Kenneth Kuttler

2102 CHAPTER 62. STOCHASTIC PROCESSES

The next task is to consider an upcrossing estimate as was done before for discretesubmartingales.

τ0 ≡ min(inf{t > 0 : X (t) = a} ,M) ,

τ1 ≡ min(inf{

t > 0 : (X (t ∨ τ0)−X (τ0))+ = b−a},M),

τ2 ≡ min(inf{

t > 0 : (X (τ1)−X (t ∨ τ1))+ = b−a},M),

τ3 ≡ min(inf{

t > 0 : (X (t ∨ τ2)−X (τ2))+ = b−a},M),

τ4 ≡ min(inf{

t > 0 : (X (τ3)−X (t ∨ τ3))+ = b−a},M),

...

If X (t) is never a, then τ0 = M and there are no upcrossings. It is obvious τ1 ≥ τ0 sinceotherwise, the inequality could not hold. Thus the evens have X (τ2k) = a and X (τ2k+1) =b.

Lemma 62.10.1 The above τ i are stopping times for t ∈ [0,M].

Proof: It is obvious that τ0 is a stopping time because it is the minimum of M and thefirst hitting time of a closed set by a continuous adapted process. Consider a stopping timeη ≤M and let

σ ≡ inf{

t > 0 : (X (t ∨η)−X (η))+ = b−a}

I claim that t→ X (t ∨η)−X (η) is adapted to Ft . Suppose α ≥ 0 and consider[(X (t ∨η)−X (η))+ > α

](62.10.36)

The above set equals([(X (t ∨η)−X (η))+ > α

]∩ [η ≤ t]

)∩([(X (t ∨η)−X (η))+ > α

]∩ [η > t]

)Consider the second of the above two sets. Since α ≥ 0, this set is /0. This is because forη > t, X (t ∨η)−X (η) = 0. Now consider the first. It equals[

(X (t ∨η)−X (η))+ > α]∩ [η ∨ t ≤ t] ,

a set of Ft∨η intersected with [η ∨ t ≤ t] and so it is in Ft from properties of stoppingtimes.

If α < 0, then 62.10.36 reduces to Ω, also in Ft . Therefore, by Proposition 62.7.5, σ

is a stopping time because it is the first hitting time of a closed set of a continuous adaptedprocess. It follows that σ ∧M is also a stopping time. Similarly t → X (η)−X (t ∨η) isadapted and

σ ≡ inf{

t > 0 : (X (η)−X (t ∨η))+ = b−a}

is also a stopping time from the same reasoning. It follows that the τ i defined above are allstopping times.

Note that in the above, if η = M, then σ = M also. Thus in the definition of the τ i, ifany τ i = M, it follows that also τ i+1 = M and so there is no change in the stopping times.Also note that these stopping times τ i are increasing as i increases.

2102 CHAPTER 62. STOCHASTIC PROCESSESThe next task is to consider an upcrossing estimate as was done before for discretesubmartingales.T = min(inf{t>0:X(t)=a},M),tT = min(inf{r>0:(X(tV t%) —X(t)), =b—a},M),m1 = ene ae ean7 = min(inf{t>0:(X(tV 1) —X(t2)), =b—a},M),4 = min int {> 0: (X (73 \_ Maven =b—a},M),If X (t) is never a, then Tt) = M and there are no upcrossings. It is obvious Tt; > To sinceotherwise, the inequality could not hold. Thus the evens have X (T2,) = a and X (T2441) =b.Lemma 62.10.1 The above 7; are stopping times for t € [0,M].Proof: It is obvious that To is a stopping time because it is the minimum of M and thefirst hitting time of a closed set by a continuous adapted process. Consider a stopping time7 <M and leto =inf {t>0:(X(tVn)—X(n)), =b-a}I claim that t > X (t V1) — X (1) is adapted to -¥,. Suppose @ > 0 and consider[(X (tVn) —X (n)), > a] (62.10.36)The above set equals([(X(tVn)—X(n)), > a] A[n <4) N([(X(tVn)—X(n)), > a] N[n > 4)Consider the second of the above two sets. Since a > 0, this set is 0. This is because for1 >t, X (tVn) —X (Nn) =0. Now consider the first. It equals[(X(¢Vn)—X(m)), >a] O[nve sd),a set of F,yy intersected with [n Vt <t] and so it is in Y; from properties of stoppingtimes.If a < 0, then 62.10.36 reduces to Q, also in ¥;. Therefore, by Proposition 62.7.5, ois a stopping time because it is the first hitting time of a closed set of a continuous adaptedprocess. It follows that o A M is also a stopping time. Similarly t > X (1) — X (t V7) isadapted ando =inf {t>0:(X(n)—X(tvn)), =b—a}is also a stopping time from the same reasoning. It follows that the t; defined above are allstopping times. ffNote that in the above, if 7 = M, then o = M also. Thus in the definition of the 7;, ifany T; = M, it follows that also t;,; = M and so there is no change in the stopping times.Also note that these stopping times 7; are increasing as i increases.