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Market Price of Risk for Log Prices and Bachelier Dynamics

Article Quant Q&A · Author: Sandu Ursu
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Summary

The document asks how to define the market price of risk when a process is written for a log price rather than for the asset price itself. It starts from geometric Brownian motion, where the drift adjustment under a risk-neutral measure is tied to the difference between the physical expected return and the risk-free rate. It then considers the log-price process, whose drift includes the Itô correction, and questions whether applying the same formula directly to that drift gives consistent risk-neutral dynamics. It also asks how the issue appears in the Bachelier model.

The supplied answer rewrites the stated process in a form proportional to the asset level and derives a state-dependent risk premium, then proposes a change of measure intended to make the discounted asset a martingale. This illustrates that the market price of risk depends on the modeled state variable and its diffusion coefficient; a log-price drift cannot simply be substituted into the price-process formula. The response's transformation and notation warrant careful verification, particularly because its resulting drift is state dependent. The document offers no broader derivation or worked Bachelier case, so the answer should not be treated as a complete resolution.

Key ideas

  • The market price of risk for an asset-price process compares its physical drift with the risk-free rate relative to its diffusion scale.
  • A log-price process has an Itô-adjusted drift, so its drift cannot be used as though it were the asset's price drift.
  • Changing the modeled state variable changes the drift and diffusion terms relevant to a measure change.
  • The answer proposes a state-dependent adjustment for the process in the question, but its algebra and assumptions need checking.
  • The document raises the Bachelier case but does not provide a full treatment of it.

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StatisticsDerivatives pricingRisk management
Full text
# Some aspects of the market price of risk


# Some aspects of the market price of risk












I am a little confused about the market price of risk.

Take the following geometric Brownian motion:

$$dS_t = \mu S_t dt+\sigma S_t dW_t$$

The market price of risk is defined as:

$$\frac{\mu-r}{\sigma}$$

And by Girsanov's theorem, we get the dynamics under $\mathbb Q$ to be:

$$ \begin{align}dS_t &= \mu S_t dt+\sigma S_t \left(dW^{\mathbb Q}_t - \frac{\mu-r}{\sigma}dt\right)\\ & = r S_t dt+\sigma S_t dW^{\mathbb Q}_t \end{align} $$

Question:

- What is the marker price of risk for an asset having the following process (physical world):

$$dX_t=\left(\mu-\frac{1}{2}\sigma^2\right)dt+ \sigma \,dW$$

I would guess that it is:

$$\frac{\mu-\frac{1}{2}\sigma^2-r}{\sigma}$$

And if so, by applying Girsanov's theorem, we can derive the dynamics under the risk-neutral measure to be:

$$dX_t=r dt+ \sigma \,dW^\mathbb{Q}$$

However, one may notice that $dX_t$ is just the $d\ln S_t$. By applying Itô's lemma, and substituting $dS_t$ under $\mathbb{Q}$ (from above), we get the risk-neutral dynamics:

$$d\ln S_t=\left(r-\frac{1}{2}\sigma^2\right)dt+ \sigma \,dW^\mathbb{Q}$$

This result suggests that my guess was wrong. Simply taking the drift of the model to be $\mu$ and the volatility of the model to be $\sigma$ to compute the market price of risk is incorrect.

A clarifying question:

- What is the market price of risk for the Bachelier model?

## Answer by Valometrics.com (score 1)

https://quant.stackexchange.com/a/50935

You can transform your process to the following:

$$dX_t= \left[(\mu-\frac{1}{2}\sigma^2) /X_t\right] \times X_t \times dt + (\frac{\sigma}{Xt}) \times X_t \times dW_t$$

So the market price of risk is equal to:

((mu-square(sigma)/2) /Xt)-r/(sigma/Xt)=((mu-square(sigma)/2)- (r * Xt))/sigma

$$(\mu-\sigma^2/2- r X_t)/\sigma$$ if you are looking for a risk neutral measure under which the discounted Xt price is a martingale you can define: $$dW_P=dW_Q-((\mu-\sigma^2/2- r X_t)/\sigma)dt$$ then: $$dX_t=rX_t dt+\sigma dW_Q$$ you can then use ito lemma to get: $$dexp(-rt)X_t=exp(-rt)\sigma dW_Q$$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.