Posts tagged probability

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Recently, I came across two problems that required the convex hull trick. Often, we have a set of lines, and given a time $t$, we want to find out which line has the maximal (or minimal value). To make this more concrete, here's the first problem in which I came across this technique, Cyclist Race.

Chef has been invited to be a judge for a cycling race.

You may think about cyclists in the race as points on a line. All the cyclists start at the same point with an initial speed of zero. All of them move in the same direction. Some cyclists may force their cycles to speed up. The change of speed takes place immediately. The rest of the time, they move with a constant speed. There are N cyclists in total, numbered from 1 to N.

In order to award points to the participants, Chef wants to know how far the cyclists have reached at certain points of time. Corresponding to the above two, there will be two types of queries.

  • Change the speed of cyclist i at some time.
  • Ask for the position of the race leader at some time.

Return the results of the second type of query.

Now, it's not too hard to see that the distance traveled by each cyclist is a piecewise linear function. For each query of the second type, we could just evaluate all these peicewise linear functions and take the max, but there's a better way.

In this particular problem, the speed is always increasing. So if you look at the thick lines in the title picture that indicate which cyclist is in the lead, it forms the bottom of a convex hull, hence the name, the convex hull trick. The distance of the lead cyclist is also piecewise linear, so the goal becomes to merge the piecewise linear functions of all the cyclist into one.

Essentially, we'll want to create a vector $\left((x_0, l_0), (x_1,l_1),\ldots,(x_{n-1},l_{n-1})\right)$, where $l_i$ is a line, and $x_i$ is the $x$-coordinate at which the line becomes relavant. That is, line $l_i$ has the maximal value on the interval $\left[x_i, x_{i+1}\right]$, where $x_n = \infty$. In this way, if we sort our vector by $x_i$, then to find the maximal value at $x$, we can do a binary search.

To build this vector, we to first have all our lines sorted in ascending order by slope. We iterate through our lines, but we only want to keep some of them. Let us call our convex hull $C$. Initialize $C = \left((-\infty, l_0)\right)$. Now, suppose we encouter line $l$ and we need to make a decision. Recall that the slope $l$ will need be greater than or equal to any line line in $C$. There are several cases.

  1. If $C$ has at least two lines, consider the previous line $l^\prime$ and the line before $l^\prime$, $l^{\prime\prime}$. $l^\prime$ becomes relevant at the intersection with $l^{\prime\prime}$ at say $x^\prime$. If $l$ intersects $l^{\prime\prime}$ at $x$ and $x \leq x^\prime$ $l$ becomes relevant before $l^\prime$, so we remove $l^\prime$. Repeat until we remove as many unnecessary lines as possible. Then, if append to our vector $(x,l)$ where $x$ is the $x$-coordinate of the intersection of $l$ and $l^\prime$.
  2. If $C$ has only one line and $l$ has the same slope as that line, then only keep the line with the bigger $y$-intercept.
  3. If $C$ has only one line and $l$ has a larger slope, then append to our vector $(x,l)$ where $x$ is the $x$-coordinate of the $l$ and the one line in $C$.

See the title picture for an example. We'd initialize with the green line $\left(-\infty, y = 0.1x\right)$. Now, we're in case 3, so we'd add the blue line $(0, y = 0.2x)$ since the two lines intersect at $(0,0)$. For the thick red line, we find ourselves in case 1. We'll pop off the blue line and find ourselves in case 3 again, so now our vector is $C = \left((-\infty, y= 0.1x),(0,y=x/3)\right)$.

The next line we encounter is the green line $y = 0.5x - 0.8$. We're in case 1, but we can't pop off any lines since its intersection with other two lines is far to the right, so we simply append this line and its intersection with the thick red line, so we have $$C = \left((-\infty, y= 0.1x), (0,y=x/3), (4.8,y=0.5x-0.8)\right).$$ Our next line is thick blue line, $y = 2x/3 - 1.4$. This intersects the thick red line at $x = 4.2$, while the thick pervious green line intersects at $x =4.8$, so we'll pop off the green line $y = 0.5x-0.5$, and push the thick blue line on to the stack and get $C = \left((-\infty, y= 0.1x), (0,y=x/3), (4.2,y=2x/3-1.4)\right)$. Finally, we encounter the thick green line and our convex hull is $$ C = \left((-\infty, y= 0.1x), (0,y=x/3), (4.2,y=2x/3-1.4),(5.4, y=2x-8.6)\right). $$

Hopefully, it's clear that the natural data structure to keep track of the lines is a vector, which we'll use as a stack. Then, adding new lines is just a constant time operation. The most time-consuming operation is the initial sorting of the lines, so constructing the convex hull is $O(N\log N)$, where $N$ is the number of lines. Evaluating the distance at some $x$ is $O(\log N)$ using binary search as we discussed earlier. If we have $M$, queries the total running time will be $O\left((N + M)\log N \right)$.

Here's the code for this problem.

#include <algorithm>
#include <iostream>
#include <utility>
#include <vector>

using namespace std;

class Line {
  long double m, b; // y = mx + b
  Line(long double m, long double b) : m(m), b(b) {}
  long double& slope() { return m; }
  long double& yIntercept() { return b; }  
  pair<long double, long double> intersect(Line other) {
    long double x = (this -> b - other.yIntercept())/(other.slope() - this -> m);
    long double y = (this -> m)*x + this -> b;
    return make_pair(x, y);
  double getY(long double x) { return m*x + b; }

vector<pair<long double, Line>> makeConvexHull(vector<Line> &lines) {
  int N = lines.size(); 
  vector<pair<long double, Line>> convexHull; convexHull.reserve(N);
  if (N == 0) return convexHull;
  convexHull.emplace_back(0, lines.front());
  for (int i = 1; i < N; ++i) {
    // pop stack while new line's interesection with second to last line is to the left of last line
    // note that slopes are strictly increasing
    while (convexHull.size() > 1 &&
           lines[i].intersect(convexHull[convexHull.size() - 2].second).first <= convexHull.back().first) {
    // check intersection with x = 0 when there's only 1 line
    if (convexHull.size() == 1 && lines[i].yIntercept() >= convexHull.back().first) convexHull.pop_back();
    convexHull.emplace_back(lines[i].intersect(convexHull.back().second).first, lines[i]);
  return convexHull;

int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false); cin.tie(NULL);
  int N, Q; cin >> N >> Q;      // number of cyclists and queries
  // current speed, time change, distance traveled so far
  vector<Line> lines; // y = mx + b, first is m, second is b
  vector<pair<int, long long>> cyclists(N, make_pair(0, 0)); // first is speed v, second is x, where x + vt is location at time t
  vector<long long> queryTimes;
  for (int q = 0; q < Q; ++q) {
    int queryType; cin >> queryType;
    long long time;
    switch(queryType) {
    case 1:                     // speed change
      int cyclist, newSpeed;
      cin >> time >> cyclist >> newSpeed;
      --cyclist;   // convert to 0 indexing
      // suppose current function is x + vt
      // new function is x + v*time + (t - time)*newSpeed
      // (x + v*time - time*newSpeed) + newSpeed*t
      // = (x + (v-newSpeed)*time) + newSpeed*t
      cyclists[cyclist].second += time*(cyclists[cyclist].first - newSpeed);
      cyclists[cyclist].first = newSpeed;      
      lines.emplace_back(newSpeed, cyclists[cyclist].second);
    case 2:                     // leader position
      cin >> time;
  // sort slopes in ascending order
  sort(lines.begin(), lines.end(),
       [](Line &a, Line &b) -> bool {
         if (a.slope() != b.slope()) return a.slope() < b.slope();
         return a.yIntercept() < b.yIntercept();
  if (lines.size() == 1) {      // there will always be at least 1 line
    for (long long t : queryTimes) {
      cout << (long long) lines.back().getY(t) << '\n';
  } else {
    // eliminate irrelevant lines, first is time where the line becomes relevant
    vector<pair<long double, Line>> convexHull = makeConvexHull(lines); 
    // since times are strictly increasing just walk through lines 1 by 1
    int cursor = 0;
    for (long long t : queryTimes) {
      while (cursor + 1 < convexHull.size() && convexHull[cursor + 1].first <= t) ++cursor;
      cout << (long long) convexHull[cursor].second.getY(t) << '\n';
  cout << flush;
  return 0;

In this particular problem, negative times make no sense, so we can start at $x = 0$ for our convex hull.

