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I'm studying stochastic differential equations, but I'm stucked in this problem and I'd like to ask if anyone can suggest me how to solve it in a rigorous way, since I am strongly tempted to simplify the differential $dt$ and then make the substitution, knowing that it is not conceptually correct.
Up to this point, I've already computed $dY_t$ thanks to Ito's lemma.
What is not clear to me is how to compute and the meaning of the derivative $$\frac{dY_t}{dt}$$ Here's the text of the problem:
Given the stochastic differential $$dX_t = \frac{1}{X_t}dt +\sigma X_tdW_t$$ let's define the processes $$F_t = e^{\frac{1}{2}\sigma^2 t-\sigma W_t}$$ and $$Y_t = X_tF_t$$ whose stochastic differential is given by $$dY_t = \frac{F_t}{X_t}dt$$ and by writing $X_t = \frac{Y_t}{F_t}$, prove that $$\frac {dY_t}{dt}= \frac{F_t^2}{Y_t}$$ with $\sigma \in \mathbb{R}$ and $X_0=x, Y_0 = x$.

Francesco Falzoni
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Aug 29 '24 at 19:12
  • Agree with AnotherUser on content. HINT: Ito's lemma –  Aug 29 '24 at 19:37
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    The rigorous way is to use Ito calculus to show $dY_t = \frac{F_t}{X_t},dt$. Then there is nothing more to prove. – Kurt G. Aug 29 '24 at 19:39
  • The stochastic differential $dY_t=\frac{F_t}{X_t}dt$ is the outcome of a previous question and I computed it thanks to Ito's lemma, but the problem now is to compute the time derivative $\frac{dY_t}{dt}$ – Francesco Falzoni Aug 29 '24 at 19:52

1 Answers1

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To be clear stochastic processes with Brownian motion are not differentiable in time because Brownian motion isn't. So $\frac{dY_t}{dt}$ is not possible to compute,instead it is only interpreted as an integral

$$Y_{t}=\int....$$

Now to compute $dY=d(XF)$, we use the product rule for Itô Product Rule for Ito Processes

$$dY=FdX+XdF+d[X,F]=\frac{F}{X}dt+FX\sigma dW_t+XF\sigma^2dt-\sigma XFdW_{t}-\sigma^2 XFdt=\frac{F}{X}dt,$$

where we used that $dF=\sigma^2 F dt-\sigma FdW_t$.

Thomas Kojar
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