Martingales, Trading Strategies, and the Limits of Beating a Random Walk

# Proof that you cannot beat a random walk


# Proof that you cannot beat a random walk












There is much speculation to what degree financial series are random (and what kind of randomness prevails).

I want to turn the question on its head and ask:

Is there a mathematical proof that whatever trading strategy you use you cannot beat a random walk (that is the expected value will always be 0 assuming no drift)?

(I found this blog post where the author used the so called "75% rule" to purportedly beat a random walk but I think he got the distinction between prices and returns wrong. This method would only work if you had a range of allowed prices (e.g. a mean reverting series). See e.g. here for a discussion.)

## Answer by 楊祝昇 (score 26, accepted)

https://quant.stackexchange.com/a/1719

I can help you beat random walk 'in the way you want', i.e. the expected value $E[\$]$ will always be positive even assuming no drift. However, I have to warn people that $E[\$] > 0$ is NOT really an adequate condition for 'beating' in reality (at least to myself).

Let's define some mathematical notations for derivation, and rephrase (simplify) vonjd's question without losing generality. Assume a trader plays a fair game, and his surplus $X(0), X(1), X(2), ... X(t)$ is a martingale.

Q: Can the trader find a stopping time $s$ such that $E[X($s$)] > X(0)$?

- A proof supporting Bootvis' answer, for comparison, consider a normal trading strategy that bets evenly. Then,

$$\begin{align*}E[X(s)] &= E[ E[X(s)|X(s-1), X(s-2),..., X(0)] ] \\ &= E[X(s-1)] = E[X(s-2)] = ... = E[X(0)] = X(0).\end{align*}$$

- Now, consider a 'double-betting' strategy. We keep doubling your losing trade until first win. Let's set the initial surplus, $X(0) = 0$ for simplicity.

Accordingly, $X(k) = X(k-1) + G(k)$, where $G(k)=\pm 2^{k}$ with probability $1/2$. Note that we get the power $(k)$ of $2$ in $G(k)$ because of 'double-betting'. Our market is still random walk.

This strategy is designed stop at a time $s = min{k}$ s.t. $G(k) > 0$ (Note that $Prob{s=infinity} = 0$)

Compute $E[X(s)]$ by conditioning on s:

$$\begin{align*} E[X(s)] &= E[E[X(s)|s]] = \sum_{k=1}^{\infty} E[X(s)|s=k] * Prob{s=k} \\ &= \sum_{k=1}^{...} (-1-2-4-8...-2^{(k-1)} + 2^{k}) * (1/2)^{k} \\ &= \sum_{k=1}^{...} 1 * (1/2)^k = 1 > 0 = X(0)\end{align*}$$

Conclusion

A trader can make $E[X(s)]>0$ for random walk using the double-betting strategy. We proved that you can beat random walk in your definition of 'beating', i.e. expected value > 0.

This is actually a simplified proof supporting Akshay's answer. Whatever it's called: volatility pumping, Kelly strategy, optimal growth portfolio, and etc. These ideas simply ask one more question: why double? Is there an optimal betting ratio because of ... (various reasons and assumptions)?



- A $E[X(s)]>0$ strategy is guaranteed to make you real fortune if and only if we have 'unlimited amount of capital'. For details (long story), see wiki: Martingale betting system.

- You might ask what should we do if we only have limited capital? The Kelly criteria actually kind of offers the effect of the double-betting strategy for limited capital. For example, if you have a very weak trading signal (close to random walk in which there is no signal at all), the Kelly criteria will recommend you to bet something like \$1 (initially) for \$1M capital, and increase/decrease your position by certain % when you lose/win. Yeah, \$1M indeed looks like unlimited capital to \$1.

- (From comment) There is no contradiction to the common sense that 'pure independence = zero E[PnL]'. $E[] > 0$ in my example and vonjd'd Parrondo's paradox are indeed exploited from sort of dependency. While the Parrondo's paradox exploits the dependency between two losing games, mine is exploiting the dependency from my losing trades (which is less obvious). But warn again: This is at the cost of ruin risk! Though Kelly and vol-pump strategies eliminate ruin risk, they still suffer from trending risk.

## Answer by Bob Jansen (score 7)

https://quant.stackexchange.com/a/1711

If the price of every asset follows an independent random walk without drift then every position has an expected return of zero. So, in expectation, there is no combination of positions that has an expectation different from zero.

## Answer by bill_080 (score 4)

https://quant.stackexchange.com/a/1713

That post has been up since March. Either he hasn't figured it out, or he's trying to get people to click through to the book.