Applications to Dynamic Programming

In certain cases, we can apply this trick to a dynamic programming problem to significantly improve the running time. Consider the problem Levels and Regions.

Radewoosh is playing a computer game. There are $N$ levels, numbered $1$ through $N$. Levels are divided into $K$ regions (groups). Each region contains some positive number of consecutive levels.

The game repeats the the following process:

  1. If all regions are beaten then the game ends immediately. Otherwise, the system finds the first region with at least one non-beaten level. Let $X$ denote this region.
  2. The system creates an empty bag for tokens. Each token will represent one level and there may be many tokens representing the same level.
    • For each already beaten level $i$ in the region $X$, the system adds $t_i$ tokens to the bag (tokens representing the $i$-th level).
    • Let $j$ denote the first non-beaten level in the region $X$. The system adds $t_j$ tokens to the bag.
  3. Finally, the system takes a uniformly random token from the bag and a player starts the level represented by the token. A player spends one hour and beats the level, even if he has already beaten it in the past.

Given $N$, $K$ and values $t_1,t_2,\ldots,t_N$, your task is to split levels into regions. Each level must belong to exactly one region, and each region must contain non-empty consecutive set of levels. What is the minimum possible expected number of hours required to finish the game?

It's not obvious at all how the convex hull trick should apply here. Well, at least it wasn't obvious to me. Let's do some math first. Consider the case where we only have 1 region first consisting of levels $1,2,\ldots,n$. Let $T_i$ be the time at which we finish level $i$. Write \begin{equation} T_n = T_1 + (T_2 - T_1) + (T_3 - T_2) + \cdots + (T_n - T_{n-1}). \label{eqn:expectation_Tn} \end{equation}

$T_1 = 1$ since we'll just put $t_i$ tokens in the bag and draw from those $t_i$ tokens, so $t_i/t_i = 1$, so we always play and beat the first level immediately. Now $T_{i} - T_{i-1}$ is the time that it takes to be level $i$ given that levels $1,2,\ldots,i-1$ are beaten. This has a geometric distribution. Every time we try to beat level $i$, we'll put in $t_1 + t_2 + \cdots + t_i$ tokens in the back and try to get one of the $t_i$ tokens labeled $i$. The probability of doing so is \begin{equation} p = \frac{t_i}{\sum_{j=1}^i t_j}. \end{equation} Thus, we'll have that \begin{align} \mathbb{E}(T_i - T_{i-1}) &= p + 2(1-p)p + 3(1-p)p + \cdots = \sum_{k=1}^\infty k(1-p)^{k-1}p \nonumber\\ &= \frac{p}{\left(1-(1-p)\right)^2} = \frac{1}{p} \nonumber\\ &= \frac{\sum_{j=1}^i t_j}{t_i}. \label{eqn:expectation_delta_T} \end{align}

Now by linearity of expectation, applying Equation \ref{eqn:expectation_delta_T} to Equation \ref{eqn:expectation_Tn} will give us that \begin{equation} \mathbb{E}T_n = \sum_{i = 1}^n\frac{\sum_{j=1}^i t_j}{t_i}. \label{eqn:expection_T_n} \end{equation}

Now, define $E_{k,n}$ denote the minimum expected time to beat $n$ levels if we split them into $k$ regions. Note that each region must have at least 1 level, so this is only well-defined if $n \geq k$. Now, the levels must be beaten in order, so imagine putting levels $t_l,t_{l+1},\ldots,t_n$ into the last region. Since each region must have at least 1 level, we'll have that $k \leq l \leq n$, which gives us the recurrence relation \begin{equation} E_{k,n} = \inf_{k \leq l \leq n}\left\{E_{k - 1, l - 1} + \sum_{i = l}^n\frac{\sum_{j=l}^i t_j}{t_i}\right\} \label{eqn:expectation_recurrence} \end{equation} by Equation \ref{eqn:expection_T_n}.

Now, suppose we define \begin{equation} S_k = \sum_{i=1}^k t_i ~\text{and}~ R_k = \sum_{i=1}^k \frac{1}{t_i}, \label{eqn:sum_dp} \end{equation} which leads to \begin{equation} E_{1,n} = \mathbb{E}T_n = \sum_{i=1}^n\frac{S_i}{t_i} \label{eqn:base_expectation} \end{equation} Equation \ref{eqn:expectation_recurrence} becomes \begin{align} E_{k,n} &= \inf_{k \leq l \leq n}\left\{E_{k - 1, l - 1} + \sum_{i = l}^n\frac{\sum_{j=l}^i t_j}{t_i}\right\} \nonumber\\ &= \inf_{k \leq l \leq n}\left\{E_{k - 1, l - 1} + \sum_{i = l}^n\frac{S_i - S_{l-1}}{t_i}\right\} \nonumber\\ &= \inf_{k \leq l \leq n}\left\{E_{k - 1, l - 1} + \sum_{i = l}^n\frac{S_i}{t_i} - S_{l-1}\left(R_n - R_{l-1}\right)\right\} \nonumber\\ &= \inf_{k \leq l \leq n}\left\{E_{k - 1, l - 1} + \left(E_{1,n} - E_{1,l-1}\right) - S_{l-1}\left(R_n - R_{l-1}\right)\right\} \nonumber\\ &= E_{1,n} + \inf_{k \leq l \leq n}\left\{\left(E_{k - 1, l - 1} - E_{1,l-1} + S_{l-1}R_{l-1}\right) - S_{l-1}R_n\right\} \label{eqn:expectation_line} \end{align} by Equations \ref{eqn:base_expectation} and \ref{eqn:sum_dp}.

Do you see the lines of decreasing slope in Equation \ref{eqn:expectation_line}, yet? It's okay if you don't. Look at the expression inside the $\inf$. The $y$-intercept is in parentheses and the slope is $-S_{l-1}$ which is decreasing with $l$. So index our lines by $l$.

Fix $k$. Imagine that we have calculated $E_{j,n}$ for all $j < k$. Now, we're going to attempt to calculate $E_{k,k},E_{k,k+1},\ldots, E_{k,N}$ in that order. To calculate $E_{k,n}$ we only need to consider the lines $l = k,\ldots,n$. The intercept does not vary as a function of $n$, so we can add lines one-by-one. Before calculating $E_k,n$, we'll add the line with slope $-S_{n-1}$. Now our answer will be $E_{K,N}$.

In this way, it simplifies a $O(KN^2)$ problem into a $O(KN\log N)$ problem. Notice that here, instead of building our convex hull before making any queries like in the previous problem, we dynamically update it. Here's the code with some differences since $0$-indexing was used in the code.

