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0

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 ...


2

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 - ...


6

$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming $X_t$ is here a continuous, square integrable stochastic process) $$ df(t,X_t) = ...


5

What can be shown is that the above expressions are equal in probability. First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments: The expected values - it is zero ... easy to see. Next what you did not specify is that the correlation between ...



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