Consider two measures $\mathbb Q^{M}$ and $\mathbb Q^{N}$, as well as the two numéraires $M$ and $N$, furthermore assume that $X\frac{N}{M}$ is a $\mathbb Q^{M}$-martingale. Furthermore, the covariation $X$ and $\frac{M}{N}$ satisfy the following:

$$dX(t)\cdot d\frac{M(t)}{N(t)} = \frac{M(t)}{N(t)}X(t)\gamma(t)dt$$

Finally we assume:

$$ dX(t) = X(t)(\mu(t)dt+\sigma(t)dW(t))\; \; \; \text{under }\mathbb Q^{M},$$

It is then stated that by using Girsanov, it follows

$$ dX(t) = X(t)(\mu(t)+\gamma(t) )dt+\sigma(t)dW(t))\; \; \; \text{under }\mathbb Q^{N} $$

Where does the term $\gamma$ come from when moving from measure $\mathbb Q^{N}$ to $\mathbb Q^{M}$?

My attempt: Since the justification was made by using Girsanov, I have attempted to reconcile this with what I know about Girsanov.

We have in general that $\mathbb E^{\mathbb Q^{N}}[X]= \mathbb E^{\mathbb Q^{M}}[X\frac{M}{N}]$, i.e. $\frac{d\mathbb Q^{N}}{d\mathbb Q^{M}}=\frac{M}{N}$

How can we write $$\frac{M(t)}{N(t)}$$ in exponential form though?

Any ideas on what I am missing?


1 Answer 1


From your assumption that \begin{align*} dX(t)\cdot d\frac{M(t)}{N(t)} &= \frac{M(t)}{N(t)}X(t)\gamma(t)dt,\\ dX(t) &= X(t)\big(\mu(t)dt+\sigma(t)dW(t)\big), \end{align*} under $\mathbb Q^{M}$, we can speculate that \begin{align*} d\frac{N(t)}{M(t)} = -\frac{N(t)}{M(t)}\frac{\gamma(t)}{\sigma(t)}dW(t) \tag{1} \end{align*} as $\frac{N(t)}{M(t)}$ should be a martingale under $\mathbb Q^{M}$ with the numeraires $M$. Then, under $\mathbb Q^{M}$, \begin{align*} d\frac{M(t)}{N(t)} = \frac{M(t)}{N(t)}\bigg(\frac{\gamma^2(t)}{\sigma^2(t)} dt + \frac{\gamma(t)}{\sigma(t)} dW_t\bigg).\tag{2} \end{align*}

However, I doubt that you actually meant \begin{align*} dX(t)\cdot d\frac{N(t)}{M(t)} &= \frac{N(t)}{M(t)}X(t)\gamma(t)dt. \end{align*}

From $(1)$, \begin{align*} \frac{N(t)}{M(t)} = \frac{N(0)}{M(0)} \exp\bigg(-\frac{1}{2}\int_0^t\frac{\gamma^2(s)}{\sigma^2(s)} ds - \int_0^t \frac{\gamma(s)}{\sigma(s)}dW(s) \bigg). \end{align*} Based on Girsanov theorem, $\hat{W}=\{\hat{W}(t), \, t \ge 0\}$, where, for $t \ge 0$, \begin{align*} \hat{W}(t) = W(t) + \int_0^t \frac{\gamma(s)}{\sigma(s)}ds, \end{align*} is a standard Brownian motion under $\mathbb Q^{N}$. Then, \begin{align*} dX(t) = X(t)\Big[\big(\mu(t)-\gamma(t) \big)dt+\sigma(t)d\hat{W}(t)\Big], \end{align*} under $\mathbb Q^{N}$.

  • $\begingroup$ Welcome back, Gordon :) Good to have you on here. Ps: maybe you find this one also interesting...? Feel free to contribute... $\endgroup$ Commented Feb 2, 2022 at 8:08
  • $\begingroup$ Thanks @JanStuller. I will have a look. $\endgroup$
    – Gordon
    Commented Feb 2, 2022 at 14:48

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