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <utility>
#include <vector>

using namespace std;

class Line {
  long double m, b; // y = mx + b
  Line(long double m, long double b) : m(m), b(b) {}
  long double slope() { return m; }
  long double& yIntercept() { return b; }  
  pair<long double, long double> intersect(Line other) {
    long double x = (this -> b - other.yIntercept())/(other.slope() - this -> m);
    long double y = (this -> m)*x + this -> b;
    return make_pair(x, y);
  double getY(long double x) { return m*x + b; }

int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false); cin.tie(NULL);
  int N, K; cin >> N >> K; // number of levels and regions
  vector<int> T; T.reserve(N);  // tokens
  for (int i = 0; i < N; ++i) {
    int t; cin >> t;
  vector<long long> S; S.reserve(N); // cumulative sum of tokens
  vector<long double> R; R.reserve(N); // cumulative sum of token reciprocals
  for (int n = 1; n < N; ++n) {
    S.push_back(S.back() + T[n]);
    R.push_back(R.back() + 1.0L/T[n]);
  /* let eV[k][n] be the minimum expected time to beat
   * levels 0,1,...,n-1 spread across regions 0,1,...,k-1
  vector<vector<long double>> eV(K, vector<long double>(N)); // only indices eV[k][n] with n >= k are valid
  eV.front().front() = 1;
  for (int n = 1; n < N; ++n) { // time to beat each level is a geometric distribution
    eV[0][n] = eV[0][n-1] + ((long double) S[n])/T[n];
  /* eV[k][n] = min(eV[k-1][l-1] + (S[l]-S[l-1])/t_l + ... + (S[n] - S[l-1])/t_{n}),
   * where we vary k <= l < n. That is, we're putting everything with index at least l
   * in the last region. Note that each region needs at least 1 level. This simplifes to
   * eV[k][n] = min(eV[k-1][l-1] + eV[0][n] - eV[0][l-1]  - S[l-1](R[n] - R[l-1])
   *          = eV[0][n] + min{(eV[k-1][l-1] - eV[0][l-1] + S[l-1]R[l-1]) - S[l-1]R[n]},
   * Thus, for each l we have a line with slope -S[l - 1].
   * -S[l-1] is strictly decreasing and R[n] is strictly increasing.
   * Use the convex hull trick.
  vector<pair<long double, Line>> convexHull; convexHull.reserve(N);
  for (int k = 1; k < K; ++k) {    
    /* in convex hull we add lines in order of decreasing slope, 
     * the the convexHull[i].first is the x-coordinate where 
     * convexHull[i].second and convexHull[i-1].second intersect
    int cursor = 0;
    for (int n = k; n < N; ++n) { // calculate eV[k][n]
      /* add lines l = k,...,n to build convex hull
       * loop invariant is that lines l = k,...,n-1 have been added, so just add line l = n
      long double slope = -S[n - 1];
      long double yIntercept = eV[k - 1][n - 1] - eV[0][n - 1] + S[n - 1]*R[n - 1];
      Line l(slope, yIntercept);
      while (convexHull.size() > 1 && 
             convexHull.back().first >= l.intersect(convexHull[convexHull.size() - 2].second).first) {
        convexHull.pop_back(); // get rid of irrelevant lines by checking that line intersection is to the left
      // check intesection with x = 0 if no lines left
      if (convexHull.size() == 1 && l.yIntercept() <= convexHull.back().second.yIntercept()) convexHull.pop_back();
      convexHull.emplace_back(convexHull.empty() ? 0 : l.intersect(convexHull.back().second).first, l); // add line
      /* binary search for the correct line since they are sorted by x intersections
       * lower bound is old cursor since x coordinate of intersection is monotonically decreasing
       * and R[n] is increasing, too
      if (cursor >= convexHull.size()) cursor = convexHull.size() - 1;
      cursor = upper_bound(convexHull.begin() + cursor, convexHull.end(), R[n], 
                           [](long double x, pair<long double, Line> l) { return x < l.first; }) - convexHull.begin();
      --cursor; // since we actually want the last index that is less than or equal to
      eV[k][n] = eV[0][n] + convexHull[cursor].second.getY(R[n]);

  cout << fixed;
  cout << setprecision(10);
  cout << eV.back().back() << endl;
  return 0;

By the way the drawing was done with GeoGebra. It's a pretty cool way to do math.

Photo URL is broken

I recently added equation numbering to my blog. To see it in action, here's one of my homework exercises. I like this problem because I actually found it useful to diagonalize to exponentiate the matrix.

Let $\{X_n\}$ be the random walk on the space $\{0,1,2,3,4\},$ with $0$ and $4$ absorbing and with $p(i,i) = 1/2,$ $p(i,i+1) = p(i,i-1)= 1/4$ for $1 \leq i \leq 3.$ Let $\tau$ denote the absorption time. The graph can be seen in the title picture.

Part A

What is the limit as $n \rightarrow \infty$ of the law $X_n$ conditioned on non-absorption, that is, $$\lim_{n \rightarrow \infty}\mathcal{L}(X_n \mid \tau > n)?$$


We have that $\displaystyle \boxed{ \lim_{n \rightarrow \infty}\mathcal{L}\left(X_n \mid \tau > n\right) = \begin{cases} 1 - \frac{1}{\sqrt{2}}, &X_n \in \{1,3\} \\ \sqrt{2} - 1, &X_n = 2. \end{cases}}$

To see this, without loss of generality, let $X_0 = 2.$ We can do this since if $X_0 = 1$ or $X_0 = 3,$ then $\mathbb{P}(X_m = X_0,~\forall m \leq n \mid \tau > n) \rightarrow 0$ as $n \rightarrow \infty.$ Then, we can apply the Markov property.

Now, define $p_n^*(x,y) = \mathbb{P}(X_{n} = y \mid \tau > n,~X_0 = 2).$ We have that
\begin{equation} p^*_1(2,y) = \begin{cases} 1/4, &y \in \{1,3\} \\ 1/2, &y = 2. \end{cases} ~\text{and, in general,}~ p_n^*(2,y) = \begin{cases} a_n, &y \in \{1,3\} \\ b_n = 1 - 2a_n, &y = 2 \end{cases} \label{eqn:3a_a_def} \end{equation} by symmetry, where $a_1 = 1/4$ and $b_1 = 1/2 = 1 - 2a_1.$

Now, we have that \begin{align} a_{n+1} = \mathbb{P}(X_{n+1} = 1 \mid \tau > n + 1,~X_0 = 2) &= p_{n+1}^*(2,1) \nonumber\\ &= \frac{p_n^*(2,1)p(1,1) + p_n^*(2,2)p(2,1)}{\mathbb{P}(\tau > n + 1 \mid X_0 = 2, \tau > n)} \nonumber\\ &= \frac{p_n^*(2,1)p(1,1) + p_n^*(2,2)p(2,1)}{1 - p_n^*(2,1)p(1,0) - p_n^*(2,3)p(3,4)} \nonumber\\ &= \frac{a_n(1/2) + b_n(1/4)}{1 - (2a_n/4)} \nonumber\\ &= \frac{a_n(1/2) + (1-2a_n)(1/4)}{(4 - 2a_n)/4} \nonumber\\ &= \frac{1}{4 - 2a_n}. \label{eqn:3a_a_recurs} \end{align} $a_n$ converges by induction because $$ |a_{n+1} - a_n| = \left|\frac{1}{4-2a_n} - \frac{1}{4-a_{n-1}}\right| = \left|\frac{4 - 2a_{n-1} - 4 + 2a_n}{(4-2a_{n})(4-2a_{n-1})}\right| \leq \frac{1}{2}\left|a_n-a_{n-1}\right| $$ since $0 \leq a_n \leq 1$ for all $n.$ Solving for $a$ in \begin{equation} a = \frac{1}{4-2a} \Rightarrow 2a^2 - 4a + 1 = 0 \Rightarrow a = 1 \pm \frac{1}{\sqrt{2}}. \label{eqn:3a_a} \end{equation}
Thus, we must have that $a_n \rightarrow 1 - \frac{1}{\sqrt{2}}$ since $a_n \leq 1$ for all $n.$ Then, we have that $b_n \rightarrow b = 1 - 2a = \sqrt{2} - 1.$

Part B

What is the limiting distribution conditioned on never getting absorbed, that is, $$ \lim_{n \rightarrow \infty}\lim_{M \rightarrow \infty} \mathcal{L}(X_n \mid \tau > M)? $$


We have that $\displaystyle\boxed{\lim_{n \rightarrow \infty}\lim_{M \rightarrow \infty}\mathcal{L}(X_n \mid \tau > M)= \begin{cases}1/4, &X_n \in \{1,3\}\\ 1/2, &X_n = 2.\end{cases}}$

