Usbrth

Let $T$ be the exit time of from the interval $[-b,a]$ of a standard Brownian Motion $X_t$, then how would we go about calculating the following two expectations:

• $E[T^2]$ (and)


• $E[int_0^T X_tds]$?

(Ideas which haven't been tied in:)

I want to use the optional stopping theorem but what martingale would I use?
Also for the second I know (by Ito's formula) I can write:
beginequation
int_0^T X_tds = 6int_0^T X_t^3 dX_t - 6int_0^T X_t^2 ds
endequation
but how can I put that to use?
I'm thinking since $X_t -X_0$ is a normal random variable, I could put that to use in calculating the second term on the RHS of the above but I'm not certain how...

First of all, recall that it follows from Wald's identities that

$$mathbbP(X_T=-b) = fracaa+b qquad mathbbP(X_T = a) = fracba+b qquad mathbbE(T)=ab, tag1$$

see e.g. this answer for a proof or Corollary 5.11 in Brownian Motion - An Introduction to Stochastic Processes (by René Schilling and Lothar Partzsch).

Part I: Calculate $mathbbE(int_0^T X_s , ds)$:

It follows from Itô's formula that $$int_0^t X_s , ds = - int_0^t X_s^2 , dX_s + fracX_t^33.$$ Since the first term on the right-hand side is a martingale, the optional stopping theorem (applied to the bounded stopping time $T wedge k$) yields

$$mathbbE left( int_0^T wedge k X_s , ds right) = frac13 mathbbE(X_T wedge k^3).$$

As $|X_s wedge T| leq maxa,b$ for all $s geq 0$, it now follows from the dominated convergence theorem that

$$mathbbE left( int_0^T X_s , ds right) = frac13 mathbbE(X_T^3).$$

Finally, using $$X_T = a 1_X_T=a -b 1_X_T=-b,$$ we obtain $$mathbbE left( int_0^T X_s , ds right) = frac(-b)^33 mathbbP(X_T=-b) + fraca^33 mathbbP(X_T=a).$$

Plugging in the results from $(1)$, we are done.

Part II: Calculate $mathbbE(T^2)$:

1. Show that $M_t := X_t^3 -3 t X_t$ is a martingale and that (with a similar argumentation as in part I) $$beginalign* mathbbE(X_T^3) &= 3a mathbbE(T 1_X_T = a) -3b mathbbE(T 1_X_T=-b) \ &= 3a mathbbE(T) - 3 (a+b) mathbbE(T 1_X_T = -b). tag2 endalign*$$
   


2. Use $(1)$ and Step 1 to calculate $mathbbE(T 1_X_T=-b)$.


3. Show that $N_t := X_t^4-6t X_t^2 + 3t^2$ is a martingale. Using the optional stopping theorem show that $$mathbbE(N_T)=0. tag3$$
   


4. Using the identity $$X_T = a 1_X_T=a -b 1_X_T=-b,$$ step 2 and $(3)$ calculate $mathbbE(T^2)$.