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?


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 2


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



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


  • $\begingroup$ If $P=\sigma_2S_1S_2 -\sigma_1 S_2S_1$, then $dP =( \sigma_2-\sigma_1)d(S_1S_2)$, which is $dP = ( \sigma_2-\sigma_1) (S_1dS_2 +S_2dS_1+dS_1dS_2$) $\endgroup$
    – ir7
    Jul 28, 2020 at 17:22
  • $\begingroup$ In your first equation you are stating the portfolio is self-financing (for bank account rate set to 0). $\endgroup$
    – ir7
    Jul 28, 2020 at 17:41
  • $\begingroup$ @ir7 Now that I think of it you seem correct. How would you correct for that? $\endgroup$
    – fes
    Jul 28, 2020 at 19:04
  • $\begingroup$ I'm struggling too.I think I did it, but with a different portfolio (see my updated answer). $\endgroup$
    – ir7
    Jul 28, 2020 at 20:12
  • $\begingroup$ I think same technique (if valid) works for the given weights too (my second update). $\endgroup$
    – ir7
    Jul 28, 2020 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


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:

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


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

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:

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

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

  • 1
    $\begingroup$ Nice argument adding the bond to make the portfolio self-financing $\endgroup$
    – StackG
    Jul 31, 2020 at 8:18
  • $\begingroup$ @StackG Thank you. $\endgroup$
    – ir7
    Jul 31, 2020 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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