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