Diffusion scaling of a limit-order book model Steven E. Shreve - PowerPoint PPT Presentation

Constraints on parameters In order to have a diffusion limit, among the six parameters λ 0 , λ 1 , λ 2 , µ 0 , µ 1 , and µ 2 , there are three degrees of freedom. Let a and b be positive constants satisfying a + b > ab . Then λ 1 = ( a − 1) λ 0 , λ 2 = ( a + b − ab ) λ 0 , µ 1 = ( b − 1) µ 0 , µ 2 = ( a + b − ab ) µ 0 , a λ 0 = b µ 0 . In addition to a and b , there is a scale parameter, which can be set by choosing µ 0 . To simplify the presentation, we set λ 1 = λ 2 = µ 1 = µ 2 = 1 , √ λ 0 = µ 0 = λ := (1 + 5) / 2 . 6 / 50

Limit-order book arrivals and departures c c 1 1 λ 1 1 λ U V W X Y Z c c λ 1 1 λ 1 1 c 1 1 λ 1 1 1 1 λ λ c λ 1 1 λ 1 1 λ 1 1 c c c 1 1 λ 1 1 λ 1 1 λ c c c λ 1 1 λ 1 1 λ 1 1 7 / 50

Transitions of ( W , X ) X 1 1 1 1 λ λ 1 1 1 1 1 1 1 1 1 λ λ W λ λ 1 1 λ 1 1 1 λ 1 1

Transitions of ( W , X ) X 1 1 ( W , X ) is null recurrent √ ⇔ λ = 1+ 5 1 1 λ λ 2 1 1 1 1 1 1 1 1 1 λ λ W λ λ 1 1 λ 1 1 1 λ 1 1 8 / 50

Split Brownian motion The diffusion scaling of a generic process Q is defined to be 1 � Q n ( t ) := √ nQ ( nt ) . 9 / 50

Split Brownian motion The diffusion scaling of a generic process Q is defined to be 1 � Q n ( t ) := √ nQ ( nt ) . Theorem Conditional on the bracketing processes V and Y remaining nonzero, ( � W n , � X n ) converges in distribution to a split Brownian motion ( W ∗ , X ∗ ) = (max { G ∗ , 0 } , min { G ∗ , 0 } ) , where G ∗ is a one-dimensional Brownian motion with variance 4 λ per unit time. � 9 / 50

Split Brownian motion X 1 1 1 1 λ λ 1 1 1 1 1 1 1 1 1 λ λ W λ λ 1 1 λ 1 1 1 λ 1 1 10 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 11 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 12 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 13 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 14 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 15 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 16 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 17 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 18 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 19 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 20 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 21 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 22 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 23 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 24 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 25 / 50

Split Brownian motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 26 / 50

The other queues X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗

The other queues Br. Motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ d dt � W ∗ , W ∗ � t = 4 λ ,

The other queues Br. Motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ Br. Motion d d dt � W ∗ , W ∗ � t = 4 λ , dt � Y ∗ , Y ∗ � t = 4 λ

The other queues Br. Motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ Br. Motion d d dt � W ∗ , W ∗ � t = 4 λ , dt � Y ∗ , Y ∗ � t = 4 λ d dt � W ∗ , Y ∗ � t = 4

The other queues Br. Motion Frozen at 1 θ X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ Frozen at − 1 θ Br. Motion d d dt � W ∗ , W ∗ � t = 4 λ , dt � Y ∗ , Y ∗ � t = 4 λ d dt � W ∗ , Y ∗ � t = 4

The other queues Br. Motion Frozen at 1 θ Frozen at 0 X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ Frozen at − 1 θ Br. Motion d d dt � W ∗ , W ∗ � t = 4 λ , dt � Y ∗ , Y ∗ � t = 4 λ d dt � W ∗ , Y ∗ � t = 4 27 / 50

The other queues X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 28 / 50

The other queues X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 29 / 50

The other queues X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 30 / 50

The other queues X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗

The other queues Frozen at 1 θ Frozen at 0 X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ Frozen at − 1 θ 31 / 50

The other queues Frozen at 1 θ Frozen at 0 X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ Frozen at − 1 θ Starts to diffuse Jumps to 1 θ X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ Jumps to 0 Jumps to − 1 θ 32 / 50

The other queues X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ V ∗ and X ∗ are in a race to zero. 33 / 50

The other queues X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ V ∗ and X ∗ are in a race to zero. A. Metzler , Stat. & Probab. Letters , 2010: “On the first passage problem for correlated Brownian motion.” � 33 / 50

