July | 2021 | Victor Chen

Suppose in an election with two candidates, we are certain of the final tally. However, as the votes are counted sequentially, the lead may switch between the candidates during the process. This was certainly the case in the 2020 U.S. presidential election, where in several swing states, different types of votes (mail, early-voting, in-person) were counted at different stages. While the final tally does not change in a recount, the lead change may. One may ask: How likely will there be lead changes in a recount? Can the winning candidate always lead from start to finish? The Ballot Theorem provides a quantitative answer to this question.

(Informal) Suppose candidates $A$ and $B$ each have $a$ and $b$ votes, respectively, with $a>b$ . If the ballots are sorted uniformly at random and counted sequentially, the probability that $A$ always leads $B$ in this sequential count is $\frac{a-b}{a+b}$ .

The theorem matches our intuition that if the election is close, i.e., the count difference $a-b$ is small, then it is very likely that in a recount, $B$ will be winning at some point, even though $A$ will eventually overtake $B$ ‘s lead and win it all at the end. Let us formalize the statement in terms of random variables.

If $X_{1}\ldots,X_{t}$ are random variables, $\overline{X_{t}}:=\frac{1}{t}\sum_{i=1}^{t}X_{i}$ is the average of $X_{1}\ldots,X_{t}.$

(Ballot Theorem) Suppose $X_{1}\ldots,X_{N}\in\{-1,1\}$ are i.i.d. random variables with $\overline{X_{N}}=\rho$ , where $\rho\in\mathbb{R}^{+}$ . Then

$P\left(\forall t\in[N]\overline{X_{t}}>0|\overline{X_{N}}=\rho\right)=\rho.$

To see that the formal statement captures the semantics of ballot counting, imagine that all $N$ votes are in a box, and we randomly sample a ballot from a box without replacement. At time $t$ , $X_{t}=1$ if the $t$ -th sample is for Candidate $A$ and $-1$ if the $t$ -th sample is for Candidate $B$ . Then $\rho=\frac{a-b}{a+b}$ . In the rest of this post, we will prove the Ballot Theorem using two commonly used techniques in probability, the reflection principle and martingale.

Reflection Principle

Reflection Principle is a common technique used in Brownian motion, which is a continuous stochastic process that can model the movement of particles or the trajectory of stock prices over time. The idea is that if at any time, the stock price goes up or down equally likely, then any upward trajectory of the stock price has a corresponding downward trajectory (“reflected” along the x-axis). This symmetry allows us to reduce certain questions about a sequence of random variables to just one. For instance, the Ballot Theorem shows the probability that all $N$ random variables $\overline{X_{t}}_{t\in[N]$ are positive is reduced to the single constant random variable $\overline{X_{N}}$ .

Before writing the proof formally, let us first describe the Ballot Theorem at a high level. Imagine the ballots are sorted, and we start counting the votes one by one. Suppose the first vote belongs to Candidate $B$ . Since $a>b$ , at some point during counting, $B$ must relinquish its lead and tie with $A$ , say with $x$ votes each. Now pause the counting (temporarily!), and pair up each of the counted votes for $A$ with a unique counted vote for $B$ . With this pairing, we can re-create another recount by just swapping every counted $A$ vote with its paired $B$ vote. Then in this other recount, the first vote belongs to Candidate $A$ and afer $2x$ votes, $A$ is tied with $B$ . In other words, whenever a recount has $B$ losing its initial lead, there is another recount where $A$ initially loses its lead as well!

Imagine plotting the vote counts. The scenario that $B$ has an initial lead is represented by a curve that is below the x-axis and eventually hits the x-axis. The curve for the imagined recount described above is simply obtained by reflecting the original curve along the x-axis.

The probability that the first vote is for $B$ and $B$ relinquish this lead is $\frac{b}{a+b}$ (since $B$ does not have enough votes to win in the final tally). By reflection, the probability that the first vote is for $A$ and $A$ relinquishes this lead is also $\frac{b}{a+b}$ . So the probability that $A$ never falls behind is $1-\frac{2b}{a+b}=\frac{a-b}{a+b}$ .

