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.


1 Answer 1


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.

  • $\begingroup$ Could you provide more detail on the Taylor expansion up to the second order. I am not understanding it. Shouldn't the first term be $$-\frac{1}{Y_t^2}dY_t$$? $\endgroup$
    – Wolfy
    Commented Apr 11, 2018 at 20:06
  • $\begingroup$ Yes, I made a mistake, I'll correct that. $\endgroup$ Commented Apr 11, 2018 at 20:10
  • $\begingroup$ Thanks, otherwise from that I understand everything. I will accept your answer. $\endgroup$
    – Wolfy
    Commented Apr 11, 2018 at 20:10

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