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.


Write $f(t) := \mathbb E \left[ \exp\left( - \int_0^t r_s \ d s \right) \right]$. Define stochastic processes $y_t = \exp \left( -\int_0^t r_s \ d s \right)$ and $z_t = r_t y_t$. Then $y_t = 1 - \int_0^t r_s y_s \ d s$ and $z_t = r_0 + \int_0^t (\mu(r_u) - r_u^2) y_u \ d t + \int_0^t \sigma(r_u) y_u \ d B_u$.

By taking expectations $$f(t) = \mathbb E \left[ y_t \right] = \mathbb E \left[ 1 - \int_0^t z_s \ d s \right] = 1 - \mathbb E \left[\int_0^t \left\{ r_0 + \int_0^s (\mu (r_u) - r_u^2) y_u \ d u \right\} \ d s \right]$$ Interchanging differentiation and integration (expectation) gives $$f'(t) = - r_0 - \int_0^t \mathbb E \left[ (\mu(r_u) - r_u^2) y_u \right] \ d u$$ and $f''(t) = \mathbb E \left[ (r_t^2 - \mu(r_t)) y_t \right]$.

Since the second derivative of $f$ exists, the second derivative of $g(t) = \log(f(t))$ exists, and we may use a standard Taylor series argument to show that

$$\log(f(t)) = \log(f(0)) + \frac{1}{f(0)} f'(0) t + O(t^2) = -r_0 t + O(t^2),$$ or equivalently $$f(t) = \exp(-r_0 t) + O(t^2).$$

It remains to fill in details regarding the interchange of integration and differentiation and the application of Fubini here and there, which should be fine using a priori estimates on the stochastic quantities using your strong linear growth assumptions.

  • $\begingroup$ Yours is a clever way to circumvent the difficulty of taking second derivatives of the function of a stochastic variable. Or rather, it is a frontal attack for the second derivative. The expectation works to smooth out the non-differentiable part. Thank you! I have corrected my typo. You left out a $y_u$ in the $dBu$ term of the integral expansion of $z_t$. I have taken the liberty to make the derivation integral expansion of $y_t$ more explicit at least for me. I hope you would like it. $\endgroup$ – Hans Jun 3 '16 at 22:59
  • 1
    $\begingroup$ @Hans, I do not think it is very polite to make such heavy edits to a user's answer, when he has not approved this. Many of your changes seemed subjective and unnecessary. If you think there are errors in the answer then it would be more polite to leave a comment notifying Joris. $\endgroup$ – Yemon Choi Jun 4 '16 at 4:10
  • $\begingroup$ @YemonChoi: You are right. I thought my edit adhered closely along his line of reasoning but was clearer at least to me. I have said as much in my first comment. But I will ask for his opinion and permission explicitly first. Thank you for reminding me. $\endgroup$ – Hans Jun 4 '16 at 8:43
  • $\begingroup$ Joris Bierkens: As I have expressed in my first comment, I like your answer very much. I took the liberty and edited your answer --- correcting a typo --- in a way I thought would be clearer at least to me. But as Yemon Choi rightly reminded me that I had overstepped my bound. He has reverted it back to your original version. Would you be so kind as to take a look at my edit and see if you like it? If you do, you may elect to use my last edit. If not, I will just write another answer which will clearly state that it is merely my rephrasing of your answer. $\endgroup$ – Hans Jun 4 '16 at 8:58
  • 2
    $\begingroup$ OK, Joris. You probably missed it, but as I have mentioned in my earlier comment, you left out a $y_t$ in the $dB_t$ term of $z_t$. The $b$ there should be $\mu$. We also need $f$ and the second derivatives of $f$ and $g$ to be bound (continuous). I was awaiting your response to my comment before I accept your answer. I have now. Thank you. $\endgroup$ – Hans Jun 6 '16 at 18:44

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.