Convexity of Risk-Neutral Measures in Incomplete Markets

# Multiple Risk-Neutral measures in incomplete market


# Multiple Risk-Neutral measures in incomplete market












This question is in regards to incomplete markets where multiple risk-neutral measures exist. I am a little bit confused by this idea. Say we have an incomplete market with only one stochastic process $X_t$ with

$$\mathrm{d}X_t = \mu(t, X_t)\mathrm{d}t + σ(t, X_t)\mathrm{d}W_t$$

Now let's say this market has two risk neutral measures ($\mathbb P_1$ and $\mathbb P_2$). Can I find another such measure? What is the relation between the two measures?

## Answer by Kevin (score 5, accepted)

https://quant.stackexchange.com/a/54678

- The first fundamental theorem of asset pricing states (basically) a market is free of arbitrage if and only if there exists at least one equivalent martingale measure (EMM)

- The second fundamental theorem of asset pricing states (basically) that if a market is free of arbitrage and complete, then the equivalent martingale measure is unique.

A market is complete (basically) if there are at least as many tradable assets than risk sources. In a discrete setting, you would need at least as many traded assets with linear independent payoffs than states of nature.

Suppose you have one stock but two (or more) risk sources (e.g. stochastic volatility, jumps, interest rates etc.). Suppose the market is free of arbitrage. Then there exists at least one EMM. But because the market is complete, it's not unique. So, there exists infinitely many EMMs.

There are only three possibilities

- There exists no EMM (if there exist arbitrage strategies)

- There exists one EMM (if the market is free of arbitrage and complete)

- There exist infinitely many EMMs (if the market is free of arbitrage but incomplete)

It's really easy to see. If $\mathbb P_1$ and $\mathbb P_2$ are EMMs, then so is $\mathbb P_\lambda=\lambda\mathbb P_1+(1-\lambda)\mathbb P_2$ for any $\lambda\in[0,1]$. So, it's impossible to only have 2 or 42 EMMs.

In reality, markets are probably (almost) free of arbitrage but incomplete. As a result, there does not exist the one martingale measure but a range of possible probability measures which give rise to infinitely many arbitrage-free prices for a derivative. Thus, you get an interval of acceptable prices at which you can trade your derivative. In order to single one measure out, you have to make further assumptions (derive a general equilibrium model, ignore some risk factors etc.). Alternatively, there's a large literature on hedging in incomplete markets.

Proof that $\mathbb P_\lambda$ is an equivalent martingale measure (EMM). Let $\mathbb P_1$ and $\mathbb P_2$ be two EMMs and $\lambda\in[0,1]$.

- The set of probability measures over a measurable set is convex. Thus, $\mathbb P_\lambda$ is a probability measure. Obviously, $\mathbb P_\lambda[\Omega]=1$.

- Again, measure theory gives us $$ \int f\mathrm{d}\mathbb P_\lambda = \lambda\int f\mathrm{d}\mathbb P_1 + (1-\lambda)\int f\mathrm{d}\mathbb P_2$$ Or, using expectations to denote integrals, $$\mathbb{E}^{\mathbb{P}_\lambda}[X]=\lambda\mathbb{E}^{\mathbb{P}_1}[X]+(1-\lambda)\mathbb{E}^{\mathbb{P}_2}[X]$$ This of course assumes that $f$ and $X$ are integrable and equally applies to conditional expectations.

- Now, we need to show that the discounted asset prices are $\mathbb{P}_\lambda$ martingales. Suppose we have $d$ risky assets, $S_t^{(d)}$ and one numeraire, $B_t$. By definition, $\frac{S_t^{(d)}}{B_t}$ is a martingale with respect to $\mathbb{P}_1$ and $\mathbb{P}_2$ (because they are EMMs). Let $t\geq s$. Then, \begin{align*} \mathbb{E}^{\mathbb{P}_\lambda}\left[\frac{S_t^{(d)}}{B_t}\Bigg|\mathcal{F}_s\right] &= \lambda \mathbb{E}^{\mathbb{P}_1}\left[\frac{S_t^{(d)}}{B_t}\Bigg|\mathcal{F}_s\right] + (1-\lambda) \mathbb{E}^{\mathbb{P}_2}\left[\frac{S_t^{(d)}}{B_t}\Bigg|\mathcal{F}_s\right] \\ &= \lambda \frac{S_s^{(d)}}{B_s} + (1-\lambda) \frac{S_s^{(d)}}{B_s} \\ &= \frac{S_s^{(d)}}{B_s}. \end{align*} Thus, discounted asset prices are martingales with respect to $\mathbb{P}_\lambda$ and thus, $\mathbb{P}_\lambda$ is another EMM.