Let $T$ be the stopping time when $\overline{X_{t}}$ first becomes $0$ , i.e.,
$T=\begin{cases} \inf\{t\in[N]:\overline{X_{t}}=0\} & \text{if the set is non-empty}\\ N & \text{otherwise.} \end{cases}$

Note that $P\left(\forall t\in[N]\overline{X_{t}}>0|\overline{X_{N}}=\rho\right)=1-P\left(\overline{X_{T}}=0|\overline{X_{N}}=\rho\right)$ , so it suffices to compute $P\left(\overline{X_{T}}=0|\overline{X_{N}}=\rho\right)$ . To this end, note that if $\overline{X_{T}}=0$ , by the minimality of $T$ , either $\forall t<T,\overline{X_{t}}>0$ or $\forall t<T,\overline{X_{t}}<0$ .

Now here is the reflection principle at work.

$P\left(\forall t<T,\overline{X_{t}}>0,\overline{X_{T}}=0|\overline{X_{N}}=\rho\right)=P\left(\forall t<T,\overline{X_{t}}<0,\overline{X_{T}}=0|\overline{X_{N}}=\rho\right).$

(of Claim) Consider $\omega\in\{-1,1\}^{N}$ such that $\overline{X_{t}}(\omega)>0,\overline{X_{T}}(\omega)=0,\overline{X_{N}}(\omega)=\rho$ . Consider its reflection $R(\omega)=(-\omega_{1},\ldots,-\omega_{T},\omega_{T+1},\ldots,\omega_{N})$ . Since the $X_{i}$ are i.i.d. and $\omega$ and $R(\omega)$ are in one-to-one correspondence, the two probabilities are equal.

Again by the minimality of $T$ ,

$P(\overline{X_{T}}=0|\overline{X_{N}}=\rho)=2P\left(\forall t<T,\overline{X_{t}}<0,\overline{X_{T}}=0|\overline{X_{N}}=\rho\right).$

Note that $P\left(\forall t<T,\overline{X_{t}}<0,\overline{X_{T}}=0|\overline{X_{N}}=\rho\right)=P\left(X_{1}=-1|\overline{X_{N}}=\rho\right)$ . To see this, if $\omega\in\{-1,1\}^{N}$ is such that $\forall t<T,\overline{X_{t}}(\omega)<0,\overline{X_{T}}(\omega)=0,\overline{X_{N}}(\omega)=\rho$ , then $X_{1}(\omega)=-1$ . Conversely, if $X_{1}(\omega)=-1$ and $\overline{X_{N}}(\omega)=\rho>0$ , there must be some $T$ such that $\forall t<T,\overline{X_{t}}(\omega)<0,\overline{X_{T}}(\omega)=0$ .

Thus, putting the above together, we have

$\begin{eqnarray*} P\left(\forall t\in[N]\overline{X_{t}}>0|\overline{X_{N}}=\rho\right) & = & 1-P\left(\overline{X_{T}}=0|\overline{X_{N}}=\rho\right)\\ & = & 1-2P\left(\forall t<T,\overline{X_{t}}<0,\overline{X_{T}}=0|\overline{X_{N}}=\rho\right)\\ & = & 1-2P\left(X_{1}=-1|\overline{X_{N}}=\rho\right). \end{eqnarray*}$

And finally, since the random variables $\{X_{i}\}_{i\in[N]}$ are i.i.d.,

$\begin{eqnarray*} 1-2P\left(X_{1}=-1|\overline{X_{N}}=\rho\right) & = & \mathbb{E}\left[X_{1}|\overline{X_{N}}=\rho\right]\\ & = & \mathbb{E}\left[\overline{X_{N}}|\overline{X_{N}}=\rho\right]\\ & = & \rho, \end{eqnarray*}$

finishing the proof.

