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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 $t\to 0^+$?

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 $$\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]\le t\int_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2). \tag{1}$$

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.

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 $t\to 0^+$?

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 $$\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]\le t\int_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2). \tag{1}$$

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.

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

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 $$\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]\le t\int_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2). \tag{1}$$

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.

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 $t\to 0^+$?

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 $$\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]\le t\int_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2). \tag{1}$$

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.

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 $$|\mu(t,x)|+|\sigma(t,x)|<C(1+|x|),\ \forall t\in[0,T],\, x\in\mathbf R$$ for some positive constant $C$. Is it true that $$\mathbf E\Big[\exp\Big(-\int_0^t r(s)ds\Big)\Big]=1+\exp(-r(0)t)+O(t^2)$$ as $t\to 0^+$?

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 $$\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]\le t\int_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2). \tag{1}$$

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.

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

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 $$\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]\le t\int_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2). \tag{1}$$

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.