Again, we can assume without loss of generality that $X_0 = 2.$ Fix $n \in \mathbb{N}.$ From Equation \ref{eqn:3a_a_def} in the previous part, we know $\mathbb{P}(X_n = 2\mid \tau > n) = 1-2a_{n},$ where we define $a_0 = 0.$ Now, suppose, we know $\mathbb{P}(X_n = 2 \mid \tau > M)$ for $M \geq n.$ Then, by Bayes' theorem, \begin{align} \mathbb{P}(X_n = 2 \mid \tau > M + 1) &= \mathbb{P}(X_n = 2 \mid \tau > M + 1, \tau > M) \nonumber\\ &= \frac{\mathbb{P}(\tau > M + 1 \mid X_n = 2, \tau > M)\mathbb{P}(X_n = 2 \mid \tau > M)}{\mathbb{P}(\tau > M + 1 \mid \tau > M)}. \label{eqn:3b_bayes} \end{align} Calculating the two unknown factors, we have that in the denominator, \begin{align} \mathbb{P}(\tau > M + 1 \mid \tau > M) &= \sum_{x=1}^3 \mathbb{P}(\tau > M + 1, X_M = x \mid \tau > M) \nonumber\\ &= \frac{3}{4}\left(\mathbb{P}(X_M = 1 \mid \tau > M) + \mathbb{P}(X_M = 3 \mid \tau > M)\right) + \mathbb{P}(X_M = 2 \mid \tau > M)\nonumber\\ &= \frac{3}{4}\left(a_{M} + a_{M}\right) + (1-2a_{M}) \nonumber\\ &= 1 - \frac{1}{2}a_{M}, \label{eqn:3b_tau} \end{align} and in the numerator, \begin{align} \mathbb{P}(\tau > M + 1 \mid X_n = 2, \tau > M) &= \sum_{x=1}^3\mathbb{P}(\tau > M + 1, X_M = x \mid X_n = 2, \tau > M)\nonumber\\ &= \sum_{x=1}^3\mathbb{P}(\tau > M + 1, X_M = x \mid X_n = 2, \tau > M)\nonumber\\ &= \frac{3}{2}\mathbb{P}(X_M = 1 \mid X_n = 2, \tau > M) + \mathbb{P}(X_M = 2 \mid X_n = 2, \tau > M) \nonumber\\ &= \frac{3}{2}a_{M - n} + 1 - 2a_{M - n} \nonumber\\ &= 1 - \frac{1}{2}a_{M - n}, \label{eqn:3b_bayes_num} \end{align} where we use that $\mathbb{P}(\tau > M + 1, X_M = 1 \mid X_n = 2, \tau > M) = \mathbb{P}(\tau > M + 1, X_M = 3 \mid X_n = 2, \tau > M),$ and the fact that $\mathbb{P}(X_M = x \mid X_n = 2, \tau > M)$ is the probability of a chain starting at $2$ that doesn't hit an absorbing state in $M - n$ transistions.

Putting together Equations \ref{eqn:3b_bayes_num}, \ref{eqn:3b_tau}, and \ref{eqn:3b_bayes}, we have that \begin{equation*} \mathbb{P}(X_n = 2 \mid \tau > M + 1) = \frac{1 - \frac{1}{2}a_{M-n}}{1 - \frac{1}{2}a_{M}}\mathbb{P}(X_n = 2 \mid \tau > M), \end{equation*} and by induction, we'll have that \begin{equation*} \mathbb{P}(X_n = 2 \mid \tau > M) = (1 - 2a_{n}) \prod_{k=n}^{M - 1} \frac{1 - \frac{1}{2}a_{k-n}}{1 - \frac{1}{2}a_{k}}, \end{equation*} where the product is just $1 - 2a_{n}$ if $M = n.$ For large enough $M,$ when $k \geq 2n$ the factors in the numerator will cancel, so we will have that for $M \geq 2n$ that \begin{align} \mathbb{P}(X_n = 2 \mid \tau > M) &= (1 - 2a_{n}) \prod_{k=0}^{n - 1}\left(1-\frac{1}{2}a_{k}\right) \prod_{k=M - n}^{M-1}\frac{1}{1-\frac{1}{2}a_{k}} \nonumber\\ &= (1 - 2a_{n}) \prod_{k=0}^{n - 1}\left(4-2a_{k}\right) \prod_{k=M - n}^{M-1}\frac{1}{4-2a_{k}} \nonumber\\ &= (1 - 2a_{n}) \prod_{k=1}^{n}\frac{1}{a_{k}} \prod_{k=M - n + 1}^{M}a_k \label{eqn:3b_p2} \end{align} by the recursive definition of $a_n$ in Equation \ref{eqn:3a_a_recurs}.

Taking the limit as $M \rightarrow \infty,$ we have that \begin{equation} \lim_{M \rightarrow \infty} \mathbb{P}(X_n = 2 \mid \tau > M) = (1 - 2a_{n})a^{n}\prod_{k=1}^{n}\frac{1}{a_{k}}. \label{eqn:3b_p_lim} \end{equation} by Equation \ref{eqn:3a_a}, where $\displaystyle a = 1 - \frac{1}{\sqrt{2}} = \frac{1}{2 + \sqrt{2}}.$

Define $c_{-1} = 0$ and $c_0 = 1/4,$ so $a_0 = 0 = \frac{c_{-1}}{c_0}.$ In general, we can write $a_n = \frac{c_{n-1}}{c_n},$ where $c_n = -2c_{n-2} + 4c_{n-1}$ for $k \geq 1.$ To see this, by definition of $a_n$ in Equation \ref{eqn:3a_a_recurs}, \begin{equation} a_{n + 1} = \frac{1}{4 - 2a_n} = \frac{1}{4 - 2\frac{c_{n-1}}{c_{n}}} = \frac{c_n}{-2c_{n-1} + 4c_n} = \frac{c_n}{c_{n+1}}. \label{eqn:3b_a_c} \end{equation}

Now, with Equation \ref{eqn:3b_a_c}, Equation \ref{eqn:3b_p_lim} becomes \begin{equation} \lim_{M \rightarrow \infty} \mathbb{P}(X_n = 2 \mid \tau > M) = (1 - 2a_{n})a^{n}\prod_{k=1}^{n}\frac{c_k}{c_{k-1}} = (1 - 2a_{n})a^{n}\frac{c_{n}}{c_0} \label{eqn:3b_p_lim_2} \end{equation}

Note, that we can rewrite $c_n$ by exponentiating a matrix: \begin{align} \begin{pmatrix} c_{n} \\ c_{n + 1} \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ -2 & 4 \end{pmatrix}^n \begin{pmatrix} c_0 \\ c_1 \end{pmatrix} \nonumber\\ &= \begin{pmatrix} 1 & 1 \\ 2 + \sqrt{2} & 2 - \sqrt{2} \end{pmatrix} \begin{pmatrix} 2 + \sqrt{2} & 0 \\ 0 & 2 - \sqrt{2} \end{pmatrix}^n \begin{pmatrix} \frac{1}{2}-\frac{1}{\sqrt{2}} & \frac{1}{2\sqrt{2}} \\ \frac{1}{2}+\frac{1}{\sqrt{2}} & -\frac{1}{2\sqrt{2}} \end{pmatrix} \begin{pmatrix} 1/4 \\ 1 \end{pmatrix} \nonumber\\ &= \begin{pmatrix} \frac{1}{8}\left((2+\sqrt{2})^n(1+\sqrt{2}) + (2-\sqrt{2})^n(1-\sqrt{2})\right) \\ \frac{1}{8}\left((2+\sqrt{2})^{n+1}(1+\sqrt{2}) + (2-\sqrt{2})^{n+1}(1-\sqrt{2})\right) \end{pmatrix} \label{eqn:3b_c_matrix} \end{align} by diagonalization. Using Equation \ref{eqn:3b_c_matrix}, Equation \ref{eqn:3b_p_lim_2} becomes \begin{equation} \lim_{M \rightarrow \infty} \mathbb{P}(X_n = 2 \mid \tau > M) = \frac{1}{2}(1 - 2a_{n})\left((1+\sqrt{2}) + \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)^n(1-\sqrt{2})\right), \label{eqn:3b_p_lim_3} \end{equation} where we recall that $a = \frac{1}{2 + \sqrt{2}}.$ $|2-\sqrt{2}| < 1$ and so, taking the limit as $n \rightarrow \infty$ of Equation \ref{eqn:3b_p_lim_3}, the second term goes to $0$: \begin{equation} \lim_{n\rightarrow 0}\lim_{M \rightarrow \infty} \mathbb{P}(X_n = 2 \mid \tau > M) = \frac{1-2a}{2}(1+\sqrt{2}) = \frac{\sqrt{2} - 1}{2}(1+\sqrt{2}) = \frac{1}{2}. \label{eqn:3b_p_lim_lim} \end{equation} And finally, by symmetry, \begin{align} \lim_{n\rightarrow 0}\lim_{M \rightarrow \infty} \mathbb{P}(X_n = 1 \mid \tau > M) &= \lim_{n\rightarrow 0}\lim_{M \rightarrow \infty} \mathbb{P}(X_n = 3 \mid \tau > M) \nonumber\\ &= \frac{1 - \lim_{n\rightarrow 0}\lim_{M \rightarrow \infty} \mathbb{P}(X_n = 2 \mid \tau > M)}{2} \nonumber\\ &= \frac{1}{4}. \label{eqn:3b_p_1_3} \end{align} Equations \ref{eqn:3b_p_lim_lim} and \ref{eqn:3b_p_1_3} combine to give us the limiting distribution.