The other queues Suppose V ∗ wins. X ∗ Y ∗ Z ∗ T ∗ U ∗ V ∗ W ∗ ◮ Reset the “bracketing processes” to be U ∗ and X ∗ . ◮ ( V ∗ , W ∗ ) begins executing a split Brownian motion. � 34 / 50

Snapped Brownian motion Let’s consider the V ∗ process in more detail. 35 / 50

Snapped Brownian motion Let’s consider the V ∗ process in more detail. As long as the “bracketing processes” V ∗ and Y ∗ remain nonzero, ( W ∗ , X ∗ ) executes a split Brownian motion: ( W ∗ , X ∗ ) = (max { G ∗ , 0 } , min { G ∗ , 0 } ) , where G ∗ is a one-dimensional Brownian motion with variance 4 λ per unit time. Frozen at 1 θ X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 35 / 50

Snapped Brownian motion Still have the split Brownian motion, ( W ∗ , X ∗ ) = (max { G ∗ , 0 } , min { G ∗ , 0 } ) , but now V ∗ is diffusing. Br. Motion X ∗ Y ∗ Z ∗ U ∗ V ∗ W ∗ 36 / 50

Snapped Brownian motion G ∗ ( t ) t

Snapped Brownian motion G ∗ ( t ) t 1 V ∗ ( t ) θ t 37 / 50

Summary of properties of the limiting model 38 / 50

Summary of properties of the limiting model ◮ At almost every time, there is a two-tick spread (i.e., one empty tick), but this happens only 24% of the time in the pre-limit model. 38 / 50

Summary of properties of the limiting model ◮ At almost every time, there is a two-tick spread (i.e., one empty tick), but this happens only 24% of the time in the pre-limit model. ◮ The queues at the best bid and best ask in the limiting model form a two-dimensional correlated Brownian motion. 38 / 50

Summary of properties of the limiting model ◮ At almost every time, there is a two-tick spread (i.e., one empty tick), but this happens only 24% of the time in the pre-limit model. ◮ The queues at the best bid and best ask in the limiting model form a two-dimensional correlated Brownian motion. ◮ The queues behind the best bid and best ask in the limiting model are frozen at 1 θ and − 1 θ . 38 / 50

Summary of properties of the limiting model ◮ At almost every time, there is a two-tick spread (i.e., one empty tick), but this happens only 24% of the time in the pre-limit model. ◮ The queues at the best bid and best ask in the limiting model form a two-dimensional correlated Brownian motion. ◮ The queues behind the best bid and best ask in the limiting model are frozen at 1 θ and − 1 θ . ◮ When the queue at the best bid or the best ask is depleted, we have a three-tick spread. 38 / 50

Summary of properties of the limiting model ◮ At almost every time, there is a two-tick spread (i.e., one empty tick), but this happens only 24% of the time in the pre-limit model. ◮ The queues at the best bid and best ask in the limiting model form a two-dimensional correlated Brownian motion. ◮ The queues behind the best bid and best ask in the limiting model are frozen at 1 θ and − 1 θ . ◮ When the queue at the best bid or the best ask is depleted, we have a three-tick spread. ◮ We transition through the three-tick spread using the concept of a snapped Brownian motion. � 38 / 50

Renewal states Y ∗ Z ∗ U ∗ V ∗ W ∗ X ∗

Renewal states Y ∗ Z ∗ U ∗ V ∗ W ∗ X ∗ X ∗ Y ∗ T ∗ U ∗ V ∗ W ∗

Renewal states Y ∗ Z ∗ U ∗ V ∗ W ∗ X ∗ X ∗ Y ∗ Z ∗ A ∗ T ∗ U ∗ V ∗ W ∗ V ∗ W ∗ X ∗ Y ∗

Renewal states Y ∗ Z ∗ U ∗ V ∗ W ∗ X ∗ X ∗ Y ∗ Z ∗ A ∗ T ∗ U ∗ V ∗ W ∗ V ∗ W ∗ X ∗ Y ∗ Which way? How long? 39 / 50

How long to transition between renewal states? Recall W ∗ = max { G ∗ , 0 } , X ∗ = min { G ∗ , 0 } . 40 / 50

How long to transition between renewal states? Recall W ∗ = max { G ∗ , 0 } , X ∗ = min { G ∗ , 0 } . Negative excursions of G ∗ : V ∗ diffuses; Y ∗ frozen at -1 /θ . 40 / 50

How long to transition between renewal states? Recall W ∗ = max { G ∗ , 0 } , X ∗ = min { G ∗ , 0 } . Negative excursions of G ∗ : V ∗ diffuses; Y ∗ frozen at -1 /θ . Positive excursions of G ∗ : Y ∗ diffuses; V ∗ frozen at 1 /θ . 40 / 50

