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(a) edge node[label] {$2$} (w)
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(a) edge node[label] {$1$} (y)
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(w) edge node[label] {$2$} (x)
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(y) edge node[label] {$2$} (z)
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(x) edge node[label] {$1$} (y)
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(x) edge node[label] {$1$} (b)
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(z) edge node[label] {$2$} (b)
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;
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\end{tikzpicture}
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\end{center}
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Solve this system of equations. \par
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\hint{Use symmetry. $P(w) = 1 - P(z)$ and $P(x) = 1 - P(y)$. Why?}
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\begin{solution}
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$P(w) = \nicefrac{3}{4}$ \par
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$P(x) = \nicefrac{2}{4}$ \par
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$P(y) = \nicefrac{2}{4}$ \par
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$P(z) = \nicefrac{1}{4}$
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\end{solution}
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\vfill
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\pagebreak
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