Graphs in $\LaTeX$

Here's the code for the graph. I used a package called tikz.




\usepackage[active, tightpage]{preview}

\renewcommand\PreviewBbAdjust {-0bp -25bp 0bp 25bp}

\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.75cm,
  thick,main node/.style={circle,draw}]

  \node[main node] (0) {$0$};
  \node[main node] (1) [right of=0] {$1$};
  \node[main node] (2) [right of=1] {$2$};
  \node[main node] (3) [right of=2] {$3$};
  \node[main node] (4) [right of=3] {$4$};

  \path[every node/.style={font=\sffamily}]
  (0) edge [loop above] node {$1$} (0)
  (1) edge [bend right] node [below] {$1/4$} (2)
  edge [loop above] node {$1/2$} (1)
  edge node [below] {$1/4$} (0)
  (2) edge [bend right] node [above] {$1/4$} (1)
  edge [loop below] node {$1/2$} (2)
  edge [bend left] node [above] {$1/4$} (3)
  (3) edge [bend left] node [below] {$1/4$} (2)
  edge [loop above] node {$1/2$} (3)
  edge node [below] {$1/4$} (4)
  (4) edge [loop above] node {$1$} (4);      

Then, to convert the graph to a png file with a transparent background, I used ImageMagick.

convert -density 300 graph.pdf -format png graph.png

In probability, math contests, and programming contests, we often need to count. Here, I'll write about a few cases that I see pretty often. Before we jump into things, recall the binomial coefficient and various ways of calculating it: $$ {n \choose k} = \frac{n!}{n!(n-k)!} = {n - 1 \choose k } + {n - 1 \choose k - 1 }, $$ where $n \geq k$ and is $0$ if $n < k.$ Thus, we compute the binomial coefficient with dynamic programming by using a triangular array: \begin{matrix} 1 \\ 1 & 1 \\ 1 & 2 & 1\\ 1 & 3 & 3 & 1\\ 1 & 4 & 6 & 4 & 1\\ \vdots & \vdots & \vdots & \ddots & \ddots & \ddots \end{matrix} where if we define $$ X_{n,k} = \begin{cases} {n \choose k}, &k \leq n \\ 0, &k > n, \end{cases} $$ then ${n \choose k} = X_{n,k} = X_{n-1,k} + X_{n-1,k-1}.$

We will see that we can count many things in this manner.

From $n$ objects with replacement

Some cases when we draw from $n$ objects with replacement are an infinite deck of cards, assigning types or categories, or drawing from $\{1,2,3,4,5,6\}$ by rolling a dice.

Ordered set of $k$ objects

If we are sampling $k$ objects from $n$ objects with replacement, each of the $k$ objects has $n$ possibilities. Thus, there are are $\boxed{n^k}$ possibilities.

For example, if $k$ distinct people are buying from a selection of $n$ ice cream flavors, there are $n^k$ different ways, this group of people can order. Another common situation is a sequence of $k$ bits. In this case $n = 2,$ so there are $2^k$ possible sequences.

Unordered set of $k$ objects

Another way to think of this is putting $k$ balls into $n$ bins, where the balls are not distinct. The method for solving this problem is also known as stars and bars.

Imagine each ball as a star, so we have $k$ of them. Along with the stars we have $n - 1$ bars. Now arrange these $(n-1) + k$ objects in any order. For example, take $$\star\star \mid \mid \star \mid \star\star\star.$$ This order would correspond to $2$ balls in the bin $1$, $0$ balls in bin $2$, $1$ ball in bin $3$, and $3$ balls in bin $4$. Thus, we have $(n-1 + k)!$ orderings if the objects were distinct. Since the $k$ balls and $n-1$ bars are identical, we divide by $(n-1)!$ and $k!$. Thus, the number of possible sets is $$ \frac{(n-1 + k)!}{(n-1)!k!} = \boxed{{n + k - 1\choose k}.} $$

From $n$ objects without replacement

This scenario common occurs where we have a finite collection of objects such as a deck of cards.

Unordered set of $k$ objects

We might see this situation when counting the number of $5$-card hands in poker for instance. The order that you draw the cards doesn't matter.

If we have $n$ cards, for the first card there are $n$ possibilities. For the next card, there are $n-1$ possibilities. Thus, if we draw $k$ cards, we have $n(n-1)\cdots(n-k+1) = n!/(n-k)!$ possible draws. Since the order doesn't matter, we divide by $k!.$ Thus, the count is $$\frac{n!}{(n-k)!k!} = \boxed{{n \choose k}.}$$

This makes the formula $${n \choose k} = {n-1 \choose k} + {n - 1 \choose k -1}$$ for computing the binomial coefficient intuitive. Imagine trying to choose $k$ objects from $n$ objects. We can either include or not include the $n$th object. If we don't include the $n$th object we choose $k$ objects from the first $n-1$ objects, which gives us the ${n-1 \choose k}$ term. If we do include the $n$th object then, we only choose $k-1$ objects from the first $n-1$ objects, which gives us the ${n-1 \choose k - 1}$ term.

Another common use of the of binomial coefficient is counting paths. Suppose we are on a grid, we can only make right and down moves, and we are trying to get from $A$ to $B$, where $B$ is $k$ moves to the right and $l$ moves downward. Our set of $n = k + l$ objects is $\{1, 2,\ldots,n\}.$ We choose $k$ indices to make a right move, and the rest of the moves will be down. Then, the number of paths is ${n \choose k}.$

Ordered set of $k$ objects

In this case, we care about the order of the cards we draw. From the discussion above, it's calculated the same way as an unordered set of $k$ objects except we don't divide by $k!$, so the number of possible draws is $$ n(n-1)\cdots(n-k+1) = \boxed{\frac{n!}{(n-k)!} = (n)_k,} $$ where the I have used the Pochhammer symbol to denote the falling factorial.

Recontres Numbers

These count the number of permutations with a certain number of fixed points. A permutation on $n$ elements is an element of the symmetric group $S_n$. For those without a background in algebra, it is essentially a way of rearranging $n$ objects. From our discussion of above, there are $n!$ ways to do so. For example, $$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4\\ 3 & 2 & 4 & 1\\ \end{pmatrix}$$ reorders $(1,2,3,4)$ as $(3,2,4,1)$. $\sigma$ can be thought of as a function $\sigma(1) = 3$, $\sigma(2) = 2$, $\sigma(3) = 4$, and $\sigma(4) = 1$. Since for only $x = 2$ do we have $\sigma(x) = x$, there is only $1$ fixed point.

