Let $X$ be a one-dimensional Ito diffusion $$X_t=x+ \int_0^t b(X_s)ds + \int_0^t \sigma(X_s)dW_s,$$ where $b,\sigma$ satisfy the usual Lipschitz continuity and linear growth conditions. Define the running maximum $M_t:=\sup_{0\le t \le T} X_t$. Is it possible to compute explicitly or estimate $E[M_t]$ and $E[M_t^2]$? The naive estimate using the linear growth condition and Gronwall's inequality yields a bound involving $e^{Kt}$. But I wish to obtain better bounds such as sub-exponential. Many thanks for your help!
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$\begingroup$ I guess my question is not properly phrased. The exponential bound is in general unavoidable. Consider the simple example of geometric Brownian motion: $X_t=x\exp\{(\mu−\frac{1}{2}\sigma^2)t+\sigma W_t\}$ with $x,\mu>0$ and $\sigma \ge 0$. Then we have $E[M_t] \ge E[X_t]=x e^{μt}$. Then can we impose conditions on the coefficients b,σ so that a sub-exponential bound can be derived? $\endgroup$– epsilonMay 16, 2014 at 3:24
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$\begingroup$ My feeling is that you cannot get sub-exponential bounds, unless your $b$ and $\sigma$ are sub-linear. Think about case where both $b$ and $\sigma$ are linear. $\endgroup$– AnandOct 21, 2014 at 20:27
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$\begingroup$ @Anand, thanks for the comment. Indeed, as in my previous comment, even in the case of geometric Brownian motion example, there is no such sub-exponential bound. $\endgroup$– epsilonOct 30, 2014 at 19:45
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$\begingroup$ so under linear growth conditions, I guess the best you can get is the exponential bound. $\endgroup$– AnandNov 6, 2014 at 22:31
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