# If $dX_{t} = X_{t}\,dt + \,dB_{t}$, why does $e^{- t}dX_{t} = e^{-t} X_{t} \,dt + e^{-t} \,dB_{t}$?

I'm taking a course in stochastic differential equations, and in order to solve $dX_{t} = X_{t}\,dt + \,dB_{t}$, the book gives a hint: to multiply both sides of this equation by $e^{-t}$. (But, as explained below, "multiplication" doesn't really mean usual multiplication because this is an integral equation.)

But the original equation $dX_{t} = X_{t}\,dt + \,dB_{t}$ really means:

$$X_{t} - X_{0} = \int \limits_{0}^{t} X_{s} \,ds + \int \limits_{0}^{t} \,dB_{s}.$$

If $X_{t}$ satisfies the above equation, then according to the hint the following equation also holds.

$$e^{-t}X_{t} - e^{-t}X_{0} = \int \limits_{0}^{t} e^{-s}X_{s}\,ds + \int \limits_{0}^{t} e^{-s} \,dB_{s}$$

But why?

• you are working with a differential equation and multiplying by an "integrating factor" so you can reduce to a form that is easier to integrate. Also, your title is wrong. – Chinny84 Feb 9 '16 at 16:37
• So, so...what It is susuggesting is to use integration by parts rule to the process $e^{-t} X_t$ i.e. compute $d(e^{-t} X_t)$ – Kolmo Feb 9 '16 at 16:37
• @user46944 : use the hint of Kolmo with Itô's lemma. Best regards – TheBridge Feb 9 '16 at 16:42
• The author probably rather suggests to compute $$d(e^{-t}X_{t})= e^{-t}dX_{t}-e^{-t}X_tdt=e^{-t}(X_tdt+dB_t)-e^{-t}X_tdt=e^{-t}dB_t$$ hence $$e^{-t}X_t=X_0+\int_0^te^{-s}dB_s.$$ – Did Feb 9 '16 at 20:47
• Have you understood what Did was saying? I am trying but cant. – user561840 Jul 25 '18 at 12:51

What the book it hinting at is the knowledge that if $$X_t=X_0+\int_0^t X_s\,ds +B_t,$$ then $X$ is a continuous semimartingale and $$\int_0^t e^{-s}\,dX_s=\int_0^t e^{-s}X_s\,ds+\int_0^t e^{-s}\,dB_s.$$ The differential form of an SDE is more than just a way to save ink (or electrons); it can (as here) serve as a guide to deducing one integral equation from another. Finally, the "product rule" for the stochastic integral gives us $$\int_0^t e^{-s}\,dX_s-\int_0^t e^{-s}X_s\,ds = e^{-t}X_t-X_0,$$ from which we deduce that $e^{-t}X_t=X_0+\int_0^t e^{-s}\,dB_s$.
Ito lemma justifies "multiplying" from the left with $$e^{-t}$$ as it gives us sufficient conditions to obtain a new Ito process. In ordinary calculus it is much easier to see what operations are allowed and where one ends up using them, here however it is less clear.
But Ito lemma should be sufficient to justify what Did does in the above comments. However calling it multiply is a poor choice of formulation(especially when one declears that $$dX$$ really is an integral) since we really apply a function to the RHS in view of Ito lemma even if this function essentially is multiplication.