Now let $D_{n,k}$ be the number of permutations of $n$ objects with $k$ fixed points. Let $D_{0,0} = 1$. Clearly, $D_{1,0} = 0$ since $\sigma(1) = 1$ is the only possible permutation for $n = 1.$ Then, we have the recursive formula for $k > 0$ $$ D_{n,k} = {n \choose k}D_{n-k,0}, $$ which can be thought of as taking an unordered set of $k$ points to be fixed from $\{1,2,\ldots,n\}$, hence the ${n \choose k}$. For the remaining $n - k$ points, we have no fixed points because we want exactly $k$ fixed points.

So if we know $D_{0,0}, D_{1,0},\ldots,D_{n,0}$, we can calculate $D_{n+1,1}, D_{n+1,2},\ldots, D_{n+1,n+1}$. Then, for $k = 0$, since there are $(n+1)!$ total permutations, we have that $$ D_{n+1,0} = (n+1)! - \sum_{k=1}^{n+1}D_{n+1,k}. $$ There's a better way to calculate $D_{n,0}$, though, which I learned at CTY. These permutations with no fixed points are called derangements. Clearly, $D_{0,0} = 1$ and $D_{1,0} = 0$.

Now assume $n \geq 2$. Focus on element $n$. A permutation can be thought of as disjoint cycles. Recall the notation in abstract algebra, where we may write $$\sigma = \begin{pmatrix}1 & 2 & 3\end{pmatrix}\begin{pmatrix}4 & 5\end{pmatrix}\in S_5,$$ which gives us $2$ cycles, one of length $3$ and the other of length $2$. A cycle of length $1$ is a fixed point, so $n$ is part of a cycle of length $2$ or more than $2$. In the case that $n$ is part of a cycle of length $2$, there are $n-1$ options for the other element in the cycle. The number of ways to permute the remaining elements is $D_{n-2,0}$. In the case that $n$ is part of a cycle of length greater than $2$, we can consider permuting the first $n-1$ elements with no fixed points. For each such permutation, we have $n - 1$ elements after which we can insert element $n$, so it becomes part of an existing cycle. In this way, we have that $$ D_{n,0} = (n-1)\left(D_{n-1,0} + D_{n-2,0}\right). $$

Again, we have a triangular array \begin{matrix} 1 \\ 0 & 1 \\ 1 & 0 & 1\\ 2 & 3 & 0 & 1\\ 9 & 8 & 6 & 0 & 1\\ \vdots & \vdots & \vdots & \ddots & \ddots & \ddots \end{matrix}

One helpful way to visualize this process that I like is to imagine a dance class with $n$ couples. After each dance everyone has to find a new partner. There are $D_{n,k}$ ways that $k$ couples stay the same.

Bell Numbers

The Bell numbers count the ways to partition a set. Consider the set $S = \{1,2,3\}$. The possible nonempty subsets are $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, $\{2,3\}$, $\{1,3\}$, and $\{1,2,3\}$. A partition would be a group of disjoint nonempty subsets such that each element of $S$ is an element of some subset in the partition. Thus, our partitions are $\left\{\{a\},\{b\},\{c\}\right\}$, $\left\{\{a\},\{b,c\}\right\}$, $\left\{\{a, c\},\{b\}\right\}$, $\left\{\{a, b\},\{c\}\right\}$, and $\left\{\{a, b,c\}\right\}$.

Let $B_n$ be number of ways to partition a set of $n$ objects. $B_0 = 1$, $B_1 = 1$, $B_2 = 2$ and $B_3 = 5$ for example. In general to calculate $B_{n+1}$, we have the recurrence relation $$ \boxed{B_{n+1} = \sum_{k=0}^n{n \choose k}B_{n-k} = \sum_{k=0}^n{n \choose k}B_{k}} $$ since ${n \choose k} = {n \choose n-k}$. To see this, consider partitions of the set $\{1,2,\ldots,n,n+1\}$. In the partition there is a subset that contains $n+1$, say $S$. We can have $|S| = k + 1 \in \{1,2,\ldots,n,n+1\}$. Clearly, $n+1 \in S$. Choosing the other $k$ elements of $S$ amounts to selecting an unordered set from $\{1,2,\ldots,n\}$, hence the ${n \choose k}$ factor in each term. For the remaining $n + 1 - (k+1) = n - k$ objects there are $B_{n-k}$ ways to partition them. Thus, we have the terms ${n \choose k}B_{n-k}$. We avoid double counting since the partitions corresponding to each term are disjoint because $n+1$ is in a subset of different size.

Catalan Numbers

Consider strings of $n$ $0$s and $n$ $1$s, so the string has length $2n$. Call this set $\Omega^{2n}$. Let $C_n$ be the number of such strings where no initial substring has more $1$s than $0$s. Formally, $$C_n = \left|\left\{ X = (x_1,x_2,\ldots,x_{2n}) \in \{0,1\}^{2n} : \sum_{i=1}^{2n} x_i = n, \sum_{i=1}^k x_i \leq \frac{k}{2}, \forall k \in \{1,2,\ldots,2n\}\right\}\right|.$$ $C_n$ is the $n$th Catalan number.

To calculate these numbers first note that $C_0 = 1$ and that every $X \in \Omega^{2(n+1)}$ for $n \geq 1$ can be written $$X = (0,X_1,1,X_2),$$ where $X_1$ and $X_2$ are elements of of $\Omega^{2k}$ and $\Omega^{2(n-k)}$, respectively for some $k \in \{0,1,\ldots,n\}$. Such a form is unique. To see this, let $X = (x_1,x_2,\ldots,x_{2n},x_{2n+1},x_{2n+2})$. Note that by the defintion, the first number in the string must be a $0$. Since the total numbers of $0$s and $1$s in the sequence must be equal, there exists an even index $j$ such that $\sum_{i=1}^j x_i = j/2$. Fix $j$ to be the smallest such index. We must have that $x_j = 1$ since otherwise the defintion would have been violated as $\sum_{i=1}^{j-1}x_i = j/2 > (j-1)/2$.

Then, we'll have $$X = (x_1 = 0,X_1,x_j = 1,X_2),$$ where $X_1$ is a string of length $2k = j-2$ and $X_2$ has length $2n + 2 - j = 2n-2k$. We show that $X_1 \in \Omega^{2k}$ and $X_2 \in \Omega^{2(n-k)}$. Since there are an equal number of $0$s and $1$s at index $j$, $X_1$ must have an equal number of $0$s and $1$s. If at any point $1 \leq l \leq 2k$, we have that $\sum_{i=2}^{l + 1}x_i > l/2$ then $\sum_{i=1}^{l+1}x_i \geq (l+1)/2$, which implies that $X \not\in \Omega^{2(n+1)}$ or that there is an index smaller than $j$ such that the initial substring has an equal number of $0$s and $1$s since $l+1 \leq j-1$. Both are a contradiction so we have $X_1 \in \Omega^{2k}$. Showing $X_2 \in \Omega^{2(n-k)}$ is similar. We have that $X_2$ must have an equal number of $0$s and $1$s in order for the whole string to have an equal number of $0$s and $1$s. If for any $1 \leq l \leq 2(n-k)$, we have that $\sum_{i=j+1}^{j+l}x_i > l/2$, then $$ \sum_{i=1}^{j+l}x_i = \sum_{i=1}^{j}x_i + \sum_{i=j+1}^{j+l}x_i = \frac{j}{2} + \sum_{i=j+1}^{j+l}x_i > \frac{j}{2} + \frac{l}{2} = \frac{j+l}{2}, $$ which implies that $X \not\in \Omega^{2(n+1)}$, which is a contradiction.

