Suggest edit — Martingale Model

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---
title: "Martingale Model"
type: concept
related: [Markovian Process, Random Variables, Martingale Model]
source: https://www.jemoka.com/posts/kbhmartingale_model/
confidence: high
status: active
---

The Martingale Model states: if we observed the closing price of the market yesterday, we expect that the market is going to open at the close price yesterday.
Formally:
\begin{equation} E\qty [X_{k}|X_{k-1}, X_{k-2},\ldots] = X_{k-1} \end{equation}
&ldquo;irrespective of what you know, no matter how long the history, the best expectation of today&rsquo;s price is yesterday&rsquo;s price.&rdquo;
This is not a for sure! modeling statement: this is simply the expected value!! That means, after \(\infty\) times of re-running the universe starting &ldquo;yesterday&rdquo;, the new opening price will converge to the last closing price.
Two important conclusions:
If we know the closing price yesterday (it is observed), the price today will be DETERMINED and not!!! a random variable If the closing price yesterday is a random variable, the price today will be IN-DETERMINED and also a random variable Therefore, the &ldquo;randomness is fair&rdquo;, and therefore the &ldquo;market is not drifting in favor/against you.&rdquo;
The Martingale Model comes from the idea that &ldquo;true gambling is true equal conditions (money, opponents, bystanders, situations, die, and dice.)&rdquo; Therefore, any amount of bias towards one direction/party is advantageous for that person.
In fact, it was theorized that an efficient market should follow exactly this behavior.
changes in history Of course, the difference between the expression:
\begin{equation} E\qty [X_{k}|X_{k-1}, X_{k-2},\ldots] = X_{k-1} \end{equation}
versus
\begin{equation} E\qty [X_{k}|X_{k-1}] = X_{k-1} \end{equation}
is pretty big. The two will only be the same if the markets is assumed to be a markovian process.
Martingale historical conditioning Ok, if we are told that the process is Martingale, but we only have two days ago, what do we have?
i.e. what if we want to know:
\begin{equation} E\qty [X_{k} | X_{k-2}] = ? \end{equation}
Turns out, there&rsquo;s a small trick you can do. Without even Martingale, we can claim that:
\begin{equation} E\qty [X_{k} | X_{k-2}] = \sum_{x} E\qty [X_{k} | X_{k-1}, X_{k-1} = x] \cdot Pr \qty(X_{k-1}=x|X_{k-2}) \end{equation}
That, the price today is just the sum of all possible prices for day \(k-1\) we name small \(x\) times the probability \(Pr\) that it actually happens given the existing \(k-2\) observation.
Of course, given the Martingale Model now, given some possible price in day \(k-1\) named \(x\), price in \(k\) is also \(x\). Therefore:
\begin{equation} E[X_{k}|X_{k-1},X_{k-1} = x] =x \end{equation}
Applying this, then, we have
\begin{equation} \sum_{x} E\qty [X_{k} | X_{k-1}, X_{k-1} = x] \cdot Pr \qty(X_{k-1}=x|X_{k-2}) = \sum_{x} x \cdot Pr \qty(X_{k-1}=x|X_{k-2}) \end{equation}
The right sum, then, is just the expected value of \(X_{k-1}\) given \(X_{k-2}\)!! Meaning:
\begin{equation} \sum_{x} x \cdot Pr \qty(X_{k-1}=x|X_{k-2}) = E[X_{k-1} | X_{k-2}] \end{equation}
Now, we are in a Martingale Model. Therefore:
\begin{equation} \sum_{x} x \cdot Pr \qty(X_{k-1}=x|X_{k-2}) = E[X_{k-1} | X_{k-2}] = X_{k-2} \end{equation}
And so, putting it all together, we have:
\begin{align} E\qty [X_{k} | X_{k-2}] &amp;= \sum_{x} E\qty [X_{k} | X_{k-1}, X_{k-1} = x] \cdot Pr \qty(X_{k-1}=x|X_{k-2}) \\ &amp;= \sum_{x} x \cdot Pr \qty(X_{k-1}=x|X_{k-2}) \\ &amp;= E[X_{k-1} | X_{k-2}] \\ &amp;= X_{k-2} \end{align}
Amazing. So Martingale holds over time