How long to transition between renewal states? Recall W ∗ = max { G ∗ , 0 } , X ∗ = min { G ∗ , 0 } . Negative excursions of G ∗ : V ∗ diffuses; Y ∗ frozen at -1 /θ . Positive excursions of G ∗ : Y ∗ diffuses; V ∗ frozen at 1 /θ . Lengths of positive excursions of G ∗ Local time of G ∗ at 0 Lengths of negative excursions of G ∗

How long to transition between renewal states? Recall W ∗ = max { G ∗ , 0 } , X ∗ = min { G ∗ , 0 } . Negative excursions of G ∗ : V ∗ diffuses; Y ∗ frozen at -1 /θ . Positive excursions of G ∗ : Y ∗ diffuses; V ∗ frozen at 1 /θ . Lengths of positive excursions of G ∗ τ V Local time of G ∗ at 0 Lengths of negative excursions of G ∗

How long to transition between renewal states? Recall W ∗ = max { G ∗ , 0 } , X ∗ = min { G ∗ , 0 } . Negative excursions of G ∗ : V ∗ diffuses; Y ∗ frozen at -1 /θ . Positive excursions of G ∗ : Y ∗ diffuses; V ∗ frozen at 1 /θ . Lengths of positive excursions of G ∗ τ V τ Y Local time of G ∗ at 0 Lengths of negative excursions of G ∗ 40 / 50

Calculation of renewal time distribution ◮ Let p ( ℓ ) be the probability V ∗ reaches zero during a negative excursion of G ∗ of length ℓ . Can be computed by adapting Metzler. 41 / 50

Calculation of renewal time distribution ◮ Let p ( ℓ ) be the probability V ∗ reaches zero during a negative excursion of G ∗ of length ℓ . Can be computed by adapting Metzler. ◮ p ( ℓ ) is also the probability Y ∗ reaches zero during a positive excursion of G ∗ . 41 / 50

Calculation of renewal time distribution ◮ Let p ( ℓ ) be the probability V ∗ reaches zero during a negative excursion of G ∗ of length ℓ . Can be computed by adapting Metzler. ◮ p ( ℓ ) is also the probability Y ∗ reaches zero during a positive excursion of G ∗ . ◮ Four independent Poisson random measures: 41 / 50

Calculation of renewal time distribution ◮ Let p ( ℓ ) be the probability V ∗ reaches zero during a negative excursion of G ∗ of length ℓ . Can be computed by adapting Metzler. ◮ p ( ℓ ) is also the probability Y ∗ reaches zero during a positive excursion of G ∗ . ◮ Four independent Poisson random measures: ◮ ν ± 0 ( dt d ℓ ) – Lengths of positive (negative) excursion of G ∗ during which Y ∗ ( V ∗ ) reaches zero. L´ evy measure is µ 0 ( d ℓ ) = p ( ℓ ) d ℓ √ 2 πℓ 3 . 2 41 / 50

Calculation of renewal time distribution ◮ Let p ( ℓ ) be the probability V ∗ reaches zero during a negative excursion of G ∗ of length ℓ . Can be computed by adapting Metzler. ◮ p ( ℓ ) is also the probability Y ∗ reaches zero during a positive excursion of G ∗ . ◮ Four independent Poisson random measures: ◮ ν ± 0 ( dt d ℓ ) – Lengths of positive (negative) excursion of G ∗ during which Y ∗ ( V ∗ ) reaches zero. L´ evy measure is µ 0 ( d ℓ ) = p ( ℓ ) d ℓ √ 2 πℓ 3 . 2 ◮ ν ± × ( dt d ℓ ) – Lengths of positive (negative) excursions of G ∗ during which Y ∗ ( V ∗ ) does not reach zero. L´ evy measure is µ × ( d ℓ ) = (1 − p ( ℓ )) d ℓ √ . 2 πℓ 3 2 41 / 50

Calculation of renewal time distribution � � ◮ τ Y = min { t ≥ 0 : ν + (0 , t ] × (0 , ∞ ) > 0. 0 42 / 50

Calculation of renewal time distribution � � ◮ τ Y = min { t ≥ 0 : ν + (0 , t ] × (0 , ∞ ) > 0. 0 � � ◮ τ V = min { t ≥ 0 : ν − (0 , t ] × (0 , ∞ ) > 0. 0 42 / 50

