Deriving Ratio Dynamics with Itô’s Rule for Numeraire Changes

# Change of numeraire and the adjustment to the growth rate in terms of the numeraire ratio


# Change of numeraire and the adjustment to the growth rate in terms of the numeraire ratio












I'm working with a process $f$ and considering how its drift adjusts when moving between different numeraires. Here’s what I have so far:

#### Process $f$ with numeraire $g$ :

- The process $f$ has a drift of $f \cdot \lambda \cdot \sigma_f \, dt$ under a real world measure.

- with numeraire $g$, the process $d\left(\frac{f}{g}\right)$ has a drift (again under a real world measure): $$ \frac{f}{g} \left( (\sigma_f - \sigma_g)(\lambda - \sigma_g) \right) dt $$

#### Process $f$ with numeraire $h$ :

- When switching to numeraire $h$, the drift of $f$ is now $f \cdot \lambda^* \cdot \sigma_f \, dt$ (under a real world measure).

- with numeraire $h$, the process $d\left(\frac{f}{h}\right)$ has a drift (under a real world measure): $$ \frac{f}{h} \left( (\sigma_f - \sigma_h)(\lambda^* - \sigma_h) \right) dt $$

#### Ratio between $h$ and $g$ :

- I know the dynamics of the numeraire ratio $d\left(\frac{h}{g}\right)$ are given by: $$ d\left(\frac{h}{g}\right) = \left( \lambda^* \sigma_h - \lambda \sigma_g + \sigma_g^2 - \sigma_h \sigma_g \right) \frac{h}{g} dt + (\sigma_h - \sigma_g) \frac{h}{g} dz $$

- From this, I deduce that: $$ \sigma_w = (\sigma_h - \sigma_g) $$ where w = h/g

#### Question:

I also know that the adjustment to the drift of $f$ when moving from numeraire $g$ to numeraire $h$ is $(\lambda^* - \lambda) \sigma_f$.

But I want to express this change in drift of f in terms of the volatility of the numeraire ratio: $$ \alpha_f = \sigma_w \cdot \sigma_f $$

How can I prove this result?

Side note: I can then show that this result is equal to the instantaneous covariance of w and f.

## Answer by Kurt G. (score 2)

https://quant.stackexchange.com/a/80976

First of all, regardless what $f$ and $g$ are, we have (integration-by-parts) $$\tag1 d\left(\frac fg\right)=\frac {df}{g}+f\,\,d\left(\frac 1g\right)+d\left\langle f,\frac 1g \right\rangle_t\,. $$ If these processes are of the form $$\tag2 df=\mu_f\,dt+\sigma_f\,dW^f_t\,,\quad dg=\mu_g\,dt+\sigma_g\,dW^g_t $$ then (by Ito) $$\tag3 d\left(\frac 1g\right)=-\frac{dg}{g^2}+\frac{d\langle g\rangle_t}{g^3} =-\frac{\mu_g\,dt+\sigma_g\,dW^g_t}{g^2}+\frac{\sigma_g^2\,dt}{g^3}\,. $$ Can you proceed to figure out the drift of $\dfrac fg\,?$

Hint: the drift of $\frac fg$ is obtained from putting the expressions (2) and (3) into (1):

\begin{align}\tag6 d\left(\frac fg\right)=\frac{\mu_f\,dt+\sigma_f\,dW^f_t}g-f\frac{\mu_g\,dt+\sigma_g\,dW^g_t}{g^2}+f\frac{\sigma^2_g\,dt}{g^3}-\frac {\sigma_f\,\sigma_g}{g^2}\,d\langle W^f,W^g\rangle_t\,. \end{align} If $W^f$ and $W^g$ are correlated with $\rho$ the drift becomes $$\tag5 \left\{ \frac{\mu_f}g-f\frac{\mu_g}{g^2}+f\frac{\sigma_g^2}{g^3}-\frac{\sigma_f\,\sigma_g\,\rho}{g^2}\right\}\,. $$

- Even if I assume $\rho=1$ and $\mu_f=f\lambda\sigma_f$ (like you do) I can for the life of me not related this to your claim that the drift of $d\left(\dfrac fg\right)$ is $$\tag6 \frac{f}{g} \left( (\sigma_f - \sigma_g)(\lambda - \sigma_g) \right)\,dt\,. $$

- What do you assume about $\mu_g\,?$ Give all your definitions and do some work.