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Let $B(t)$ be a Wiener process. We know that $X^{(1)}(t) = B(t)$ and $X^{(2)}(t) = -B(t)$ both weak solutions to the SDE \begin{equation} dX(t) = dB(t), \quad X(0) = 0. \end{equation}

Now let $B(t)$ and $W(t)$ be independent Wiener processes, and let $X(t)$ and $Y(t)$ be two Ito processes satisfying the SDEs \begin{equation} d X(t) = dB(t) + dW(t) \quad \text{and}\quad d Y(t) = dB(t) - dW(t). \end{equation} The solution of both SDEs is \begin{equation} X(t) = X(0) + B(t) + W(t) \quad \text{and} \quad Y(t) = Y(0) + B(t) - W(t) \end{equation} My understanding is that if $X(0) = Y(0)$, the distribution of these two processes is the same, and then that would mean that I could have written the SDE for $X(t)$ as $d X(t) = dB(t) - dW(t)$ and nothing would change.

Ultimately this leads me to believe that the SDEs \begin{equation} dX(t) = X(t) dt + X(t) dB(t), \quad \text{and} \quad dX(t) = X(t) dt -X(t) dB(t) \end{equation} describe the same process. Or are they the same in a weak sense, but not in a strong sense?

perojov
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  • What you write about the weak solutions $X^1=B$ and $X^2=-B$ of $dX=dB$ is correct. But I firmly believe that from the introduction of "the" Wiener processes $B$ and $W$ onwards you are getting onto slippery ice because it is unclear if your solutions are weak or strong. – Kurt G. Jun 29 '24 at 05:37
  • Thanks! The solutions should be weak for this to hold. – perojov Jul 15 '24 at 14:22

1 Answers1

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The SDEs \begin{equation} dX(t) = X(t) dt + X(t) dB(t), \quad \text{and} \quad dY(t) = Y(t) dt -Y(t) dB(t) \end{equation} have the following strong solutions Solution to General Linear SDE

$$X(t)=X_0 exp(\frac{t}{2}+B_{t})\text{ and }Y(t)=Y_0 exp(\frac{t}{2}-B_{t}).$$

So they are only equal only up to distribution, not almost surely.

Thomas Kojar
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