Applying Itô’s Lemma to the Reciprocal of a Geometric Brownian Motion
Summary
The document works through the reciprocal of a stochastic process that follows geometric Brownian motion, as a step toward finding the process for a ratio of two correlated processes. It identifies a likely notation error in the stated reciprocal formula: the diffusion term should use the Brownian motion driving the denominator process.
The derivation applies Itô’s lemma to the reciprocal function. Its second derivative contributes a correction because the square of a Brownian increment has quadratic variation proportional to time. Substituting the process dynamics into the differential therefore produces both drift and diffusion terms; ordinary first-order differentiation would omit part of the drift. The response gives a brief Taylor-expansion rationale for retaining second-order terms. It does not complete the ratio calculation or explore the effect of correlation between the two Brownian motions, so those steps must be handled separately.
Key ideas
- The reciprocal process must retain the Brownian motion that drives the original denominator process.
- Itô’s lemma includes a second-derivative term that contributes to the drift.
- Brownian quadratic variation satisfies a differential rule proportional to time.
- The reciprocal calculation is an intermediate step, and correlation matters when forming the full ratio.
Tags
Full text
# Finding the process of $X/Y$
# Finding the process of $X/Y$
This comes from Mark Joshi's concepts of mathematical finance exercise 4 chapter 11.
> If $$dX_t = \alpha X_t dt + \beta X_t dW_t$$ $$dY_t = \alpha Y_t dt + \gamma Y_t d\tilde{W}_t$$ with $W$ and $\tilde{W}$ correlated Brownian motions with correlation $\rho$. Find the process of $X/Y$
I do not understand how the author gets
$$d\left(\frac{1}{Y_t}\right) = \frac{1}{Y_t}\left[(\gamma^2 - \alpha)dt - \gamma dW_t)\right]$$
shouldn't the $W_t$ be $\tilde{W}_t$? Please provide multiple steps for understanding.
## Answer by Raskolnikov (score 5, accepted)
https://quant.stackexchange.com/a/39240
You are right about the dropped $\sim$, it's probably just a typo. Furthermore, remember that in stochastic calculus, you have to take into account second order derivatives, i.e.
$$d\left(\frac{1}{Y_t}\right) = -\frac{1}{Y_t^2}dY_t + \frac{1}{2}\frac{2}{Y_t^3}dY_t^2$$
which is the Taylor expansion up to second order. Then you substitute $dY_t$ in the right hand side and take into account that
$$dY_t^2 = \gamma^2 Y_t^2 d\tilde{W}_t^2 = \gamma^2 Y_t^2 dt \; . $$
The reason you keep second order terms is because they might contain terms with quadratic variation proportional to $dt$. This is the case of Brownian motion itself which has $dW_t^2=dt$.
Extra on Taylor expansion:
Provided a function is sufficiently differentiable in some point of its domain, it is possible to approximate it by a polynomial in some neighborhood of that point. Say $f(x)$ around the point $a$ is approximately equal to
$$f(x) \approx f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2$$
Or if we put $x-a=dx$ and $f(x)-f(a)=df$ we can write this as
$$df \approx f'(a)dx+\frac{1}{2}f''(a)dx^2$$
This is true if the function is twice differentiable and some additional conditions which are a bit too technical to expand upon here.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.