63.8. LEVY’S THEOREM 2173
By the optional stopping theorem,{
X (τε ∧ t j)}
j is a martingale as is also{X (τε ∧ t j)− τε ∧ t j
}j .
Thus for A ∈Fτε∧t j−1 ,∫A
E(
y2j |Fτε∧t j−1
)dP =
∫A
E((
X (τε ∧ t j)−X(τε ∧ t j−1
))2 |Fτε∧t j−1
)dP
=∫
AE(
X (τε ∧ t j)2 |Fτε∧t j−1
)+X
(τε ∧ t j−1
)2
−2X(τε ∧ t j−1
)E(
X (τε ∧ t j) |Fτε∧t j−1
)dP
=∫
AE(
X (τε ∧ t j)2− τε ∧ t j|Fτε∧t j−1
)dP+
∫A
E(
τε ∧ t j|Fτε∧t j−1
)dP
+∫
AX(τε ∧ t j−1
)2 dP−2∫
AX(τε ∧ t j−1
)2 dP
=∫
AX(τε ∧ t j−1
)2 dP−∫
Aτε ∧ t j−1dP+
∫A
E(
τε ∧ t j|Fτε∧t j−1
)dP
+∫
AX(τε ∧ t j−1
)2 dP−2∫
AX(τε ∧ t j−1
)2 dP
=∫
AE(
τε ∧ t j|Fτε∧t j−1
)dP−
∫A
τε ∧ t j−1dP
=∫
A
(τε ∧ t j− τε ∧ t j−1
)dP≤
∫A
t j− t j−1dP.
Thus, since A is arbitrary,
σ2j ≡
∫A
E(
y2j |Fτε∧t j−1
)dP =
E((
X (τε ∧ t j)−X(τε ∧ t j−1
))2 |Fτε∧t j−1
)≤ t j− t j−1 = δ n(ε) (63.8.44)
Also,
E(
y j|Fτε∧t j−1
)= E
(X (τε ∧ t j)−X
(τε ∧ t j−1
)|Fτε∧t j−1
)= 0. (63.8.45)
Now it is time to find E(
eiλX(τε∧t j))
.
E(
eiλX(τε∧t j))= E
(eiλ(X(τε∧t j−1)+y j)
)= E
(eiλX(τε∧t j−1)E
(eiλy j |Fτε∧t j−1
)). (63.8.46)