The proof used a subtle property of the stopping time $T$ . When $A$ always leads $B$ , $T$ is defined to be $N$ , and so $\overline{X}_{T}>0$ by assumption. At first glance, the definition that $T=N$ if $A$ always leads seems arbitrary. However, the semantic of stopping time requires that the event $\{T=t\}$ must be “observable” at time $t$ . The information that $A$ always leads $B$ is only determined at time $N$ , so $T$ must be defined to be $N$ . In the language of measure theory, $\{T=N\}\in\sigma(X_{1}\ldots,X_{N})$ . See the next section where stopping time is formally defined in terms of $\sigma$ -algebra.

Martingale

In a betting system, a gambler often formulates a strategy at time $t$ based on previous observations, say the first $t-1$ card draws. For example, if we see a sequence of coin tosses being head, it is tempting to bet big on the 11th toss being a tail. A martingale is essentially a strategy where the expectation valuation at time $t$ given the observations up to $t-1$ does not differ from previous valuation. Continuing the coin example, if the coin tosses are fair, then the expected winning on the 11th toss is the same as previous coin tosses.

A sequence of integrable random variables $\{X_{t}\}_{t\in\mathbb{Z}}$ is a martingale if $\mathbb{E}X_{t+1}|X_{1},\ldots,X_{t}=X_{t}$ .

At time $t$ , a gambler may have access to other information computed from $X_{1}\ldots,X_{t-1}$ , not just the values of these previous random variables. Thus, we use a larger set, a $\sigma$ -algebra $\mathcal{F}_{t}$ that semantically captures all the information available up to time $t$ (as opposed to just $\{X_{1},\ldots,X_{t}=x_{1},\ldots,x_{t}\}$ . More generally,

A sequence of integrable random variables $\{X_{t}\}_{t\in\mathbb{Z}}$ is a martingale with respect to the $\sigma$ -algebras $\{\mathcal{F}_{t}\}_{t\in\mathbb{Z}}$ if

$\{\mathcal{F}_{t}\}_{t\in\mathbb{Z}}$ is a filtration, i.e., $\forall s<t$ , $\mathcal{F}_{s}\subset\mathcal{F}_{t}$ ,

$X_{t}$ is $\mathcal{F}_{t}$ -measurable,

$\forall s<t$ , $\mathbb{E}X_{t}|\mathcal{F}_{s}=X_{s}$ .

Semantically, the first criterion just says that information at time $t$ is cumulative: it includes information available previous time. For the second criterion, by definition $\forall x\in\mathbb{R}$ , $\{X_{t}\leq x\}\in\mathcal{F}_{t}$ , so it guarantees that $\mathcal{F}_{t}$ actually captures the information at time $t$ . Lastly, the third criterion enforces that the betting system is fair as discussed.

In probability, many problems have a hidden martingale, and finding it often illuminates the problem more. Here is roughly how martingale is embedded inside the Ballot Theorem. Imagine we have randomly sorted $N$ ballots for a recount. At time $-t$ (the negative indices will be explained soon), we have $t$ ballots uncounted. Since we knew the final tally, and we have just counted $N-t$ ballots, we know how many of the remaining $t$ ballots are for Candidates $A$ and $B$ , just not the specific order that they will come out. Thus, if we draw a random ballot from the remaining $t$ ballots, its distribution is exactly $\overline{X}_{-t}:=\frac{1}{t}\sum_{i=1}^{t}X_{i}$ , implying that $\overline{X}_{-t}$ is a martingale (precise calculation within the proof below).

We make the curious choice to index our sample average $\overline{X}_{-t}$ with negative indices, even though we could have shifted the entire index sequence by $N$ to the right. However, in theory, the number of random variables may be infinite (the setting for the Law of Large Numbers). In this case, when the average is taken over just one random variable or zero variable, the martingale stops. It is simply more instructive to let $X_{-1}$ or $X_{0}$ be the last random variable in this martingale. In fact, martingale with negative indices are also called backward martingale. Backward martingales are easier to work with as they have stronger convergence guarantees. The reason is that a backward martingale has a last random variable, which by definition is bounded and therefore bounds the entire sequence of random variables via the martingale property.

