Using Itô’s Lemma to Convert Log-Ratio Dynamics

Summary

The note shows how to derive the dynamics of a ratio of two processes after first finding the dynamics of its logarithm. Define the log ratio as a process X and the ratio itself as its exponential. Applying Itô’s formula to the exponential adds a second-order term involving the quadratic variation of X.

Substituting the stated log-ratio dynamics produces a drift correction and a Brownian shock term. If the driving process is standard Brownian motion, its quadratic variation is elapsed time, so the second-order term cancels the negative drift correction. This explains the transition from the log ratio to the ratio. The conclusion depends on the assumption about the driver: if it is not Brownian motion, its quadratic variation must be specified before simplifying.

Key ideas

• Set the ratio equal to the exponential of its logarithm.
• Applying Itô’s formula to the exponential adds a quadratic-variation term.
• For a standard Brownian driver, quadratic variation equals elapsed time.
• The resulting second-order term cancels the stated drift correction.

Tags

# How to derive equivalent martingale measure using Ito's Lemma


# How to derive equivalent martingale measure using Ito's Lemma












Can someone explain how to get equation 27.14 below? I understand the first usage of Ito's Lemma to get $d(\ln f-\ln g)$ but I do not understand how to use Ito's Lemma to go from $d(\ln \frac{f}{g})$ to $d(\frac{f}{g})$. Can someone help to elucidate this process?

## Answer by math (score 2, accepted)

https://quant.stackexchange.com/a/11334

You have two processes, $X_t:=\log{\frac{f}{g}}$ and $Y_t=\frac{f}{g}$. Note, I use $\log$ for the natural logarithm. Hence we have $Y_t=\exp{(X_t)}$. Therefore, applying Itô:

$$dY_t=\exp{(X_t)}dX_t + \frac{1}{2}\exp{(X_t)}d\langle X,X\rangle_t$$

Using the dynamics of $X_t$, we get

$$dY_t=\frac{f}{g}[-\frac{(\sigma_f-\sigma_g)^2}{2}dt+(\sigma_f-\sigma_g)dz]+\frac{f}{g}\frac{(\sigma_f-\sigma_g)^2}{2}d\langle z, z\rangle_t$$

I think $z$ is a Brownian Motion, such that $d\langle z,z\rangle_t=dt$, which yields the desired result. If $z$ is not a BM, please provide additional information.

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