LINKÖPING UNIVERSITY
Department of Mathematics
Mathematical Statistics
John Karlsson
TAMS29
Stochastic Processes with
Applications in Finance
2. Modes of convergence, Jensen’s inequality
2.1a
ε<1
P (|Xn − 0| > ε) = P (Xn = 1) + P (Xn = −1) =
1
1
1
+
=
−→ 0.
2n 2n
n n→∞
p
Thus by the definition Xn → X.
b
1
1
1
1
1
r
·|(−1)−0| + 1 −
·|0−0|r + |1−0|r = 2· ·1r =
−→ 0.
E[|Xn −0| ] =
2n
n
2n
2n
n n→∞
r
r
Thus by the definition Xn → X.
2.2a
We note
sup |Xm − X|
P
> ε ≥ P ((|Xn − X|) > ε) .
m≥n
a.s.
The left hand side tends to 0 and n → ∞ since Xn → X. It follows that the right
p
hand side tends to 0 as well. Thus by the definition we have Xn → X.
b
p
We show that Xn → 1 by noting
P (|Xn − 1| > ε) =
.
.
1
n > 1, ε < 1 = P (Xn = n) =
−→ 0.
n n→∞
a.s.
We show that Xn 9 1. Assuming independence of X1 , X2 . . . . we obtain
∞ Y
1
< ε = P (Xn = 1, Xn+1 = 1, Xn+2 = 1, . . .) =
1−
m
m≥n
m=n
( ∞
)
( ∞
)
X
X 1
1
1
1
(∗)
Taylor
= exp
ln 1 −
= exp
− −
+O
= 0.
2
m
m 2m
m3
m=n
m=n
P
sup |Xm − 1|
It follows that P
supm≥n |Xm − 1| > ε = 1 for all n which gives
lim P
n→∞
sup |Xm − 1| > ε = 1 6= 0,
m≥n
1/3
a.s.
and thus Xn 9 1 by the definition of almost sure convergence.
NOTE: Another approach at (∗) if one does not want to use Taylor expansion is
N N
∞ Y
Y
Y
1
m−1
1
= lim
= lim
1−
1−
N →∞
N →∞
m
m
m
m=n
m=n
m=n
N
Y
n−1
n
N −1
n−1
·
· ... ·
= lim
= 0.
N →∞
N →∞ N
n
n+1
N
m=n
= lim
2.3a
E[|Xn − X|r ]
−→ 0,
n→∞
εr
r
where we used E[|Xn − X| ] by the assumption. Thus by the definition of converp
gence in probability Xn → X.
P (|Xn − X| > ε) = P (|Xn − X|r > εr )
Markov ineq.
≤
b
p
We show Xn → 1
ε<1
P (|Xn − 1| > ε) = P (Xn = n) =
1
−→ 0.
nα n→∞
We now study convergence to 1 in r-mean.
E[|Xn − 1|r ] = P (Xn = 1) · |1 − 1|r + P (Xn = n) · |n − 1|r
1
1
= 1 − α · 0 + α · (n − 1)r
n
n


∞, r > α
(n − 1)r
=
−→
1, r = α
n→∞ 
nα

0, r < α.
r
r
Thus it follows that Xn → 1 if r < α and Xn 9 1 if r ≥ α.
2.4
We have
P (Xn ≤ a) = P (Xn ≤ a, X ≤ a + ε) + P (Xn ≤ a, X > a + ε)
≤ P (X ≤ a + ε) + P (Xn − X ≤ a − X, a − X < −ε)
≤ P (X ≤ a + ε) + P (Xn − X < −ε)
(1)
≤ P (X ≤ a + ε) + P (|Xn − X| > ε).
In the same way but switching Xn with X and replacing a with a − ε we obtain
(2)
P (X ≤ a − ε) ≤ P (Xn ≤ a) + P (|Xn − X| > ε).
(2) and (1) gives
P (X ≤ a − ε) − P (|Xn − X| > ε) ≤ P (Xn ≤ a) ≤ P (X ≤ a + ε) + P (|Xn − X| > ε).
2/3
p
Since Xn → X we have P (|Xn − X| > ε) −→ 0. Letting n → ∞ in the above
n→∞
equation yields
P (X ≤ a − ε) ≤ lim P (Xn ≤ a) ≤ P (X ≤ a + ε).
(3)
n→∞
If a ∈ C(FX ) we have
(4)
lim P (Xn ≤ a − ε) = lim P (Xn ≤ a + ε) = P (X ≤ a) = FX (a).
n→∞
n→∞
Now assume a ∈ C(FX ) and let ε → 0 in (3) to obtain
FX (a) ≤ lim P (Xn ≤ a) ≤ FX (a),
n→∞
d
i.e. we have limn→∞ FXn (a) = FX (a) for all a ∈ C(FX ) and thus Xn → X.
2.5
We observe that for each point (x0 , f (x0 )) on the graph of a convex function we can
find a straight line y(x) = k · (x − x0 ) + f (x0 ) that goes through the point (touches
it) such that the line is below the graph in every other point. I.e. we have
(5)
y(x) = k · (x − x0 ) + f (x0 ) ≤ f (x)
for all x.
Assume that X takes values in some set (an interval) I. We have E[X] ∈ I.
Applying (5) we get
k · (x − E[X]) + ϕ(E[X]) ≤ ϕ(x)
for all x ∈ I.
Integrating (taking expectation of) both sides of this inequality yields
RHS = E[ϕ(X)]
LHS = E[k · (X − E[X]) + ϕ(E[X])] = E[kX − kE[X] + E[ϕ(E[X])]
= kE[X] − kE[X] + ϕ(E[X]) = ϕ(E[X])
i.e. ϕ(E[X]) ≤ E[ϕ(X)] and the proof is completed.
3/3