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Let $B$ be a standard Brownian motion. I want to prove that the SDE \begin{equation} X_t = \int_0^t 1_{\{ X_s \geq 0 \}} \, dB_s \end{equation} has no solution on any set-up.

A hint I was given was to consider $f(x) = -x^3 1_{\{ x < 0 \}}$ and then $f(X_t)$. If we apply Ito's formula to $f$ we get \begin{equation} -X_t^3 \, 1_{\{ X_t <0 \}}= f(X_t) = -3 \left( \int_0^t X_s^2 1_{\{ X_s < 0 \}} dX_s + \int_0^t X_s 1_{\{ X_s < 0\}} ds \right) \end{equation} and so by plugging-in $X_s$ from the SDE \begin{equation} X_t^3 \, 1_{\{ X_t <0 \}} = 3 \int_0^t X_s 1_{\{ X_s < 0\}} ds, \end{equation} since the integral over $X_s$ will have $X_s^2 1_{\{X_s < 0 \}} 1_{\{X_s \geq 0 \}}$ as the integrand and $B_s$ as the integrator. Having this I don't really know how to continue. I also don't know what ways there are to show that a SDE does not have a solution.

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1 Answer 1

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You didn't apply Itô's formula correctly; it should read

$$-X_t^3 1_{\{X_t<0\}} = -3 \left( \int_0^t X_s^2 1_{\{X_s<0\}} \, dX_s + \int_0^t X_s 1_{\{X_s<0\}} \, d\color{red}{\langle X \rangle_s} \right) \tag{1}$$

where $\langle X \rangle$ denotes the quadratic variation. From $dX_s = 1_{\{X_s>0\}} \, dB_s$, it follows that $$d\langle X \rangle_s = (dX_s)^2= 1_{\{X_s \geq 0\}} \, ds.$$ Plugging this into $(1)$ and using that $1_{\{X_s \geq 0\}} 1_{\{X_s<0\}}=0$, we get

$$-X_t^3 1_{\{X_t<0\}}=0.$$

Since $-X_t^3$ is strictly positive on $\{X_t<0\}$, this means that $\mathbb{P}(X_t<0)=0$. Hence, $$X_t = \int_0^t 1_{\{X_s \geq 0\}} \, dB_s = \int_0^t dB_s = B_t,$$

which contradicts $X_t \geq 0$ (...since Brownian motion takes negative values).

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  • $\begingroup$ Thank you very much. Bad mistake of me there! Is the implication: $P(X_t < 0) = 0$, then $\int_0^t 1_{\{ X_s \geq 0\}} dB_s = \int_0^t dB_s$ trivial? Since I only have $X_t \geq 0$ a.s. for all $t$. Might be a stupid question but I am always suspicious when it comes to stochastic integrals. $\endgroup$
    – Noah
    Commented Dec 4, 2019 at 8:57
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    $\begingroup$ @Noah By Itô's formula and Tonelli's theorem, \begin{align*} \mathbb{E} \left( \left| \int_0^t 1_{\{X_s \geq 0\}} \, dB_s - \int_0^t dB_s \right|^2 \right) &= \mathbb{E} \left( \int_0^t |1_{\{X_s \geq 0\}}-1|^2 \, ds \right) \\ &= \int_0^t \mathbb{P}(X_s<0) \, ds=0,\end{align*} and so $\int_0^t 1_{\{X_s \geq 0\}} \, dB_s = \int_0^t \, dB_s$ (almost surely). Alternatively you can argue that $(X_t)_{t \geq 0}$ has (a modification with) continuous sample paths, and so $X_q \geq 0$ a.s. for all rationals implies $X_t \geq 0$ a.s. for all $t$. $\endgroup$
    – saz
    Commented Dec 4, 2019 at 9:01
  • $\begingroup$ Thank you very much! $\endgroup$
    – Noah
    Commented Dec 4, 2019 at 9:14

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