# Derive SDE for $Z_t = \frac {X_t}{Y_t}$

Derive SDE for $$Z_t = \frac {X_t}{Y_t}$$ using the 2 SDE below:

$$dX_t = rX_tdt + \sigma_XX_tdW_t$$
$$dY_t = rY_tdt + \sigma_XY_td\tilde {W}_t$$

I got $$dZ_t = Z_t\sigma_X(dW_t - d\tilde{W}_t)$$, is this correct or should I use product rule but I'm quite uncertain? Thank you

Given two stochastic processes $$X_t$$ and $$Y_t$$ and letting $$Z_t = \frac{X_t}{Y_t}$$, then $$dZ_t$$ is given by the quotient rule $$dZ_t = \frac{Y_tdX_t - X_tdY_t - dX_tdY_t}{Y_t^2} + \frac{X_t}{Y^3_t}(dY_t)^2$$

This can be found by using Ito's multidimensional formula using $$f(x, y) = \frac{x}{y}$$.

Now Letting $$dX_t = rX_tdt + \sigma_x X_t dW_t$$ and $$dY_t = rY_tdt + \sigma_x Y_t d\bar{W}_t$$ we have

$$(dY_t)^2 = r^2Y_t^2(dt)^2 + 2r\sigma_x Y_t^2dtd\bar{W}_t + \sigma^2_xY_t^2(d\bar{W}_t)^2 = \sigma_x^2Y_t^2dt$$

$$dX_tdY_t = r^2X_tY_t(dt)^2 + r\sigma_x X_tY_tdtd\bar{W} + r\sigma_xX_tY_tdtdW_t + \sigma_x^2X_tY_tdW_td\bar{W_t}$$ $$= \sigma_x^2X_tY_t\rho dt = 0$$ This is assuming $$dW_t$$ and $$d\bar{W_t}$$ are independent, $$\rho = 0$$. Otherwise if $$\rho \ne 0$$ and we would still have $$\sigma_x^2X_tY_t\rho dt$$. Using what we have above, we get

$$dZ_t = \frac{Y_t(rX_tdt + \sigma_xX_tdW_t) - X_t(rY_tdt + \sigma_xY_td\bar{W_t})}{Y_t^2} + \frac{X_t}{Y_t^3}\sigma_x^2Y_t^2dt$$ $$= \frac{\sigma_xX_tdW_t - \sigma_xX_td\bar{W_t} + \sigma_x^2X_tdt}{Y_t}$$ $$dZ_t = \sigma_x^2Z_tdt + \sigma_xZ_t(dW_t - d\bar{W_t})$$

• Thank you. What if the correlation is = 1 ? @user1
– ya23
Nov 23, 2020 at 4:17
• Then $\rho = 1$ and we have $\sigma_x^2X_tY_tdt$ Nov 23, 2020 at 4:35