We do not need the full machinery of backward martingale since in our application, $N$ , the number of ballots, is bounded. But for pedagogical purposes we will retain the use of negative indices to remind ourselves that $\overline{X}_{-t}$ is actually a backward martingale with stronger tools should we need to avail ourselves to them.

Similar to the reflection principle, we will also make use of stopping time, but this time we will use it more extensively, so we define it carefully now.

A random variable $T:\Omega\rightarrow\mathbb{Z}$ is a stopping time with respect to $\{\mathcal{F}_{t}\}_{t\in\mathbb{Z}}$ if for all $t\in\mathbb{Z}$ , $\{T=t\}\in\mathcal{F}_{t}$ , where $\{\mathcal{F}_{t}\}_{t\in\mathbb{Z}}$ is a filtration.

Semantically, when $\mathcal{F}_{t}$ represents the cumulative information at time $t$ , any stopping time criterion (bet after five heads are tossed, fold when money is below a threshold, etc) can only depend on the information available up to time $t$ . Stopping time is often used in conjunction with martingales. Often it is useful to stop a sequence of random variables, say the first time that $\overline{X}_{-t}$ hits $0$ . If the original sequence forms a martingale, then the sequence up to the stopping time — which itself is a random variable — remains a martingale if $T$ is bounded. There are more general statements, but the following suffices for our purpose.

(Martingale Optional Stopping Time) Let $\{Y_{t}\}_{t=0}^{M}$ be a martingale and $T:\Omega\rightarrow{0,1,\ldots,M}$ a stopping time with respect to the filtration $\{\mathcal{F}_{t}\}_{t=0}^{M}$ , where $M \in \mathbb{Z}^+$ . Then $\mathbb{E}Y_{T}|\mathcal{F}_{0}=Y_{0}$ .

By the definition of conditonal expectation, it suffices to show that

$Y_{0}$ is $\mathcal{F}_{0}$ -measurable and

for all $A\in\mathcal{F}_{0}$ , $\int_{A}Y_{T}=\int_{A}Y_{0}$ .

The first condition is satisfied by definition. For the second, by telescoping, we have

$\int_{A}Y_{T}-Y_{0}=\sum_{t=0}^{M}\int_{A}Y_{\min(T,t+1)}-Y_{\min(T,t)}.$

For each term, we have

$\begin{eqnarray*} \int_{A}Y_{\min(T,t+1)}-Y_{\min(T,t)} & = & \int_{A\cap\{T\leq t\}}Y_{\min(T,t+1)}-Y_{\min(T,t)}+\\ & & \int_{A\cap\{T>t\}}Y_{\min(T,t+1)}-Y_{\min(T,t)}\\ & = & \int_{A\cap\{T\leq t\}}Y_{T}-Y_{T}+\int_{A\cap\{T>t\}}Y_{t+1}-Y_{t}\\ & = & 0, \end{eqnarray*}$

since $A\cap\{T>t\}\in\mathcal{F}_{t}$ and $\{Y_{t}\}_{t=0}^{M}$ is a martingale with respect to $\{\mathcal{F}_{t}\}_{t=0}^{M}$ .

(Of Ballot Theorem) For $t\in[N]$ , Define $\overline{X}_{-t}=\frac{1}{t}\sum_{i=1}^{t}X_{i}$ . Let $\mathcal{F}_{-t}$ denote the $\sigma$ -algebra $\sigma(\overline{X}_{-t},X_{t+1},\ldots,X_{N})$ . We first show that the sample averages $\overline{X}_{-t}$ form a martingale.

$\mathbb{E}\overline{X}_{-(t-1)}|\mathcal{F}_{-t}=\overline{X}_{-t}$ .

