# Market price of risk on two assets

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

## 2 Answers

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

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

• If $P=\sigma_2S_1S_2 -\sigma_1 S_2S_1$, then $dP =( \sigma_2-\sigma_1)d(S_1S_2)$, which is $dP = ( \sigma_2-\sigma_1) (S_1dS_2 +S_2dS_1+dS_1dS_2$) – ir7 Jul 28 '20 at 17:22
• In your first equation you are stating the portfolio is self-financing (for bank account rate set to 0). – ir7 Jul 28 '20 at 17:41
• @ir7 Now that I think of it you seem correct. How would you correct for that? – fesman Jul 28 '20 at 19:04
• I'm struggling too.I think I did it, but with a different portfolio (see my updated answer). – ir7 Jul 28 '20 at 20:12
• I think same technique (if valid) works for the given weights too (my second update). – ir7 Jul 28 '20 at 21:37

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:

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

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

• Nice argument adding the bond to make the portfolio self-financing – StackG Jul 31 '20 at 8:18
• @StackG Thank you. – ir7 Jul 31 '20 at 14:13