Solution

The stopped representation from part (a)(ii) and the Itô isometry give the exact second-moment identity

E [M_{t \land T}^{2}] = 1 + E \int_{0}^{t \land T} \frac{e ^{2 X_{s}}}{( 1 + X _{s}^{2} ) ^{2}} d s .

The cross term with the initial value is zero because the stopped stochastic integral has zero mean. This is the useful finite-time computation; evaluating the remaining occupation integral explicitly is unnecessary for proving integrability of

T

For

∣ x ∣ \leq L

\frac{e ^{2 x}}{( 1 + x ^{2} ) ^{2}} \geq c_{L}, c_{L} = \frac{e ^{- 2 L}}{( 1 + L ^{2} ) ^{2}} > 0.

Combining the second-moment identity with part (i) yields

1 + c_{L} E [t \land T] \leq E [M_{t \land T}^{2}] \leq e^{2 L} .

Let

t \to \infty

E T \leq e^{2 L} (1 + L^{2})^{2} (e^{2 L} - 1) < \infty.

In particular

T < \infty

almost surely, a conclusion obtained without assuming finite exit in the stopping argument.

One can also evaluate the limiting second moment exactly. At exit

M_{T}

is either

e^{- L}

e^{L}

, so

M_{T}^{2} = (e^{L} + e^{- L}) M_{T} - 1.

E M_{T} = 1

. Therefore

t \to \infty lim E [M_{t \land T}^{2}] = E M_{T}^{2} = e^{L} + e^{- L} - 1.

The same bound also holds at every finite time, by applying

(y - e^{- L}) (y - e^{L}) \leq 0

y = M_{t \land T}

and taking expected values. It improves the preceding bound to

E T \leq c_{L}^{- 1} (e^{L} + e^{- L} - 2)

if desired.

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