A Geometric Brownian Model Can Permit Vanishing-Risk Free Lunches

# Violation of No Free Lunch with Vanishing Risk


# Violation of No Free Lunch with Vanishing Risk












In this question I construct a very simple violation of the No Free Lunch with Vanishing Risk property (NFLVR). This should not be possible. Have I made a mistake? To be as detailed as possible, I will follow the notation of Delbaen and Schachermayer, Chapter 8.

Let $S\in\mathbb R^d$ be a semimartingale representing the market, discounted under the risk-free interest rate. Let $X:=S^1$ be the component of $S$ representing the discounted SPDR S&P 500 ETF Trust. We know that the S&P 500 outperforms the risk-free interest rate over the long-term. Assume for now that its path (under the real-world measure) is given by Geometric Brownian Motion: $X_t = X_0 \exp((\mu-\frac12\sigma^2-r)t+\sigma W_t)$. Then we have $\mu-\frac12\sigma^2-r>0$. Now consider the following sequence of trading strategies: in strategy $H^N$, we buy $\frac1N$ shares of the SPDR S&P 500 ETF Trust and hold this for $N$ years. Mathematically, $H^N(t)=(\frac1N,0,\dots,0)^T$ for $t\in[0,N]$ and $H^N(t)=(0,\dots,0)^T$ for $t>N$. To check that $H^N$ is admissible, we have to verify that $(H^N\cdot S)_t$ is bounded from below: $$(H^N\cdot S)_t = \int_0^t H^NdS = \int_0^t\frac1NdX = \frac1N(X_N-X_0)\geq -\frac1N X_0.$$ Since the last value is independent of $t$, the strategy $H^N$ is indeed admissable.

Now the (discounted) payoff of $H^N$ is $$(H^N\cdot S)_{\infty}=\int_0^{\infty} H^NdS = \int_0^{N}\frac1N d X = \frac1N (X_N-X_0).$$ Take a real number $\epsilon\in(0,\mu-\frac12\sigma^2-r)$. For $N$ large enough we have $$X_N>X_0\exp(\epsilon N);$$ this follows from the law of the iterated logarithm. In case that $X_N>X_0\exp(\epsilon N)$, we have for the payoff of $H^N$:

$$\frac1N (X_N-X_0)> \frac1N X_0 (\exp(\epsilon N)-1)> \epsilon X_0.$$ This last value is positive and independent of $N$ (free lunch). The maximal potential loss of $H^N$ occurs when the S&P 500 loses all its value, and equals $-\frac1N X_0$. This maximal loss goes to $0$ as $N\rightarrow\infty$ (vanishing risk). We have thus constructed a Free Lunch with Vanishing Risk.

To make this fully rigorous, we follow page 131 of Delbaen and Schachermayer. Let $\Omega_{100}\subset \Omega$ be the subset of the sample space on which $X_N>X_0\exp(\epsilon N)$ for all $N\geq100$ (the number 100 can be replaced with any other value). We have $\mathbb P[\Omega_{100}]>0$. Now define random variables $g_N\leq (H^N\cdot S)_{\infty}$ by $$g_N:= \begin{cases} \epsilon X_0\mathbb 1_{\Omega_{100}}\quad\text{ whenever } X_N>X_0\exp(\epsilon N),\\ -\frac1N X_0 \qquad\text{elsewise.}\end{cases} $$

Now the $g_N$ converge uniformly to the nonnegative random variable $f$ defined by $$f:=\epsilon X_0 \mathbb 1_{\Omega_{100}}.$$ In fact, we have $|g_N-f|_{\infty}\leq \frac1N X_0$ for all $N\geq100$. Since $\mathbb P[f>0]=\mathbb P[\Omega_{100}]>0$, this is a FLVR.

Now if $X$ would follow a different model than Geometric Brownian motion, then we can still assume that for $N$ large enough, its average growth is bounded from below by a certain percentage, for example $1\%$. Mathematically, we only need that $\mathbb P[\Omega_{C}]>0$ for some $C$. This is enough to construct the FLVR.

## Answer by Riemann (score 0, accepted)

https://quant.stackexchange.com/a/82383

For an asset that follows Geometric Brownian Motion, there is indeed a Free Lunch with Vanishing Risk. This is shown in the OP, and we can also verify it indirectly: I proved on Math SE that there is no Equivalent local Martingale Measure for Geometric Brownian Motion with drift larger than $\frac12\sigma^2$. Now the Fundamental Theorem of Asset Pricing tells us that there has to be a Free Lunch with Vanishing Risk.

However, the Geometric Brownian Motion model is not realistic. As @Trader453 mentioned in the comments, assets can lose all their value at some point in time. It seems realistic to assume that this happens with probability 1. This completely destroys my example of the Free Lunch with Vanishing Risk: in the example, we would have $\mathbb P[\Omega_{100}]=0$, and similarly when we replace 100 by any other value.