Topics In Analysis

61.8 Right Continuity Of Submartingales

The following theorem is an attempt to consider the question of right continuity. It turns out that you can always assume right continuity of a submartingale by going to a suitable version and this theorem is a first step in this direction.

Theorem 61.8.1 Let

be a real valued submartingale adapted to the filtration ℱ_t. Then there exists a set of measure zero N,P

= 0, such that if ω

N then,

r\totli+m,r\inℚX (r,ω ),r\toltim-,r\inℚX (r,ω)

both exist. There also exists a set of measure zero N such that for ℚ⁺ the nonnegative rationals and ω

su+p |X (t,ω)| < \infty t\inℚ \cap[0,M ]

is bounded for each M ∈ ℕ. ℚ can be replaced with any countable dense subset of ℝ.

Proof: Let

_k=1^∞ be an enumeration of the nonnegative rationals. Let t > 0 be given. Then let

be such that these are in order and

are the first n − 2 rationals less than t listed in order and s₁ = 0 while s_n = t. Then let Y _k ≡ X

. It follows

is a submartingale and so from the maximal inequality in Theorem 59.5.4,

([ m ]) 1 P 1m\leqakx\leqn |Yk| \geq 2 \leq 2m-(2E (|Yn |+ |Y1|)) - m = 2 (2E (|X (t)|+ |X (0)|))

Then letting n →∞, it follows upon letting C_t = 2E

([ ]) P sup |X (r)| \geq 2m \leq 2-mC . r\inℚ+\cap [0,t] t

By the Borel Cantelli lemma, there exists a set of measure 0, N_t such that for ω

N_t,ω is contained in only finitely many of the sets

[ ] m su+p |X (r)| \geq 2 r\inℚ \cap[0,t]

which shows that for ω

N_t,sup_{r∈ℚ⁺∩[0,t]}

is bounded. Now let N = ∪_j=1^∞N_j. This proves the second claim.

Next consider the first claim. By the upcrossing estimate, Theorem 59.5.9 or Lemma 59.2.6, and letting a < b and U

ⁿ

the upcrossings of Y _k from a to b on

for d ≤ t and c ≥ 0,

( ) 1 ( ) E Un[a,b][0,t] \leq -----E (Yn - a)+ b-1a ( ) = -----E (X (t)- a)+ . b- a

Hence

( ) P ([U n [0,t] \geq M ]) \leq 1- -1--E ((X (t) - a)+) . (61.8.27) [a,b] M b- a

(61.8.27)

Suppose for some s < t,

lim sup X (r,ω ) > b > a > lim r\toisn+f,r\inℚ X (r,ω). (61.8.28) r\tos+,r\inℚ

(61.8.28)

If this is so, then in

∩ ℚ there must be infinitely many values of r ∈ ℚ such that X

≥ b as well as infinitely many values of r ∈ ℚ such that X

≤ a. Note this involves the consideration of a limit from one side. Thus, since it is a limit from one side only, there are an arbitrarily large number of upcrossings between s and t. Therefore, letting M be a large positive number, it follows that for all n sufficiently large,

U n [0,t](ω ) \geq M [a,b]

which implies

[ n ] ω \in U [a,b][0,t] \geq M

which from 61.8.27 is a set of measure no more than

1 ( 1 ( + )) M-- b--aE (X (t) - a) .

This has shown that the set of ω such that for some s ∈ [0,t) 61.8.28 holds is contained in the set

[ ] N [a,b] \equiv \cap \inftyM=1 \cup\inftyn=1 U[na,b][0,t] \geq M

Now the sets,

[ ] Un[a,b][0,t] \geq M

are increasing in n and each has measure less than

( ) 1-- --1--E( (X (t)- a)+) M b- a

and so

( [ ]) ( ( )) P \cup\inftyn=1 U n[a,b][0,t] \geq M \leq 1-- --1-E (X (t)- a)+ . M b- a

which shows that

( ) 1 ( 1 ( )) P N [a,b] \leq --- -----E (X (t)- a)+ M b - a

for every M and therefore, P

= 0.

Therefore, corresponding to a < b, there exists a set of measure 0, N

such that for ω

N_[a,b] 61.8.28 is not true for any s ∈ [0,t). Let N ≡∪_a,b∈ℚN_[a,b], a set of measure 0 with the property that if ω

N, then 61.8.28 fails to hold for any pair of rational numbers, a < b for any s ∈ [0,t). Thus for ω

lim X (r,ω) r\tos+,r\inℚ

exists for all s ∈ [0,t). Similar reasoning applies to show the existence of the limit

lim X (r,ω). r\tos -,r\inℚ

for all s ∈ (0,t] whenever ω is outside of a set of measure zero. Of course, this exceptional set depends on t. However, if this exceptional set is denoted as N_t, one could consider N ≡∪_n=1^∞N_n. It is obvious there is no change if ℚ is replaced with any countable dense subset. This proves the theorem.

Of course the above theorem does not say the left and right limits are equal, just that they exist in some way for ω not in some set of measure zero. Also it has not been shown that lim_{r→s+,r∈ℚ}X

= X

for a.e. ω.

