Market price of risk on two assets

Question

Under the assumptions of the Black--Scholes model, I read that the market price of risk of two assets $S_1$ and $S_2$ are the same, if they both follow Geometric Brownian motion driven by the same Brownian motion.

The claim is that if \begin{align*} dS_1(t)&=\mu_1S_1(t)dt+\sigma_1S_1(t)dW(t),\qquad\text{and} \\ dS_2(t)&=\mu_2S_2(t)dt+\sigma_2S_2(t)dW(t) \end{align*} then $$\frac{\mu_1-r}{\sigma_1}=\frac{\mu_2-r}{\sigma_2}$$ where $r$ is the risk-free rate. The 'proof' of this relies on constructing a portfolio of $\sigma_2S_2$ units of $S_1$ and $-\sigma_1S_1$ units of $S_2$ and assuming that this portfolio is self-financing, then using Ito's formula on the value of this portfolio to show that it only has a drift term. I don't believe the assumption that this portfolio is self-financing holds.

Does the claim hold, and if so is there a proof of this result?

EDIT:

Thought about this a bit more and realised it falls out of the Second Fundamental Theorem of Asset Pricing where the risk-neutral measure is unique if and only the market is arbitrage-free and complete.

Assuming that the market is arbitrage-free and complete, we can construct measures $\mathbb{Q}_1$ and $\mathbb{Q}_2$ such that $$W_1(t)=W(t)+\frac{\mu_1-r}{\sigma_1}t,\qquad W_2(t)=W(t)+\frac{\mu_2-r}{\sigma_2}t$$ are $\mathbb{Q}_1$ and $\mathbb{Q}_2$ Brownian motions respectively. Both these measures give rise to a measure such that discounted asset prices are martingales. By uniqueness, $\mathbb{Q}_1=\mathbb{Q}_2$ and so $$\frac{\mu_1-r}{\sigma_1}=\frac{\mu_2-r}{\sigma_2}.$$

Answer 1

Here is a simple solution using the equivalence of no arbitrage and the existence of a stochastic discount factor. Let the SDF be $\Lambda(t)$. This evolves as

$$\frac{d\Lambda(t)}{\Lambda(t)}=-rdt-\varphi(t) dW(t),$$

where we used the fact that the drift of the SDF is the risk-free rate and that there is only one source of uncertainty. The standard pricing conditions for the stocks are

$$(\mu_1-r)dt=-\frac{dS_1(t)}{S_1(t)}\frac{d\Lambda(t)}{\Lambda(t)}=\sigma_1\varphi(t)dt$$

$$(\mu_2-r)dt=-\frac{dS_2(t)}{S_2(t)}\frac{d\Lambda(t)}{\Lambda(t)}=\sigma_2\varphi(t)dt.$$

That is the market price of risk $\varphi(t)$ is given by

$$\varphi(t)=\frac{\mu_1-r}{\sigma_1}=\frac{\mu_2-r}{\sigma_2}$$

Answer 2

Another way to look at it, is that we have a one-dimensional Brownian motion process driving the market but two risky assets. The market price of risk process (giving the equivalent martingale measure), $\lambda$, must then respect two conditions:

$$ \lambda \sigma_1 =\mu_1 -r $$ $$ \lambda \sigma_2 =\mu_2 -r $$

which implies

$$\frac{\mu_1-r}{\sigma_1}=\frac{\mu_2-r}{\sigma_2}.$$

Update: One other way (same strategy as in the question, but different portfolio).

For a self-financing portfolio $(\gamma^1, \gamma^2,\beta) $, we have:

$$ P_t = \gamma^1_tS_t^1 + \gamma^2_tS_t^2 + \beta_tB_t $$

and

$$ dP_t = \gamma^1_t dS_t^1 + \gamma^2_t dS_t^1 +\beta_tdB_t $$

which is the same as

$$ dP_t = \gamma^1_t dS_t^1 + \gamma^2_t dS_t^1 +r(P_t - \gamma^1_tS_t^1 - \gamma^2_tS_t^2) dt $$

(used $dB_t = rB_t dt$ in the last step)

It turns out that $\beta_t$ needs to be risky, function of assets. We take:

$$ \gamma_t^1 = (\sigma_1 S_t^1)^{-1} $$

$$ \gamma_t^2 = (\sigma_2 S_t^2)^{-1} $$

and $\beta$ defined by equation:

$$ d\beta_t = B_t^{-1}(\gamma_t^1 dS_t^1 + \gamma_t^2 dS_t^2 )$$

This is self-financing because:

$$ dP_t = d(\gamma^1_tS_t^1 + \gamma^2_tS_t^2 + \beta_tB_t) $$ $$ = d(\sigma_1^{-1} + \sigma_2^{-1} + \beta_tB_t) $$ $$ = B_t d\beta_t + \beta_tdB_t $$ $$ = \gamma_t^1 dS_t^1 + \gamma_t^2 dS_t^2 + \beta_tdB_t.$$

(we used the fact that quadratic covariation between $\beta_t$ and $B_t$ is $0$)

Finally, some straightforward calculations take us now to:

$$ dP_t= \gamma^1_t dS_t^1 + \gamma^2_t dS_t^1 +r(P_t - \gamma^1_tS_t^1 - \gamma^2_tS_t^2) dt $$

$$ = \left(rP_t + \frac{\mu_1-r}{\sigma_1} -\frac{\mu_2-r}{\sigma_2} \right)dt $$

Update 2: For the weights in the question, we can choose $\beta$ such that

$$d \beta = - B^{-1}(\sigma_2 S^1 dS^2 - \sigma_1 S^2 dS^1 + (\sigma_2 -\sigma_1)dS^1dS^2) $$

For $$ P = \sigma_2 S^2S^1 - \sigma_1 S^1S^2 + \beta B$$

we then have:

$$ dP = (\sigma_2 -\sigma_1)d(S^1S^2) + Bd\beta + \beta dB$$ $$ = (\sigma_2 -\sigma_1)(S^1dS^2 + S^2dS^1 + dS^1dS^2) + Bd\beta + \beta dB $$ $$ = \sigma_2S_2 dS^1 -\sigma_1 S^1 dS^2 + \beta dB $$

So, the final portfolio dynamics is:

$$ dP= \sigma_2 S^2dS^1 - \sigma_1 S^1dS^2 +r(P_t - \sigma_2 S^2S^1 + \sigma_1 S^1S^2 ) dt $$

$$ = \left(rP + \sigma_2(\mu_1-r)S^1S^2 - \sigma_1(\mu_2-r)S^1S^2\right) dt$$