Follow-up questions on Ch4 Q7
An asymmetric random walk is defined by S0 = x and Sn = x +
for n > 1, where
(
+1, with probability p
Xn =
−1, with probability q
Pn
k=1
Xk
with p 6= q. Fix θ = q/p, we have shown that Mn := θSn is a martingale
x
−θ b
and P(Ha < Hb ) = θθa −θ
b , where Ha = inf(n > 0 : Sn = a) and
Hb = inf(n > 0 : Sn = b).
Some follow-up questions:
1. Does limn→ Mn exist? What can be said about this limit?
2. Let T = Ha ∧ Hb . Find E(ST ).
3. Show that Hn = Sn − n(p − q) is a martingale. Hence find E(T ).
Part 1
Note that Mn = (q/p)Sn is always non-negative. Hence by Martingale
Convergence Theorem (MCT), limn→ Mn exists.
Let µ be the common mean of Xn . By definition of Xn we have
µ = p − q. Since Xn ’s are i.i.d random variables with finite variance,
strong law of large numbers gives Sn /n → µ = p − q almost surely as
n → ∞.
Now consider two cases:
Case 1: Suppose p > q. We have Sn ≈ n(p − q) → +∞ almost surely as
n → ∞. Moreover, under p > q we have θ < 1. Hence Mn = θSn → 0.
Case 2: Suppose q > p. We have Sn ≈ n(p − q) → −∞ almost surely as
n → ∞. Moreover under q > p we have θ > 1. Hence Mn = θSn → 0.
In summary, Mn converges to 0 almost surely.
Remark: the argument via Sn ≈ n(p − q) for large n is quite loose. But it
can be formally proved that Sn → ±∞ when p ≷ q by tracing the formal
definitions from analysis. See also Ch3 Q6 where a similar idea is used.
Part 2
We know that ST is a binary random variable, which takes value of a or b
with probability P(Ha < Hb ) and P(Hb < Ha ) respectively. Then
E(ST ) = aP(Ha < Hb ) + bP(Hb < Ha )
=
a(θx − θb ) + b(θa − θx )
.
θa − θb
Part 3
Verify that Hn = Sn − n(p − q) satisfies the three required properties of a
martingale:
1. Hn is clearly adapted.
2. E|Hn | 6 E|Sn | + n|p − q| 6 x + n + n|p − q| < ∞.
3. Check that
E(Hn+1 |Fn ) = E(Sn+1 |Fn ) − (n + 1)(p − q)
= E(Sn + Xn+1 |Fn ) − (n + 1)(p − q)
= Sn + E(Xn+1 ) − (n + 1)(p − q)
= Sn + (p − q) − (n + 1)(p − q)
= Sn − n(p − q) = Hn .
We have a martingale H. Now it’s tempting to directly use optimal
stopping theorem (OST) on HT to conclude E(HT ) = H0 . But it can’t
be done in this way since Hn is not bounded for n 6 T so the condition
for OST fails.
Part 3 (cont‘)
Remember our trick on Ch4 Q6. If we don’t have a bounded martingale
which allows direct use of OST, the second best attempt is to consider a
capped (bounded) stopping time, and see if we can take limit to get the
results we want.
Let N > 0 be some fixed constant and set TN = min(T , N). Then TN is
a bounded stopping time. OST can now be used which gives
E(HTN ) = H0 = S0 = x. Then x = E(STN − TN (p − q)) and in turn
E(TN ) =
1
(E(STN ) − x).
p−q
Part 3 (cont‘)
Take limit N → ∞ on both side.
I
The left-hand-side becomes
lim E(TN ) = E
lim TN
N→∞
= E(T ).
N→∞
The swap of E and lim is valid due to monotone convergence
theorem (MON) as TN is positive and increasing to T as N → ∞.
I
The right-hand-side becomes
lim E(STN ) = E
N→∞
lim STN
N→∞
= E(ST ).
The swap of E and lim is valid due to bounded convergence
theorem (BDD) as STN is bounded between a and b for all N.
We thus obtain:
E(T ) =
=
1
1
E(ST − x) =
p−q
p−q
a(θx − θb ) + b(θa − θx )
−x
θa − θb
(a − x)((q/p)x − (q/p)b ) + (b − x)((q/p)a − (q/p)x )
.
(p − q)((q/p)a − (q/p)b )