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Consider the one-dimensional SDE $$ dX_t = f(X_t)dt + \sigma(X_t)dW_t,\quad X_0 = 1, $$ where $W_t$ is a Brownian motion under the measure $\mathbb{P}$, in its natural filtration. Suppose that $f(x)-\sigma^2(x)/2x=0$ and $\sigma^2(x)\le x^2$. Show that $X_t>0$ almost surely.

My initial attempt was to consider the log-process $\log(X_t)$. Using Ito's lemma one finds $$ d\log(X_t) = \frac{1}{X_t}\left(f(X_t) - \frac{\sigma^2(X_t)}{2X_t}\right)dt + \frac{\sigma(X_t)}{X_t}dW_t = \frac{\sigma(X_t)}{X_t}dW_t. $$ Integrating now leads to $$ \log(X_t)=\int_0^t\frac{\sigma(X_s)}{X_s}dW_s. $$ By the Ito isometry, the variance of the above process is bounded by $t$: $$ \text{Var}\big[\log(X_t)\big] = \mathbb{E}\left[\int_0^t\frac{\sigma^2(X_s)}{X_s^2}ds\right] \le t. $$ I feel like this is heading in the right direction, but I'm struggling to see what the next step is. Any suggestions would be very welcome.

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1 Answer 1

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Define stopping times by

$$T_k := \inf\{t \geq 0; X_t \leq k^{-1}\}, \qquad k \in \mathbb{N}$$

and

$$T := \inf\{t \geq 0; X_t \leq 0\}.$$

Following the reasoning in your question we find that

$$\mathbb{E} \left( \left| \log(X_{t \wedge T_k}) \right|^2 \right) \leq t.$$

By the continuity of the sample paths of $(X_t)_{t \geq 0}$, we have $X_{T_k}(\omega) = k^{-1}$ for $\omega \in \{T_k<\infty\}$, and therefore we get that

$$\mathbb{E} \bigg( 1_{\{T_k \leq t\}} \underbrace{|\log(X_{T_k})|^2}_{=|\log(k^{-1})|^2} \bigg) \leq t.$$

Since $\{T \leq t\} \subseteq \{T_k \leq t\}$ this shows, in particular, that

$$\mathbb{E}\left(1_{\{T \leq t\}} |\log(k^{-1})|^2 \right) \leq t,$$

i.e.

$$\mathbb{P}(T \leq t) \leq \frac{t}{|\log(k^{-1})|^2}.$$

As $|\log(k^{-1})|^2 \to \infty$ as $k \to \infty$ we conclude that $\mathbb{P}(T \leq t)=0$. As $t>0$ is arbritrary, this proves the assertion.

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    $\begingroup$ How can I be half as good as you at this $\endgroup$
    – user515599
    Jun 7, 2018 at 6:39
  • $\begingroup$ Brilliant, thanks a lot saz! Could I just confirm that here $T:=\inf\{t\ge0 : X_t = 0\}$? $\endgroup$
    – Alex W
    Jun 7, 2018 at 8:28
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    $\begingroup$ @AlexW Yes, exactly; it seems that I got lost in my thoughts (somehow I thought that you had already introduced $T$ in your question). $\endgroup$
    – saz
    Jun 7, 2018 at 8:29

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