Corollary 61.8.2 In the situation of Theorem 61.8.1, let s > 0 and let D₁ and D₂ be two countable dense subsets of ℝ. Then

r\tosli-m,r\inD X (r,ω) = r\tosl-im,r\inD X (r,ω) a.e. ω 1 2

r\tosli+m,r\inD1X (r,ω) = r\tosl+im,r\inD2X (r,ω) a.e. ω

Proof: Let

be an increasing sequence from D_i converging to s and let N be the exceptional set corresponding to the countable dense set D₁ ∪ D₂. Then for ω

N, and i = 1,2,

lim X (r,ω ) = lim X (ri,ω) = lim X (r,ω) r\tos-,r\inD1\cupD2 n\to\infty n r\tos -,r\inDi

The other claim is similar. This proves the corollary.

Now here is an impressive lemma about submartingales and uniform integrability.

Lemma 61.8.3 Let X

be a submartingale adapted to a filtration ℱ_t. Let

⊆ [s,t) be a decreasing sequence converging to s. Then

_j=1^∞ is uniformly integrable.

Proof: First I will show the sequence is equiintegrable. I need to show that for all ε > 0 there exists λ large enough that for all n

\int |X (rn)|dP < ε. [|X(rn)|\geq λ]

Let ε > 0 be given. Since

_r≥0 is a submartingale, E

is a decreasing sequence bounded below by E

. This is because for r_n < r_k,

E (X (rn)) \leq E (E (X (rk)|ℱn)) = E (X (rk))

Pick k such that

E (X (rk))- lim E (X (rn)) | n\to\infty | = ||E (X (rk)) - lim E (X (rn))|| < ε∕2. n\to \infty

Then for n > k,

\int \int \int |X (rn)|dP = X (rn)dP + - X (rn)dP [|X(rn)|\geqλ] [X(rn)\geqλ] [X (rn)\leq -λ]

\int \int \int = X (r )dP + X (r )dP - X (r )dP [X(rn)\geqλ] n [X(rn)>- λ] n Ω n \int \int \int \leq X (rn)dP + E (X (rk)|ℱn )dP - X (rn)dP \int [X(rn)\geqλ] \int[X(rn)>- λ] \int Ω \leq X (rk)dP + X (rk)dP - X (rk)dP + ε∕2 \int [X(rn)\geqλ] \int[X(rn)>- λ] Ω = [X(rn)\geqλ]X (rk)dP + [X(rn)\leq- λ](- X (rk))dP + ε∕2 \int = |X (rk)|dP +ε∕2 \int [|X (rn)|\geqλ] \leq |X (rk)|dP + ε∕2 (61.8.29) [sup{|X(r)|\geq λ:r\in{rj}\inftyj=1}]

From maximal inequalities of Theorem 59.5.4

( [ ]) P sup |X (r)| \geq λ \leq 2E-(|X-(t)|+|X-(0)|) \equiv C- r\in{rn,rn-1,\cdot\cdot\cdot,r1} λ λ

and so, letting n →∞,

( [ ]) C P r\ins{urp}\infty |X (r)| \geq λ \leq -λ. n n=1

It follows that for λ sufficiently large the first term in 61.8.29 is smaller than ε∕2 because k is fixed. Now this shows there is a choice of λ such that for all n > k,

\int |X (rn)|dP < ε [|X(rn)|\geqλ]

There are only finitely many r_n for n ≤ k and by choosing λ sufficiently large the above formula can be made to hold for these also, thus showing

is equi integrable.

Now this implies the sequence of random variables is uniformly integrable as well. Let ε > 0 be given and choose λ large enough that for all n,

\int |X (r )|dP < ε∕2 [|X(rn)|\geqλ] n

Then let A be a measurable set.

\int \int \int |X (rn)|dP = |X (rn)|dP + |X (rn)|dP A A\cap[|X(\intrn)|\geq λ] A\cap[|X(rn)|<λ] < ε∕2 + |X (rn)|dP \leq ε+ λP (A) A\cap[|X(rn)|<λ] 2

and now you see that if P

is sufficiently small then for all n,

\int |X (rn)|dP < ε A

which shows the set of functions is uniformly integrable as claimed. This proves the lemma.

You can often consider a submartingale to be right continuous. This is the importance of the following theorem.

Theorem 61.8.4 Let

be a submartingale adapted to a normal filtration ℱ_t. There exists a right continuous submartingale having left limits,

such that Y

= X

a.e. for every t ∈ ℚ⁺. Furthermore

has a right continuous left limits version if and only if

t \to E (X (t))

is right continuous. More generally, Y

= X

a.e. at every point where the above function is right continuous.

Proof: From Theorem 61.8.1, there exists a set of measure zero, N such that for ω

N, left and right limits of the following form exist.

lim X (r,ω), lim X (r,ω). r\tot+,r\inℚ r\tot- ,r\inℚ

Then define for each t and ω

Y (t,ω) \equiv r\toltim+,r\inℚ X (r,ω).

and for ω ∈ N,

Y (t,ω) \equiv 0

Thus Y

= X

a.e. for t ∈ ℚ⁺. For each ω

N, there exists δ > 0 such that if r ∈ ℚ, t < r < t + 2δ, then

|Y (t,ω)- X (r,ω)| < ε∕2.

Now suppose s ∈

. Then there exists δ₁ < δ such that if s < r < s + δ₁ then

|Y (s,ω)- X (r,ω)| < ε∕2.

pick r ∈ ℚ ∩