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Let $B(t)$ be the standard Brownian motion, $\mu(t,x)$ and $\sigma(t,x)$ are continuous functions, and $$dr(t) = \mu(t,r(t))dt+\sigma(t,r(t))dB(t).$$ $(\mu,\sigma)$ obeys the linear growth condition $$|\mu(t,x)|+|\sigma(t,x)|<C(1+|x|),\ \forall t\in[0,T],\, x\in\mathbf R$$ for some positive constant $C$. By the Cauchy-Schwarz inequality and the Gronwall inequality, $$\mathbf E[r^2]<3\mathbf E[r(0)^2]e^{a(1+T)t}, \forall t\in[0,T]$$ for some positive constant $a$. We conclude there $$\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]\le t\int_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2). \tag{1}$$

How do we proceed to show $$\mathbf E\Big[\exp\Big(-\int_0^t r(s)ds\Big)\Big]=1+\exp(-r(0)t)+O(t^2)$$ as $t\to 0^+$?

I have tried Taylor expanding $e^{x}$ around $x=0$ in the following way. $$I:=\exp\Big(-\int_0^t r(s)ds\Big)=1-\int_0^t r(s)ds+\frac{e^{\theta(x)}}{2}\Big(\int_0^t r(s)ds\Big)^2 \tag{2}$$ for some $\theta(x)\in[0,x]$ and $x:=-\int_0^t r(s)ds$. Because $r(u)$ is continuous so is $\int_0^u r(s)ds$, $\exists\text{ stopping time }\tau(t,\omega),\ni-\int_0^{\tau(t,\omega)} r(s)ds=\theta(x)$ where $\omega$ is the sample under consideration. Take expectation of Equation (2), we have $$\mathrm E[I] = 1-\int_0^t\mathbf E[r(s)]ds+\frac12\mathbf E\Big[\exp\Big(-\int_0^{\tau(t,\omega)} r(s)ds\Big)\Big(\int_0^t r(s)ds\Big)^2\Big].$$ I intend to use Equation (1). In the case $r\ge 0$, $\exp\Big(-\int_0^{\tau(t,\omega)} r(s)ds\Big)\le 1$ and we can proceed easily. What do we do when $r$ can assume both signs?

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  • $\begingroup$ I wonder if under some conditions it would be possible to view $\mathbf E\Big[\exp\Big(-\int_0^{\tau(t,\omega)} r(s)ds\Big)\Big(\int_0^t r(s)ds\Big)^2\Big]$ as something like $\mathbf E^Q\Big[\Big(\int_0^t r(s)ds\Big)^2\Big]/\mathbf E\Big[\exp\Big(-\int_0^{\tau(t,\omega)} r(s)ds\Big)\Big]$ under a change of probability that would be governed by $\tau$, if under this measure you can derive bound for the second moment under $Q$ you would be done I think. Best regards $\endgroup$ – TheBridge May 30 '16 at 12:11
  • $\begingroup$ @TheBridge: Are you suggesting to use, say, the Cameron-Martin-Girsanov theorem? Any ideas how to proceed? $\endgroup$ – Hans May 30 '16 at 17:56
  • $\begingroup$ :Well sorry not really this was only a hunch looking at the form of this term, another approach could be investigating if Large Deviation Principle could lead somewhere but it would usually apply to bound a probability not an expectation so it might not work at all. Best regards $\endgroup$ – TheBridge May 30 '16 at 19:42
  • $\begingroup$ @TheBridge: In case you are interested, here is Joris Bierkens' answer: mathoverflow.net/a/241583/32660. The answer is simple and direct: Taylor expand the $\mathbf E[I]$ directly. $\endgroup$ – Hans Jun 6 '16 at 19:14

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