Thus, we have our desired result that $X = (x_1 = 0,X_1,x_j = 1,X_2)$, where $X_1 \in \Omega^{2k}$ and $X_2 \in \Omega^{2(n-k)}$, where $k \in \{0,1,\ldots,n\}$. Varying $k$, we come upon the recurrence relation $$\boxed{C_{n+1} = \sum_{k=0}^nC_kC_{n-k}.}$$ This is a pretty nice solution, but we can actually do better and find a closed-form solution.

Consider the generating function $$ c(x) = \sum_{n=0}^\infty C_nx^n = 1 + \sum_{n=1}^\infty C_nx^n = 1 + x\sum_{n=0}^\infty C_{n+1}x^n. $$ Substituting in the recurrence relation, for $C_{n+1}$, we have that \begin{align*} c(x) &= 1 + x\sum_{n=0}^\infty C_{n+1}x^n = 1 + x\sum_{n=0}^\infty \sum_{k=0}^nC_kC_{n-k}x^n \\ &= 1 + x\sum_{n=0}^\infty x^n\sum_{k=0}^nC_kC_{n-k} = 1 + x\left[\sum_{n=0}^\infty C_n x^n\right]^2 \\ &= 1 + x[c(x)]^2. \end{align*} Solving for $c(x)$ with the quadratic formula, we find that $$ c(x) = \frac{1 \pm \sqrt{1-4x}}{2x} = \frac{2}{1\mp\sqrt{1-4x}}. $$ Since $c(0) = 1$, $$\displaystyle c(x) = \frac{1 - \sqrt{1-4x}}{2x} = \frac{1}{2x}\left(1 - \sqrt{1-4x}\right).$$

Consider the Taylor series of $f(y) = \sqrt{1+y}.$ By induction, $$f^{(n)}(y) = (-1)^{n+1}\frac{\prod_{k=0}^{n-1}(2k-1)}{2^n}(1+y)^{-(2n-1)/2} \Rightarrow f^{(n)}(0) = (-1)^{n+1}\frac{\prod_{k=0}^{n-1}(2k-1)}{2^n}.$$ Moreover, \begin{align*} f^{(n)}(0) &= (-1)^{n+1}\frac{\prod_{k=0}^{n-1}(2k-1)}{2^n} = (-1)^{n+1}\frac{\prod_{k=0}^{n-1}(2k-1)}{2^n}\cdot \frac{2^n n!(2n-1)}{2^n n!(2n-1)} \\ &= \frac{(-1)^{n+1}}{4^n(2n-1)}\frac{(2n)!}{n!}. \end{align*}

Thus, we have that $$ f(y) = \sqrt{1+y} = 1 + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{4^n(2n-1)}\frac{(2n)!}{n!n!}y^n, $$ so we have \begin{align*} c(x) &= \frac{1}{2x}(1 + f(-4x)) = \frac{1}{2x}\left(\sum_{n=1}^\infty \frac{(-1)^{n}}{4^n(2n-1)}\frac{(2n)!}{n!n!}(-4x)^n\right) \\ &= \frac{1}{2x}\left(\sum_{n=1}^\infty \frac{(-1)^{2n}}{(2n-1)}\frac{(2n)!}{n!n!}x^n\right) = \sum_{n=1}^\infty \frac{1}{2n(2n-1)}\frac{(2n)!}{(n-1)!n!}x^{n-1} \\ &= \sum_{n=1}^\infty \frac{1}{n}\frac{(2n-2)!}{(n-1)!(n-1)!}x^{n-1} = \sum_{n=1}^\infty \frac{1}{n}{2(n-1) \choose n-1 }x^{n-1} \\ &= \sum_{n=0}^\infty \frac{1}{n+1}{2n \choose n}x^{n} = \sum_{n=0}^\infty C_nx^{n}, \end{align*} so $\displaystyle \boxed{C_n = \frac{1}{n+1}{2n \choose n}.}$

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One of the more interesting problems that I've come across recently is to calculate the distribution of the last time a simple random walk is at $0.$

Let $X_1,X_2,\ldots,X_{2n}$ be independent, indentically distributed random variables such that $P(X_i = 1) = P(X_i = -1) = 1/2.$ Define $S_k = \sum_{i=1}^k X_i.$ Then, we have a path $$(0,0) \rightarrow (1,S_1) \rightarrow (2,S_2) \rightarrow \cdots \rightarrow (2n,S_{2n}).$$ Define the random variable $L_{2n} = \sup\{ k \leq 2n : S_k = 0\}.$ We want the distribution of $L_{2n}.$

Note that we have that \begin{equation} \mathbb{P}(S_{2n} = 0) = 2^{-2n}{2n \choose n} \end{equation} since we have $n$ positive steps and $n$ negative steps.

Let $\displaystyle N_{n,x} ={n \choose (n+x)/2}$ denote the number of paths from $(0,0)$ to $(n,x)$ since $(n+x)/2$ positive steps implies there are $(n-x)/2$ negative steps, and $(n+x)/2 - (n-x)/2 = x$. Note that $n + x$ must be even for this to be well defined. If $n + x$ is not even, then $N_{n,x} = 0$ since $x$ must have the same parity as $n.$ First, we prove the reflection principle.

Reflection Principle

If $x,y > 0,$ the number of paths that from $(0,x)$ to $(n,y)$ that are $0$ at some time, that is, they touch the $x$-axis is equal to the total number of paths from $(0,-x)$ to $(n,y),$ which is $N_{n,y+x}.$ Therefore, the number of paths from $(0,x)$ to $(n,y)$ that do not touch $0$ is $N_{n,|y-x|} - N_{n,y+x}.$

We can establish a one-to-one correspondence between the set $A$, the paths from $(0,x)$ to $(n,y)$ that are $0$ at some time and the set $B$ the paths from $(0,-x)$ to $(n,y)$.

Consider any path $P$ in $A$. $P$ must include the point $(m,0),$ where $0 < m < n$. Fix $m$ to be the greatest such integer. We construct a path $Q_1$ from $(0,-x)$ to $(m,0)$ by going in the opposite direction as $P.$ We construct a path $Q_2$ from $(m,0)$ to $(n,y)$ by mirroring $P$. Thus, we have that $Q = Q_1 \cup Q_2 \in B.$

Now consider any path $Q$ in $B$. Since paths are continuous, $Q$ must cross the $x$-axis, so $Q$ includes a point $(m,0)$, where $0 < m < n.$ Fix $m$ to be the greatest such integer. We construct $P_1,$ a path from $(0,x)$ to $(m,0)$ by going in the opposite direction as $Q$. We construct $P_2$ by mirroring $Q$. Thus, we have that $P = P_1 \cup P_2 \in A.$

So, we have established a one-to-one correspondence, and therefore, we have proven $|A| = |B|.$

Symmetry of Zeroes

$\mathbb{P}(S_1 \neq 0, S_2 \neq 0,\ldots,S_{2n} \neq 0) = \mathbb{P}(S_{2n} = 0).$

First note that $$ \mathbb{P}(S_1 \neq 0,\ldots,S_{2n} \neq 0) = \mathbb{P}(S_1 > 0,\ldots,S_{2n} > 0) + \mathbb{P}(S_1 < 0,\ldots,S_{2n} < 0) $$ since we can never have the path touch $0.$ Also note that the two terms are equal, so \begin{equation} \mathbb{P}(S_1 \neq 0, S_2 \neq 0,\ldots,S_{2n} \neq 0) = 2\mathbb{P}(S_1 > 0, S_2 > 0,\ldots,S_{2n} > 0). \end{equation} Now, note that \begin{equation} \mathbb{P}(S_1 > 0, S_2 > 0,\ldots,S_{2n} > 0) = \sum_{r=1}^{n}\mathbb{P}(S_1 > 0, S_2 > 0,\ldots,S_{2n} = 2r) \end{equation} since we have taken an even number of steps.

