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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?

layman
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  • 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 09 '16 at 16:37
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    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 09 '16 at 16:37
  • @user46944 : use the hint of Kolmo with Itô's lemma. Best regards – TheBridge Feb 09 '16 at 16:42
  • yes, but multiplying by the "integrating factor" allows for the form that @Kolmo mentions. I was trivializing the problem to the context why it is used. Though, looking at your question its more about the validity of the approach. – Chinny84 Feb 09 '16 at 16:42
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    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 09 '16 at 20:47
  • It is quite valid to multiply the SDE in its differential form by any factor one wants. Recall that $dU_t=dV_t$ means that $U_t-U_0=V_t-V_0$, then $A_tdU_t=A_tdV_t$, that is, $\int_0^tA_sdU_s=\int_0^tA_sdV_s$. On the other hand, what you did is equivalent to state that $A_tX_t=A_0X_0+\int_0^tA_sdX_s$, which might not hold, already for integrals with no stochastic part. – Did Feb 09 '16 at 21:51
  • See previous comment (did you actually read it?). – Did Feb 09 '16 at 22:56
  • Of course $X_tdt$ is some $dV_t$, for $V_t=\int_0^tX_sds$. Likewise, $X_tdt+dB_t$ is $dV_t$ for some process $(V_t)$, for $V_t=\int_0^tX_sds+\int_0^tdB_s$. These are the definition of the differential notations in SDE. – Did Feb 10 '16 at 06:25
  • Nothing breaks down, thank you. Once again (but for the last time), of course $X_tdt+dB_t=dV_t$ for $V_t=\int_0^tX_sds+\int_0^tdB_s$ hence $A_tdV_t=A_tX_tdt+A_tdB_t$, that is, $\int_0^tA_sdV_s=\int_0^tA_sX_sds+\int_0^tA_sdB_s$. – Did Feb 10 '16 at 11:41
  • Nobody "moves things under integrals",as you keep insisting on doing. Everything I wrote is justified. Enough is enough. – Did Feb 10 '16 at 14:19
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    Have you understood what Did was saying? I am trying but cant. –  Jul 25 '18 at 12:51
  • I think I might know whats going on..in your question you are multiplying the equation in integral form and at the same time you pull the $e^{t}$ inside. This I think is not permitted. Neither is $e^{t}dX_{t}=e^{t}X_{t}-e^{0}dX_{0}$ in general which I think he suggests that you are saying. On the other hand one can multiply the differential form by $e^{t}$ to obtain via Ito $d(e^{t}X_{t})=dB$, which can be integrated according to the above rule $de^{t}X_{t}=e^{t}X_{t}-X_{0}$. In essence I think the two of you talk past eachother, but I am not sure. What do you think of my explanation? –  Jul 29 '18 at 12:04
  • I have no experience of quadratic variation..ill look into it when I have time and maybe ill come up with something. –  Jul 30 '18 at 04:32

2 Answers2

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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 $e^{-t}X_t=X_0+\int_0^t e^{-s}\,dB_s$.

John Dawkins
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  • While this justification is valid, I think Ito lemma also gives sufficient conditions for what Did does in the comments(feb 9 16' 20:47), he goes on to justify it using integration with respect of a semimartingale tho. But as my answer says below I think Ito lemma is enough to justify it. – user123124 Apr 29 '19 at 05:48
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I am writing here a clear solution since there seems to be a confusion (I was confused as well).

Following Did's calculations in the first comments (by Itô's product rule and the given SDE):

$$d(e^{-t}X_t) = e^{-t}dX_t-e^{-t}X_tdt = e^{-t}dB_t$$

and therefore $e^{-t}dX_t = e^{-t}X_tdt + e^{-t}dB_t$.

user_tam
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  • All of these years later, I can clarify my question and the answer I was ultimately looking for. Did was rude and didn't try to understand. I questioned how to prove that if $dX_t = a_t ,dt + b_t ,dB_t$, then why is it true that we can "multiply" both sides of this by $e^{-t}$. The answer is really simple: we actually $\textbf{define}$ $\int e^{-t} ,dX_t$ as $\int e^{-t} a_t ,dt + \int e^{-t} b_t ,dB_t$. It's not something we have to prove. It's something we $\textbf{define}$. – layman Oct 19 '23 at 14:18