Stochastic Product Rule Example

Question

When given a function I wanted to know the how the application relates to: $Y_t = X_tdY_t + Y_tdX_t + (dX_t)(dY_t)$

(I am okay with using the 2 variable 2 equation function form).

For example consider: $Y_t = X_t e^{-rt}$

The answer supplied in the text is $dYt=d(Xt e^{-rt}) = e^{-rt}(dX_t) - re^{-rt}(X_tdt)$.

I'm assuming you break the initial $Yt$ down into '$Y_t$' and $X_t$ components. ie $X_t = X_t$ and $Yt=e^{-rt}$ to fit the equation.

This would match the first part $e^{-rt}(dX_t)$ as being $(Y_t)(dX_t)$. But I'm not sure were the $X_tdt$ came from in the second part.

(Im assuming since the process is deterministic $\sigma x =0$ to get rid of the $(dX_t)(dY_t)$

Any help would be really appreciated

Please type the $t$ consistently as $X(t)$ or $X_t$ (use _ for a subscript). When written as $Xt$ they all look like $X$ multiplied by $t$. — Nate Eldredge, Oct 04 '21 at 02:16
Part of the confusion may be that we're using $Y_t$ for two different processes. Can we rename one of them? — Nate Eldredge, Oct 04 '21 at 02:29
It seems like all you really need is that when $Y_t = e^{-rt}$, we have $dY_t = -re^{-rt},dt$, which is just freshman calculus. — Nate Eldredge, Oct 04 '21 at 02:30

score 1 · Answer 1 · answered Aug 19 '23 at 04:26

Actually the correct version is stochastic integration by parts

$$d(X_{t}Y_t) = X_tdY_t + Y_tdX_t + d[X, Y]_t.$$

So for $ X_t$ and $Y_{t}=e^{-rt}$, we have

$$d(X_{t}Y_t) = X_t(-r)Y_{t}dt + Y_tdX_t + d[X, Y]_t,$$

where for the dt term we just used chain rule. It remains to compute the cross variation. Since $Y_{t}$ is a monotone process and assuming $X$ is continuous, this will always be zero Quadratic Variation of Increasing Process?.

Stochastic Product Rule Example

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