To calculate $\mathbb{P}(S_1 > 0, S_2 > 0,\ldots,S_{2n} = 2r),$ we note that $X_1 = 1$ since $S_1 > 0$. Then, the number of paths from $(1,1)$ to $(2n,2r)$ that do not touch $0$ by the Reflection Principle is $N_{2n-1,2r-1} - N_{2n-1,2r+1}$. Thus, \begin{align*} \mathbb{P}(S_1 > 0, S_2 > 0,\ldots,S_{2n} = 2r) &= \left(\frac{1}{2}\right)\left(\frac{1}{2^{2n-1}}\right)\left(N_{2n-1,2r-1} - N_{2n-1,2r+1}\right) \\ &= \left(\frac{1}{2^{2n}}\right)\left(N_{2n-1,2r-1} - N_{2n-1,2r+1}\right). \end{align*}

So, we have that \begin{align*} \mathbb{P}(S_1 > 0, S_2 > 0,\ldots,S_{2n} > 0) &= \left(\frac{1}{2^{2n}}\right) \sum_{r=1}^n\left(N_{2n-1,2r-1} - N_{2n-1,2r+1}\right) \\ &= \frac{1}{2^{2n}}N_{2n-1,1} = \frac{1}{2^{2n}}{2n - 1 \choose n}\\ &= \frac{1}{2^{2n}}\frac{(2n-1)!}{n!(n-1)!} = \frac{1}{2}\frac{1}{2^{2n}}\frac{(2n)!}{n!n!} \\ &= \frac{1}{2}\mathbb{P}(S_{2n} = 0) \end{align*} by telescoping since $N_{2n-1,2n+1} = 0$ and substituting in the previous equations.

Recalling the earlier equation, we have the desired result, $$ \mathbb{P}(S_1 \neq 0, S_2 \neq 0,\ldots,S_{2n} \neq 0) = 2\mathbb{P}(S_1 > 0, S_2 > 0,\ldots,S_{2n} > 0) = \mathbb{P}(S_{2n} = 0). $$

Distribution of Last Zero Visit

Let $L_{2n}$ be the random variable whose value is the last time the simple random walk visited $0.$ Formally, $$L_{2n} = \sup\left\{2k : k \in \{0,\ldots,n\},~\sum_{i=1}^{2k} X_i = 0\right\}.$$ Then, we have that $$\mathbb{P}(L_{2n} = 2k) = 2^{-2n}{2k \choose k}{2n-2k \choose n-k}.$$

To see this, we have that \begin{align*} \mathbb{P}(L_{2n} = 2k) &= \mathbb{P}(S_{2k} = 0, S_{2k+1} \neq 0, \ldots,S_{2n} \neq 0) \\ &= \mathbb{P}(S_{2k} = 0, S_{2k+1} - S_{2k} \neq 0, \ldots,S_{2n} - S_{2k} \neq 0) \\ &= \mathbb{P}(S_{2k} = 0)\mathbb{P}(S_1 \neq 0, \ldots,S_{2n-2k} \neq 0) \\ &= \mathbb{P}(S_{2k} = 0)\mathbb{P}(S_{2n-2k} = 0), \end{align*} where the last equaility is by Symmetry of Zeroes. Thus, we have that $$\mathbb{P}(L_{2n} = 2k) = \mathbb{P}(S_{2k} = 0)\mathbb{P}(S_{2n-2k} = 0) =2^{-2n}{2k \choose k}{2n-2k \choose n-k}$$ as desired.


Let's look at what this distribution looks like when $n = 15.$ We have symmetric distribution about $15$, so the mean is $15.$ However, the distribution is U-shaped, and the most likely values are very far from the mean. Thus, to take an analogy from sports, if two evenly matched teams were to played against each other multiple times over the course of a season, the most likely scenario is for one team team to lead the other team the entire season. So, saying that team A led team B the entire season is an almost meaningless statistic.

One popular model for classification is nearest neighbors. It's broadly applicable and unbiased as it makes no assumptions about the generating distribution of the data. Suppose we have $N$ observations. Each observation can be written as $(\mathbf{x}_i,y_i)$, where $0 \leq i \leq N$ and $\mathbf{x}_i = (x_{i1},x_{i2},\ldots,x_{ip})$, so we have $p$ features, and $y_i$ is the class to which $\mathbf{x}_i$ belongs. Let $y_i \in C$, where $C$ is a set of possible classes. If we were given a new observation $\mathbf{x}$. We would find the $k$ closest $(\mathbf{x}_i, y_i)$, say $(\mathbf{x}_{j_1}, y_{j_1}), (\mathbf{x}_{j_2}, y_{j_2}),\ldots, (\mathbf{x}_{j_k}, y_{j_k})$. To classify $\mathbf{x}$, we would simply take a majority vote among these $k$ closest points. While simple and intuitive, as we will see, nearest neighbors runs into problems when $p$ is large.

Consider this problem:

Consider $N$ data points uniformly distributed in a $p$-dimensional unit ball centered at the origin. Find the median distance from the origin of the closest data point among the $N$ points.

Let the median distance be $d(p, N)$. First to keep things simple consider a single data point, so $N = 1$. The volume of a $p$-dimensional ball of radius $r$ is proportional to $r^p$, so $V(r) = Kr^p$. Let $d$ be the distance of the point, so $P(d \leq d(p,1)) = 0.5$. Viewing probability as volume, we imagine a smaller ball inside the larger ball, so \begin{align*} \frac{1}{2} &= P(d \leq d(p, 1)) \\ &= \frac{V(d(p,1))}{V(1)} \\ &= d(p,1)^p \\ &\Rightarrow d(p,1) = \left(\frac{1}{2}\right)^{1/p}, \end{align*} and in general, $P(d \leq t) = t^p$, where $0 \leq t \leq 1$. For example when $p = 1$, we have

-1 0 1 For $p=2$,

Now, consider the case when we have $N$ data points, $x_1, x_2, \ldots, x_N$. The distance of the closest point is $$d = \min\left(\Vert x_1 \Vert, \Vert x_2 \Vert, \ldots, \Vert x_N \Vert\right).$$ Thus, we'll have \begin{align*} \frac{1}{2} &= P(d \leq d(p,N)) \\ &= P(d > d(p,N)),~\text{since $P(d \leq d(p,N)) + P(d > d(p,N)) = 1$} \\ &= P\left(\Vert x_1\Vert > d(p,N)\right)P\left(\Vert x_2\Vert > d(p,N)\right) \cdots P\left(\Vert x_N\Vert > d(p,N)\right) \\ &= \prod_{i=1}^N \left(1 - P\left(\Vert x_i \Vert \leq d(p,N)\right)\right) \\ &= \left(1 - d(p,N)^p\right)^N,~\text{since $x_i$ are i.i.d and $P(\Vert x_i\Vert \leq t) = t^p$}. \end{align*} And so, \begin{align*} \frac{1}{2} &= \left(1 - d(p,N)^p\right)^N \\ \Rightarrow 1 - d(p,N)^p &= \left(\frac{1}{2}\right)^{1/N} \\ \Rightarrow d(p,N)^p &= 1 - \left(\frac{1}{2}\right)^{1/N}. \end{align*} Finally, we obtain $$\boxed{d(p,N) = \left(1 - \left(\frac{1}{2}\right)^{1/N}\right)^{1/p}}.$$

So, what does this equation tell us? As the dimension $p$ increases, the distance goes to 1, so all points become far away from the origin. But as $N$ increases the distance goes to 0, so if we collect enough data, there will be a point closest to the origin. But note that as $p$ increases, we need an exponential increase in $N$ to maintain the same distance.

Let's relate this to the nearest neighbor method. To make a good prediction on $\mathbf{x}$, perhaps, we need a training set point that is within distance 0.1 from $\mathbf{x}$. We would need 7 data points for there to be a greater than 50% chance of such a point existing if $p = 1$. See how $N$ increases as $p$ increases.

p N
1 7
2 69
3 693
4 6932
5 69315
10 6,931,471,232
15 $\scriptsize{6.937016 \times 10^{14}}$

The increase in data needed for there to be a high probability that there is a point close to $\mathbf{x}$ is exponential. We have just illustrated the curse of dimensionality. So, in high dimensions, other methods must be used.