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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?
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  • $\begingroup$ $\ln S_t$ isn't a tradable asset, so the Ito expansion of it doesn't suggest anything wrong. $\endgroup$
    – Vim
    Jan 31, 2020 at 3:57
  • $\begingroup$ Not sure whether you have understood my question. $\endgroup$
    – Sandu Ursu
    Jan 31, 2020 at 14:42
  • $\begingroup$ Ok I see. Still it doesn't seem to be a contradiction, because the dynamics of $\ln S_t$ you arrrived at matches that of $S_t$. $\endgroup$
    – Vim
    Jan 31, 2020 at 18:19
  • $\begingroup$ For your Bachelier asset, if you really want to find the market price of risk, you'll need to check $dX_t/X_t$, find the drift $\mu(X, t)dt$ and diffusion $\sigma(X,t)dW$ and then the result will just be $$(\mu(X,t)-r)/\sigma(X,t)$$ which I'm sure doesn't match your result. (Note that the definition of MPR uses instantaneous growth rate, so you have to divide the whole dynamics equation by $X_t$ before starting) $\endgroup$
    – Vim
    Jan 31, 2020 at 18:33

1 Answer 1

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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: $$dexp(-rt)X_t=exp(-rt)\sigma dW_Q$$

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  • $\begingroup$ It is not readable. Can you please use latex? $\endgroup$
    – Gordon
    Jan 31, 2020 at 16:15
  • $\begingroup$ The $\mathbb Q$-dynamics of $X_t$ will then be $r X_t dt + \sigma dW_t^{\mathbb Q}$. Right? Which seems incorrect. $\endgroup$
    – Sandu Ursu
    Jan 31, 2020 at 17:16
  • $\begingroup$ The market price of risk is also known as th Sharpe ratio. It is the ratio of (reward above risk free) relative to risk. It has nothing to do with stochastic equation under risk neutral measure. For black & scholes model, one can use the sharp ratio for girsanov theorem but you can't do this for your Xt process. $\endgroup$ Jan 31, 2020 at 17:36
  • $\begingroup$ What is the dynamics of $X_t$ under $\mathbb Q$? $\endgroup$
    – Sandu Ursu
    Jan 31, 2020 at 17:52
  • $\begingroup$ replace dW by dWq - (mu-r)/sigma * dt and you will get the same equation with r instead of mu! $\endgroup$ Jan 31, 2020 at 18:01

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