**Question:**


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 
$$\left|\mu(t,x)\right|+\left|\sigma(t,x)\right|<C(1+|x|),\ \forall t\in[0,T],\, x\in\mathbf R$$
for some positive constant $C$. Is it true that
$$\mathbf E\Big[\exp\Big(-\int\_0^t r(s)ds\Big)\Big]=\exp\big(-r(0)t+O(t^2)\big)$$
as $t\to 0^+$?




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**What I have obtained so far:**


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](https://math.stackexchange.com/a/1802854/64809)
$$\mathbf E \Big[\Big(\int\_0^t r(s)ds\Big)^2\Big]\le t\int\_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2). \tag{1}$$


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? 


Perhaps bounding the quadratic moment is not enough and we need more accurate estimate of the probability distribution. I am considering using the heat kernel expansion to estimate the probability distribution of $r$. But I suspect there is a more elegant solution for this short time asymptotics. 



A Stochastic Taylor Expansion/Asymptotics
Define stochastic processes $\displaystyle y\_t := \exp\Big(-\int\_0^t r\_s \,ds \Big)$. The subscript denotes the time variable dependency. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.


Apply Ito's Lemma twice
\begin{align}
-(y\_t-y\_0) &=\int\_0^tr\_sy\_sds \\
&=\int\_0^tds\Big(r\_0y\_0+\int\_0^s(y\_udr\_u+r\_udy\_u)\Big) \\
&=r\_0t+\int\_0^tdu\ y\_u(\mu\_u-r\_u^2)(t-u)+\int\_0^tdB\_u\,y\_u\sigma\_u(t-u) \tag1
\end{align}
since $dr\_tdy\_t=-r\_ty\_tdr\_tdt=0$. The last integral comes from exchanging the order of two integrations.


Apply Ito's lemma again
\begin{align}
d\big(y(\mu-r^2)\big) &= y\left[\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}+\sigma^2\Big(\frac12\frac{\partial^2\mu}{\partial r^2}-1\Big)\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB\right] \\
&=:\alpha\, dt+\beta\, dB\_t.
\end{align}
Then interchange the order of integration,
\begin{align}
&2\int\_0^tdu\ y\_u(\mu\_u-r\_u^2)(t-u) \\
= & y\_0(\mu\_0-r\_0^2)t^2+\int\_0^tds\ \alpha\_s(t-s)^2 + \int\_0^t dB\,\beta\_s(t-s)^2.
\end{align}


Substitute the above equation into Eq. (1), then take the expectation of Eq. (1). As $\mathbb E[\alpha]$ is bounded,
$$\mathbb E[y\_t] = 1-r\_0t-\frac12(\mu\_0-r\_0^2)t^2+O(t^3)=\exp\Big(-r\_0t-\frac12\mu\_0t^2+O(t^3)\Big)$$
as $t\searrow 0$.


It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.


