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I have a stochastic differential equation of the type: $$ dX(t) = \mu(t) X(t)dt + \sigma(t) X(t) dW(t) \tag{1} $$ However, my $\mu(t)$ is a complicated function of $t$ as well as $W(t)$, somewhat like:

$$ \mu(t) = t \cdot a(t) + b(t) + \sigma W(t) \tag{2} $$

I feel like $W(t)dt$ that I get by substituting $(2)$ into $(1)$ doesn't look "right". And I'm not sure how to get the $X(t)$ from that. Edit What I mean, is that my equation $(1)$ becomes: $$ dX(t) = (t \cdot a(t) + b(t))X(t)dt + \sigma W(t)X(t)dt + \sigma(t) X(t) dW(t) \tag{3} $$ My questions are:

  1. Assuming that I obtain a $X(t)$, does that single-handedly prove that $(1)$ is uniquely solvable? I am not sure what conditions will demonstrate that something is uniquely solvable.

  2. Is the $ W(t)dt $ term that I get in equation $(3)$ sensible?

coffee-raid
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    There is a lot of theory about uniqueness of solutions to SDEs. Check Chapter 5 of Brownian Motion and Stochastic Calculus by Karatzas. In general, it turns out that if your functions mu and sigma are Lipschitz, then this is typically enough to guarantee some sort of uniqueness – Alan Chung Feb 26 '24 at 15:44
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    For the second question, yes, $W(t)dt$ is perfectly sensible. The paths of $W$ are continuous, so $\int_0^t W(s)ds$ is even defined pathwise. – user6247850 Feb 26 '24 at 19:00

1 Answers1

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The equation

$$ dX(t) = \left[(t \cdot a(t) + b(t)+ \sigma W(t)\right] X(t)dt + \sigma(t) X(t) dW(t)=\mu(t)X(t)dt + \sigma(t) X(t) dW(t) $$ is a linear equation and it has a general solution give here Solution to General Linear SDE, and to be clear all you need from these various functions is continuity.

Thomas Kojar
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