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Let $B(t)$ be the standard Brownian motion, $\mu(t,x)$ and $\sigma(t,x)$ are continuous functions, and $$dr(t) = \mu(t,r(t))dt+\sigma(t,r(t))dB(t).$$ Is there a pair $(\mu,\sigma)$ such that $$\infty>\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]> ct, \quad \forall t\in(0,T]$$ for some positive constants $c$ and $T$?


1) Consider $\mu(t,x)=\mu(t)$ and $\sigma(t,x)=\sigma(t)$.

Assume these functions are bounded on $t\in[0,T_0]$ for some positive $T_0$, the inequality does not hold. The derivation is as follows. $$I :=\int_0^t r(s)ds=r(0)t+\int_0^t\mu(s)(t-s)ds+\int_0^t \sigma(s)(t-s)dB(s),$$ and \begin{align} \mathbf E[I^2] &= \Big(r(0)t+\int_0^t\mu(s)(t-s)ds\Big)^2+\int_0^t \sigma(s)^2(t-s)^2ds \\ &<\Big(r(0)t+\frac{C_1}2t^2\Big)^2+\frac{C_2^2}{3}t^3 \\ &=o(t), \quad t\to 0^+ \end{align} for some positive $C_1, C_2$ as we take the bounds of $\mu(t)$ and $\sigma(t)$. That violates the required inequality.

So we have to consider $\mu(t)$ and $\sigma(t)$ unbounded at $t=0^+$. $\sigma(t)=t^{-a}$, for some $1>a>0$, does not work though.

2) What about general $\mu(t,x)$ and $\sigma(t,x)$?

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  • $\begingroup$ Well, sure, just take $\mu := 0$ and $\sigma := 1$. $\endgroup$ – saz May 18 '16 at 15:12
  • $\begingroup$ @saz: I have tried this case first and I got a high order $t$, so I rejected it. The expectation on the left hand side $\displaystyle r(0)^2\frac{t^4}4+\frac{t^3}3=o(t)$ as $t\to 0^+$, which contradicts the required inequality. Do you agree? $\endgroup$ – Hans May 18 '16 at 17:59
  • $\begingroup$ Sorry, my mistake. Yes, I agree with you that the expression on the left-hand side converges faster to $0$ than $t$. If you would like to have feedback on the approach you have used, then please add your thoughts to the question. $\endgroup$ – saz May 18 '16 at 18:35
  • $\begingroup$ @saz: I just added a solved bounded deterministic case. $\endgroup$ – Hans May 18 '16 at 19:35
  • $\begingroup$ @saz: I have answered my question under the linear growth condition in $x$. What now happens with superlinear growth? $\endgroup$ – Hans May 28 '16 at 1:39
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Suppose $(\mu,\sigma)$ obeys the linear growth condition $$\left|\mu(t,x)\right|+\left|\sigma(t,x)\right|<C(1+\left|x\right|),\ \forall t\in[0,T],\, x\in\mathbf R$$ 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 $a$. $$\mathbf E \Big[\Big(\int_0^t r(s)ds\Big)^2\Big]\le t\int_0^t \mathbf E[r(s)^2]ds\le 3\mathbf E[r(0)^2]e^{a(1+T)T}t^2 = O(t^2)$$ by Cauchy-Schwartz inequality and commutation of expectation and integration. Therefore the desired inequality does not hold.

Now the question is what happens when $(\mu,\sigma)$ has superlinear growth.

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