Multivariate Ito problem $M_t=\frac{X_t}{Y_t}$

Question

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 1

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 2

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

$$\frac{dM_t}{M_t}= \big{(} \mu_x + \mu_y - \rho \sigma_x \sigma_y + \sigma_y^2 \big{)} \ dt + \sigma_x dZ_t^1 - \sigma_y dZ_t^2 \quad \quad (5)$$