(Of Claim) Since the random variables $X_{i}$ are i.i.d., we have

$\begin{eqnarray*} \mathbb{E}\overline{X}_{-(t-1)}|\mathcal{F}_{-t} & = & \frac{1}{t-1}\sum_{i=1}^{t-1}\mathbb{E}X_{i}|\overline{X_{-t}},X_{t+1},\ldots,X_{N}\\ & = & \frac{1}{t-1}\sum_{i=1}^{t-1}\mathbb{E}X_{i}|\overline{X_{-t}}\\ & = & \mathbb{E}X_{1}|\overline{X_{-t}}. \end{eqnarray*}$

On the other hand, by i.i.d. again, we have

$\begin{eqnarray*} \mathbb{E}X_{1}|\overline{X_{-t}} & = & \frac{1}{t}\sum_{i=1}^{t}X_{i}|\overline{X_{-t}}\\ & = & \mathbb{E}\overline{X_{-t}}|\overline{X_{-t}}\\ & = & \overline{X_{-t}}. \end{eqnarray*}$

Now let $T$ be the stopping time when $\overline{X_{-t}}$ first becomes $0$ , i.e.,
$T=\begin{cases} \inf\{t\in\{-N,\ldots,-1\}:\frac{1}{t}\sum_{i=1}^{t}X_{i}=0\} & \text{if the set is non-empty}\\ -1 & \text{otherwise.} \end{cases}$

Since $T$ is a stopping time, in the else-case above, $T$ must be $-1$ since $\{T=-1\}\in\mathcal{F}_{-1}$ . Since $\overline{X}_{-t}$ form a martingale, by the Martingale Optional Stopping Time Theorem,
we have $\overline{X_{-N}}=\mathbb{E}\overline{X_{T}}|\mathcal{F}_{-N}.$

Since $\{\overline{X_{-N}}=\rho\}\subset\mathcal{F}_{-N}$ , by the Law of Total Expectation, we have

$\begin{eqnarray*} \mathbb{E}\left[\overline{X_{-N}}|\overline{X_{-N}}=\rho\right] & = & \mathbb{E}\left[\mathbb{E}\left[\overline{X_{T}}|\mathcal{F}_{-N}\right]|\overline{X_{-N}}=\rho\right]\\ & = & \mathbb{E}\left[\overline{X_{T}}|\overline{X_{-N}}=\rho\right]. \end{eqnarray*}$

Evaluating both sides, and writing $\mathcal{E}=\{\forall t\in[N], \overline{X_{t}}>0\}$ , we have

$\begin{eqnarray*} \rho & = & \mathbb{E}\left[\overline{X_{T}}|\overline{X_{-N}}=\rho\right]\\ & = & P(\mathcal{E}|\overline{X_{-N}}=\rho)\mathbb{E}\left[\overline{X_{T}}|\overline{X_{-N}}=\rho,\mathcal{E}\right]+\\ & & P(\mathcal{E}^{c}|\overline{X_{-N}}=\rho)\mathbb{E}\left[\overline{X_{T}}|\overline{X_{-N}}=\rho,\mathcal{E}^{c}\right]\\ & = & P(\mathcal{E}|\overline{X_{-N}}=\rho)\mathbb{E}\left[\overline{X_{T}}|\overline{X_{-N}}=\rho,\mathcal{E}\right]\\ & = & P(\mathcal{E}|\overline{X_{-N}}=\rho)\mathbb{E}\left[\overline{X_{-1}}|\overline{X_{-N}}=\rho,\mathcal{E}\right]\\ & = & P(\mathcal{E}|\overline{X_{-N}}=\rho), \end{eqnarray*}$

where the third through fifth line follow from $\overline{X_{T}}=0$ if $\mathcal{E}^{c}$ , the definition of $T$ , and that $X_{1}=1$ if $\mathcal{E}$ , respectively.

Victor Chen

mathematics blog

Monthly Archives: July 2021

The Ballot Theorem

Reflection Principle

Martingale