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Why is the stochastic exponential of a Brownian motion positive?

It's apparent when looking at the explicit representation $Z_t = \exp(B_t - t/2)$ but my intuition fails when I look at it as the solution of the SDE $$ dZ_t = Z_tdB_t.$$

I know stochastic integration isn't Lebesgue Stieltjes integration, but we still have $$ \sum U_{t_i}(B_{t_{i+1}} - B_{t_i}) \to \int_0^t U_sdB_s $$ in some cases, so please help me with my intuition that finds it weird for the stochastic exponential staying positive while some paths along which we integrate should be able to "pull" it into the negative region? See e.g. the discrete stochastic exponential that can be negative. So what changes in the continuous-time setting?

Ramen
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    Well, in first place $Z_t=\exp\left(B_t-t/2\right)$ is an exponential super-martingale, but you can proove that it is also a martindale, i.e $\mathbb{E}[Z_t|\mathcal{F_s}]=Z_s$ where $0\leq s\leq t$ and $\mathcal{F_s}$ is the filtration generated by the Brownian motion. You can see here for the proof of martingale https://quant.stackexchange.com/questions/50787/steven-shreve-stochastic-calculus-and-finance Thus, as a martingale it has a constant mean and thtis is $$\mathbb{E}[Z_t|\mathcal{F_s}]=Z_0=1>0$$ – Hunger Learn Apr 28 '20 at 08:49
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    Adding to the comment by @HungerLearn the exponential is martingale because it satisfies the Novikov's condition https://en.wikipedia.org/wiki/Novikov%27s_condition – Chaos Apr 28 '20 at 09:49
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    @HungerLearn Thank you. Actually that's clear to me, I just fail to align it with my intuition that, for now, I just base on the discrete-time case and the finite-variation case. – Ramen Apr 28 '20 at 10:27
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    Well, in maths you can see intuitively some problem, by following different paths...The easiest path for me to understand what you are searching for, is the aforementioned thought...I hope it is helpful... – Hunger Learn Apr 28 '20 at 10:59
  • @HungerLearn Thank you, I think you are right; some things just need getting used to! – Ramen Apr 29 '20 at 14:06

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By Ito's formula, the process $Y_t:=\log(Z_t)$ satisfies $$ dY_t=dB_t-(t/2)\,dt $$ for $0\le t<\tau:=\inf\{s: Z_s=0\}$. That is, assuming $Z_0=x>0$, $$ \log(Z_t) = \log x +B_t-t/2,\qquad 0\le t<\tau. $$ If $P(\tau<\infty)$ were strictly positive, then on the event $\{\tau<\infty\}$ the left side of the last display would converge to $-\infty$ as $t\uparrow \tau$, while the right side would converge to $B_\tau-\tau/2$, which is finite. Consquently, $P(\tau<\infty)=0$.

John Dawkins
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  • Well, you are employing optimal stopping times, this is the part that I understand, but I can not understand why ``the left side of the last display'' would converge to $-\infty$ as you say. I am missing something... – Hunger Learn Apr 29 '20 at 14:39
  • Because of the definition of the stopping time, is like taking the limit of the logarithm when the argument tends to zero – Chaos Apr 29 '20 at 15:05
  • Ahh, do you mean that $log(Z_t)\rightarrow -\infty$ as $t\rightarrow \tau$? since $Z_s=0$ the logarithmic function tends to minus infinity if $t$ tends close to $\tau$... – Hunger Learn Apr 29 '20 at 16:18
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    Yes, I meant that $\lim_{t\uparrow\tau}\log(Z_t) = -\infty$. – John Dawkins Apr 29 '20 at 16:33