2
$\begingroup$

For a standard geometric Brownian motion model of stock prices: $$ dS = a S dt + \sigma S dZ$$ we can transform the process to be under risk neutral measure: $$ dS = r S dt + \sigma S d \tilde{Z}$$ and from the references I found, this risk neutral measure is "unique".

If we make a transform, say $$ dS = r S dt + \tau S d \hat{Z}$$ where $\tau$ is different from $\sigma$, this equation gives the correct price of stock. but Black-Scholes equation will fail as we have changed volatility.

However, for a discrete model, e.g. a tree model, if there are $n$ states of world, then we need $n-1$ assets plus cash to uniquely pin down risk neutral measure.

Question:The Brownian motion model in effect has infinite number of states and only one asset, then where does uniqueness of risk neutral measure come from?

$\endgroup$
3
$\begingroup$

The uniqueness of the risk-neutral measure comes from the abundance of tradable assets. Let $B_t$ be the money-market account at time $t$. Let $Q_1$ and $Q_2$ be two risk-neutral measures. Then, for any tradable asset $X$ with maturity $T$, \begin{align*} E^{Q_1}\left(\frac{X_T}{B_T}\right) &= E^{Q_2}\left(\frac{X_T}{B_T}\right)\\ &=\frac{X_0}{B_0}. \end{align*} For any $A\in \mathcal{F}_T$, we define an asset with payoff $$\mathbb{I}_{A} B_T.$$ Note that, this deal may not be exchange traded, however, it can be made over-the-counter. Then \begin{align*} Q_1(A) &= E^{Q_1}\left(\frac{\mathbb{I}_{A} B_T}{B_T}\right)\\ &= E^{Q_2}\left(\frac{\mathbb{I}_{A} B_T}{B_T}\right)\\ &= Q_2(A). \end{align*} That is, $Q_1=Q_2$.

$\endgroup$
  • $\begingroup$ But did B-S model assume the indicator function asset is traded? The original proof involves replicating time differential of option with differentials of cash and stock and invoking non-arbitrage. $\endgroup$ – user3785097 Jun 26 '15 at 2:41
  • 1
    $\begingroup$ @user3785097, $E(\mathbb{I}_{A} \mid \mathcal{F}_t)$ is a martingale, on which the differential is taken. $\endgroup$ – Gordon Jun 26 '15 at 12:35
1
$\begingroup$

essentially it comes down the fact that the dyadic quadratic variation of $W_t$ is $t$ with probability 1 and any measure change has to preserve this fact. Changing volatility would violate this invariance.

$\endgroup$
  • 1
    $\begingroup$ any equivalent measure change $\endgroup$ – Ulysses Jun 22 '15 at 9:02
1
$\begingroup$

Let M denote the number of underlying traded assets in the model excluding the risk free asset, and let R denote the number of random sources. Generically we then have the following relations: 1. The model is arbitrage free if and only if M ≤ R. 2. The model is complete if and only if M ≥ R. 3. The model is complete and arbitrage free if and only if M = R. Black Scholes Model is Complete and arbitrage free then risk-neutral measure is unique.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.