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I often come across similar statements:

  • Let $X_t$ be a stochastic process that satisfies the stochastic differential equation: $$dX_t = \mu_t dt + \sigma_t dB_t$$

Where

  • $B_t$ is a Weiner process
  • $\mu_t$ , $\sigma_t$ are deterministic functions of time.

My Question: What does it actually mean to "satisfy" a stochastic differential equation?

It seems to me that these are all just definitions. We are defining $X_t$ such that the derivative of $X_t$ with respect to time is $dX_t = \mu_t dt + \sigma_t dB_t$. Therefore, it is satisfied by definition. We have chosen $X_t$ such that its derivative satisfies $\mu_t dt + \sigma_t dB_t$.

Thus, is the word "satisfy" really necessary here?

Could the above statement simply have been written as: Let $X_t$ be a stochastic process such that: $dX_t = \mu_t dt + \sigma_t dB_t$

Or does "satisfying a stochastic differential equation" have some other criteria that need to be met?

Thanks!

J. W. Tanner
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stats_noob
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  • What would you say to a deterministic function $x(t)$ satisfying the ODE $x'(t) = f(x, t)$? Seems no different, no? SDEs are a bit more delicate. – Sean Roberson Nov 19 '23 at 02:25
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    It means the following equality of stochastic integrals for all $t \geq 0$: $$\int_{0}^{t}dX_s = \int_{0}^{t}\mu_s ,ds + \int_{0}^{t}\sigma_s ,dB_s.$$ Note the left hand side is $X_t - X_0$. – Mason Nov 19 '23 at 02:29
  • @SeanRoberson: thank you so much for your reply! can you please elaborate on this? – stats_noob Nov 19 '23 at 02:48
  • @Mason: thank you so much for your reply! Is this equality that you have posted referred to as Ito's Lemma? – stats_noob Nov 19 '23 at 02:49
  • @stats_noob No, I just wrote what it means to satisfy that SDE. – Mason Nov 19 '23 at 02:59
  • @Mason: thank you for your reply! does this equality have a specific name? is it only true for stochastic integrals? example if Xt and Bt were not stochastic processes, would this equality still be expected to hold? – stats_noob Nov 19 '23 at 03:03
  • Whether the equality holds or not depends on what the values of $X$, $\mu$ and $\sigma$ are. If you have $\mu$ and $\sigma$ already, then that equation defines the process $X$. It's a trivial SDE for $X_t$. – Mason Nov 19 '23 at 03:10
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    It is "Wiener", like a man from Vienna, Wien. A "Weiner" is someone feeling sad and shedding tears. People producing "Wein", wine, are called "Winzer", thus not the same word. – Lutz Lehmann Nov 20 '23 at 23:58

1 Answers1

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First to be clear, the process

$$X_t =x_{0}+ \int_{0}^{t}\mu_s ds + \int_{0}^{t}\sigma_s dB_s$$

is called an Itô process. So I agree "satisfying..." doesn't make much sense because $X_{t}$ is already given by the above integrals. It is akin to calling $f'(x)=0, f(0)=a$ an ODE, which has trivially the solution $f(x)=a$. It is not really thought of as differential equation where one tries to identify $X_{t}$ from a recursive-type SDE equation

$$dX_{t}=\mu(t,X_{t})dt+\sigma(t,X_{t})dB_{t}$$

eg. see Solution to General Linear SDE.

Thomas Kojar
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