In the following statement, isn't he implying that "rw" is a return (as in....random walk)?

```
rw <- rnorm(100)
```

In the following statements, isn't he calling a "trade" the DIFFERENCE IN RETURNS? Isn't that meaningless?

```
if(rw[i] < m) trade[i] <- (rw[i+1]-rw[i])
if(rw[i] > m) trade[i] <- (rw[i]-rw[i+1])
```

From there on, isn't the whole thing a waste of time? Likewise, when I opened that .pdf file, the first thing I saw was Sornette's name. There's no need to read further.

As far as "proofs" go, how are you going to agree on the properties of the market? If you can't, then the idea of a "proof" goes poof.

## Answer by olaker (score 4)

https://quant.stackexchange.com/a/1741

Here is a more or less formal proof of the fact that "the system can't be beaten". The argument works whenever the underlying process is a martingale. In particular, it is valid for a random walk without drift.

Let $S=\{S_n\}$ be a discrete-time martingale which represents a series of games played at times $n=1,2,...$. Assume that $S_0=0$ (no game at time $n=0$). Let $\mathcal A=\{\mathcal A_n\}$ be a filtration of $\sigma$-algebras $\mathcal A_0\subset \mathcal A_1\subset\ ...$, such that the process $S$ is adapted with respect to $\mathcal A$, i.e. $S_n$ is $\mathcal A_n$-measurable for each $n$. Intuitively speaking, this implies that $\mathcal A_n$ contains all information about the outcomes of the game after the first $n$ rounds, and the value of $S_n$ is known to the player at time $n$.

We may think of the difference $S_n-S_{n-1}$ as the net winnings per unit stake in game $n$. Since $S$ is a martingale, $$\mathbb E(S_n-S_{n-1}|\mathcal A_{n-1})=0,\qquad n=1,2,... $$ Now let $C=\{C_n\}$ be a previsible bounded process, i.e. $C_n$ is $\mathcal A_{n-1}$-measurable and $$\sup|C_n(\omega)|\leq K $$ for each $n=1,2,...$ and some constant $K$. $C_n$ represents the player's stake in game $n$. Obviously, the value of $C_n$ must be determined based on the history up to time $n-1$ (the player has no information about the value of $S_n$ before the n-th round is played). Thus the total winnings up to time $n$ are $$X_0=0,\qquad X_n=\sum\limits_{k=1}^{n}C_k(S_k-S_{k-1}),\quad n\geq1.$$ Since the process $C$ is previsible and bounded, using the standard properties of the conditional mean, we have that $$\mathbb E(X_n-X_{n-1}|\mathcal A_{n-1})=\mathbb E(C_n(S_n-S_{n-1})|\mathcal A_{n-1})= C_n\mathbb E(S_n-S_{n-1}|\mathcal A_{n-1})=0$$ for each $n=1,2,...$. In other words, $X=\{X_n\}$ is an adapted integrable process such that $$\mathbb E(X_n|\mathcal A_{n-1})=X_{n-1}$$ for all $n\geq 1$, i.e. it's a martingale itself. It immediately follows that $\mathbb E(X_n|\mathcal A_{m})=X_m$ when $m<n$ and that the unconditional mean $\mathbb E(X_n)=\mathbb E(X_0)=0$ does not depend on $n$.

"Probability with Martingales" by Williams is a good and fairly standard reference for this stuff.

## Answer by Akshay (score 2)

https://quant.stackexchange.com/a/1714

I think you mean to say any self-financing trading strategy would give you zero returns - which it will - the principle of no-arbitrage ensures that.

For example, say you have a stock whose returns follow a random walk pattern with expected return equal to zero. The only way this can happen is (a) the stock stays fixed at a value K or (b) the stock has an equal probability of being +x% or -x% tomorrow or (c) the stock can go +x% with probability p1 and -y% with probability p2 with p1*x + p2*y = 0.

Now, for all of these cases, I can have an options-based long straddle centered around K. With non-zero volatility (stochastic as well as determinate), it is far more likely that my straddle will earn me a positive return than not. The down-side is - other traders also realize this fact and so I can't have that straddle for free (and thus earning a clean expected non-zero return).

So it is not the iid process by itself which is non-exploitable (sure it is, as in the strategy above if arbitrage exists or, even simpler, in the case of a "buy-low, sell-high" strategy which can earn a positive return from an iid). The "mathematical structure"which can be exploited is known as the process volatility. It is no-arbitrage which disallows an iid process from being exploited for returns in excess of those commensurate with its volatility - not any property of the process itself.