MAT 902 STOCHASTIK II
HOMEWORK 4
Solution 1. Let X1 , . . . , Xn be independent and identically distributed
Pn random variables such
that P [Xi = 1] = 1 − P [Xi = −1] = 1 − q = p. Denote Sn = S0 + i=1 Xi and suppose that it
stops if it hits 0 or N (0 < S0 < N ).
(1) By definition of stopping times, we want to show {T ≤ n} ∈ Fn for every n ≥ 0. Let
A = {0} ∪ {N }. Then
C
{T ≤ n} = {T > n}C = {S0 ∈ AC , S1 ∈ AC , . . . , Sn ∈ AC } ∈ Fn
(2) We first check integrability of Yn .
E[|Yn |] = E[(q/p)Sn ] = (q/p)k E[(q/p)X1 ]n = (q/p)k
It remains to prove the martingale property.
E[Yn+1 |Fn ] = E[(q/p)Sn+1 |Fn ] = (q/p)Sn E[(q/p)Xn+1 |Fn ]
q
p
Sn
Xn+1
Sn
] = (q/p)
p +q
= (q/p) E[(q/p)
p
q
= Yn .
(3) We have if p > q, then Sn → +∞ a.s. and if p < q, then Sn → −∞ a.s. Hence, T < ∞
a.s.. Note Y{n∧T } is bounded, hence also uniform integrable, so we have E[YT ] = E[Y0 ].
On the other hand, E[YT ] = P [ST = 0] + (1 − P [ST = 0])(q/p)N and E[Y0 ] = (q/p)k .
Hence,
k 1− q N −k
(p)
P [ST = 0] = pq
N
1−( pq )
(k)
Solution 2. We first check integrability of Xn .
X
E[|Xn(k) |] = E[|
ξi1 · · · ξik |] ≤
1≤i1 <···<ik ≤n
X
ξi1 · · · ξik−1 |Fn ] = E[ξn+1 |Fn ]
1≤i1 <···<ik−1 ≤n
= E[ξn+1 ]
ξi1 · · · ξik |Fn ]
1≤i1 <···<ik ≤n
1≤i1 <···<ik ≤n+1
= E[ξn+1
E[|ξi1 |] · · · E[|ξik |] < ∞
1≤i1 <···<ik ≤n
It remains to prove the martingale property:
X
(k)
ξi1 · · · ξik −
E[Xn+1 − Xn(k) |Fn ] = E[
X
X
X
X
ξi1 · · · ξik−1
1≤i1 <···<ik−1 ≤n
ξi1 · · · ξik−1 = 0
1≤i1 <···<ik−1 ≤n
Solution 3 (Note that there was a change in what this problem is asking for.). We will use
the fact that
Eφ(min(X1 , X2 , X2 ))E(max(X1 , X2 , X3 )| min(X1 , X2 , X3 ))
Z
=
φ(min(x, y, z)) max(x, y, z)P (x)P (y)P (z) dxdydz
R3
Z
=6
φ(x)zP (x)P (y)P (z) dxdydz
x≤y≤z
Z
Z
= 6 φ(x)
zP (y)P (z) dydz dx,
R
x≤y≤z
2
MAT 902 STOCHASTIK II HOMEWORK 4
as R3 can be divided up into 6 regions: {x ≤ y ≤ z}, {x ≤ z ≤ y}, {y ≤ x ≤ z}, {y ≤ z ≤ x},
{z ≤ x ≤ y}, and {z ≤ y ≤ x}; and the first integral is symmetric with respect to reordering.
If g is defined by E(max(X1 , X2 , X3 )| min(X1 , X2 , X3 )) = g(min(X1 , X2 , X3 )) (and there must
exist such a function g), then we also have
Eφ(min(X1 , X2 , X2 ))E(max(X1 , X2 , X3 )| min(X1 , X2 , X3 ))
Z
φ(min(x, y, z))g(min(x, y, z))P (x)P (y)P (z) dxdydz
=
R3
Z
=6
φ(x)g(x)P (x)P (y)P (z) dxdydz
x≤y≤z
Z
Z
= 6 φ(x)g(x)
P (y)P (z) dydz dx.
R
Hence
x≤y≤z
R
x≤y≤z
g(x) = R
zP (y)P (z) dydz
x≤y≤z
P (y)P (z) dydz
,
and E(max(X1 , X2 , X3 )| min(X1 , X2 , X3 )) = g(min(X1 , X2 , X3 )) for such a function g.
Solution 4. We need to show that E(eSn −n/2 |Fn−1 ) = eSn−1 −(n−1)/2 . Because Xi are independent, this reduces to showing that EeXn −1/2 = 1. But
Z ∞
Z ∞
dx
2
x −x2 /2 dx
Xn
√ =
e1/2 e−(x−1) /2 √ = e1/2 ,
e e
Ee =
2π
2π
−∞
−∞
and the claim follows.