Applying Multivariate Itô’s Lemma to a Ratio of Diffusions
Summary
The document works through the ratio M = X/Y when X and Y follow correlated geometric Itô processes. Applying the multivariate Itô formula requires the first and second derivatives of the ratio, including its mixed partial derivative, and the quadratic variation and cross variation terms. These terms contribute to the drift as well as the diffusion, so simply subtracting the proportional changes in X and Y misses corrections.
The accepted derivation gives the ratio’s diffusion as the difference of the two Brownian shocks and its drift as the numerator drift minus the denominator drift, plus the denominator variance correction, minus the correlation covariance correction. The question’s proposed drift has a sign error on the denominator drift; one reply also repeats that error, despite calling its expression correct. The topic is mathematically useful, but readers should verify signs directly from Itô’s formula. The result assumes the stated diffusion model and constant volatilities and correlation.
Key ideas
- The ratio function has a nonzero mixed partial derivative that contributes through cross variation.
- The denominator’s variance adds a positive correction to the ratio drift.
- Correlation between the two processes contributes a covariance correction to the drift.
- The ratio’s diffusion contains the numerator shock minus the denominator shock.
- The drift should subtract the denominator drift, so the question’s proposed sign is inconsistent.
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Full text
# Multivariate Ito problem $M_t=\frac{X_t}{Y_t}$
# Multivariate Ito problem $M_t=\frac{X_t}{Y_t}$
I am analyzing a problem given in the lecture slides published here (Slide 7-8 Example of Multivariate Ito’s Lemma).
Can anybody explain how the $M_t$ was calculated out of the Ito formula. I cannot get the same results.
Summary of the problem: There are two Ito processes given:
$$\frac{dX_t}{X_t}=\mu_{x} dt + \sigma_{x} dZ^{1}_{t} \quad \quad (1)$$
$$\frac{dY_t}{Y_t}=\mu_{y} dt + \sigma_{y} dZ^{2}_{t} \quad \quad (2)$$
so that $M_t=\frac{X_t}{Y_t}$ and the instantaneous volatility of $\frac{dM_t}{M_t}$ needs to be found.
Applying Ito Lemma to the function $M_t = f(X_t,Y_t) = X_t/Y_t$ gives
$$dM_t=\frac{\partial f(X_t,Y_t)}{\partial X_t} dX_t + \frac{\partial f(X_t,Y_t)}{\partial Y_t} dY_t + \frac{1}{2} \left(\frac{\partial^2 f(X_t,Y_t)}{\partial X_t^2} (dX_t)^2+ 2 \frac{\partial^2 f(X_t,Y_t)}{\partial X_t \partial Y_t} dX_t \ dY_t + \frac{\partial^2 f(X_t,Y_t)}{\partial Y_t^2}(dY_t)^2 \right) \quad \quad (3)$$ This should result in something like this:
$$dM_t=M_t \big{(} \mu_x + \mu_y - \rho \sigma_x \sigma_y + \sigma_y^2 \big{)} \ dt + M_t \sigma_x dZ_t^1 - M_t \sigma_y dZ_t^2 \quad \quad (4)$$
$$\frac{dM_t}{M_t}= \big{(} \mu_x + \mu_y - \rho \sigma_x \sigma_y + \sigma_y^2 \big{)} \ dt + \sigma_x dZ_t^1 - \sigma_y dZ_t^2 \quad \quad (5)$$
Plugging the partial derivatives into eq(3)
$$\frac{\partial f}{\partial X_t}=\frac{1}{Y_t}, \quad \frac{\partial^2 f}{\partial X_t^2}=0, \quad \frac{\partial f}{\partial Y_t}=\frac{-X_t}{Y_t^2}, \quad \frac{\partial^2 f}{\partial Y_t^2}=\frac{2 X_t}{Y_t^3}, \quad \frac{\partial^2 f}{\partial X_t \partial Y_t}=\frac{-1}{Y_t^2}$$
and substituting with (1) and (2) should give me the equation (4) or (5), but I cannot get it, I am getting something like
$$\frac{dM_t}{M_t} = \frac{dX_t}{X_t} - \frac{1}{2} \frac{dy}{y} X_t - \frac{1}{2} \frac{dy}{y} X_t dx \quad \quad (6)$$
Can anybody explain the final transition for the $\frac{dM_t}{M_t}$ equation?
## Answer by M. Jeunesse (score 3, accepted)
https://quant.stackexchange.com/a/25705
What is written in attached slides is correct.
However, what you have written is not correct.
Setting $M_t=\frac{X_t}{Y_t}$, and applying Ito formula will lead to :
$$dM_t=\frac{dX_t}{X_t} M_t -\frac{dY_t}{Y_t} M_t + M_t \frac{d<Y>_t}{Y^2_t}-\frac{d<X,Y>_t}{Y^2_t}$$
which gives you in your case :
$$dM_t = (\mu_x dt+\sigma_x dZ^1_t)M_t - (\mu_y dt+\sigma_y dZ^2_t)M_t+M_t\sigma^2_y dt - \frac{\rho \sigma_x X_t \sigma_y Y_t dt}{Y_t^2}$$
which leads you after simplification to :
$$\frac{dM_t}{M_t} = (\mu_x -\mu_y -\rho\sigma_x\sigma_y +\sigma_y^2) dt + \sigma_x dZ^1_t - \sigma_y dZ^2_t$$
## Answer by Michal (score 0)
https://quant.stackexchange.com/a/25720
I found the problem ,the partial derivatives were incorrectly derived.
$$dM_t = \frac{1}{Y_t} dX_t - \frac{-X_t}{Y_t^2} dY_t + \frac{-1}{Y_t^2} dX_t dY_t + \frac{X_t}{Y_t^2} dY_t \quad / : \frac{Y_t}{X_t} \quad \quad (6)$$
$$\frac{dM_t}{M_t} = \frac{dX_t}{X_t} - \frac{dY_t}{Y_t} - \frac{dX_t dY_t}{X_t Y_t} + \frac{(dY_t)^2}{(Y_t)^2} \quad \quad \quad (7)$$
after substituting with (1) and (2) and doing all the algebra I get the correct final equation
$$\frac{dM_t}{M_t}= \big{(} \mu_x + \mu_y - \rho \sigma_x \sigma_y + \sigma_y^2 \big{)} \ dt + \sigma_x dZ_t^1 - \sigma_y dZ_t^2 \quad \quad (5)$$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.