Calculation of renewal time distribution � � ◮ τ Y = min { t ≥ 0 : ν + (0 , t ] × (0 , ∞ ) > 0. 0 � � ◮ τ V = min { t ≥ 0 : ν − (0 , t ] × (0 , ∞ ) > 0. 0 ◮ τ Y and τ V are independent. 42 / 50

Calculation of renewal time distribution � � ◮ τ Y = min { t ≥ 0 : ν + (0 , t ] × (0 , ∞ ) > 0. 0 � � ◮ τ V = min { t ≥ 0 : ν − (0 , t ] × (0 , ∞ ) > 0. 0 ◮ τ Y and τ V are independent. ◮ We want to know the distribution of (i) the chronological time T 1 corresponding to local time τ Y ∧ τ V , 42 / 50

Calculation of renewal time distribution � � ◮ τ Y = min { t ≥ 0 : ν + (0 , t ] × (0 , ∞ ) > 0. 0 � � ◮ τ V = min { t ≥ 0 : ν − (0 , t ] × (0 , ∞ ) > 0. 0 ◮ τ Y and τ V are independent. ◮ We want to know the distribution of (i) the chronological time T 1 corresponding to local time τ Y ∧ τ V , (ii) plus the chronological elapsed time T 2 in the “last excursion” beginning at local time τ Y ∧ τ V before Y ∗ or V ∗ reaches zero. 42 / 50

Calculation of renewal time distribution � � ◮ τ Y = min { t ≥ 0 : ν + (0 , t ] × (0 , ∞ ) > 0. 0 � � ◮ τ V = min { t ≥ 0 : ν − (0 , t ] × (0 , ∞ ) > 0. 0 ◮ τ Y and τ V are independent. ◮ We want to know the distribution of (i) the chronological time T 1 corresponding to local time τ Y ∧ τ V , (ii) plus the chronological elapsed time T 2 in the “last excursion” beginning at local time τ Y ∧ τ V before Y ∗ or V ∗ reaches zero. ◮ (i) is � ∞ � τ Y ∧ τ V � ∞ � τ Y ∧ τ V ℓν + ℓν − T 1 := × ( dt d ℓ ) + × ( dt d ℓ ) . ℓ =0 t =0 ℓ =0 t =0 42 / 50

Calculation of renewal time distribution � � ◮ τ Y = min { t ≥ 0 : ν + (0 , t ] × (0 , ∞ ) > 0. 0 � � ◮ τ V = min { t ≥ 0 : ν − (0 , t ] × (0 , ∞ ) > 0. 0 ◮ τ Y and τ V are independent. ◮ We want to know the distribution of (i) the chronological time T 1 corresponding to local time τ Y ∧ τ V , (ii) plus the chronological elapsed time T 2 in the “last excursion” beginning at local time τ Y ∧ τ V before Y ∗ or V ∗ reaches zero. ◮ (i) is � ∞ � τ Y ∧ τ V � ∞ � τ Y ∧ τ V ℓν + ℓν − T 1 := × ( dt d ℓ ) + × ( dt d ℓ ) . ℓ =0 t =0 ℓ =0 t =0 ◮ For (ii), we observe that the distribution of the length of the “last excursion” is µ 0 ( d ℓ ) /µ 0 ((0 , ∞ )). 42 / 50

Calculation of renewal time distribution � � ◮ τ Y = min { t ≥ 0 : ν + (0 , t ] × (0 , ∞ ) > 0. 0 � � ◮ τ V = min { t ≥ 0 : ν − (0 , t ] × (0 , ∞ ) > 0. 0 ◮ τ Y and τ V are independent. ◮ We want to know the distribution of (i) the chronological time T 1 corresponding to local time τ Y ∧ τ V , (ii) plus the chronological elapsed time T 2 in the “last excursion” beginning at local time τ Y ∧ τ V before Y ∗ or V ∗ reaches zero. ◮ (i) is � ∞ � τ Y ∧ τ V � ∞ � τ Y ∧ τ V ℓν + ℓν − T 1 := × ( dt d ℓ ) + × ( dt d ℓ ) . ℓ =0 t =0 ℓ =0 t =0 ◮ For (ii), we observe that the distribution of the length of the “last excursion” is µ 0 ( d ℓ ) /µ 0 ((0 , ∞ )). ◮ Adapt Metzler again to compute the distribution of the elapsed time T 2 in the “last excursion,” conditioned on its length. � 42 / 50

Calculation of renewal time distribution The moment-generating function of T 1 + T 2 is � e − α ( T 1 + T 2 ) � E � �� α � ∞ � ℓ � ∞ � e − α s p ( s , ℓ ) e − αℓ p ( ℓ ) d ℓ = √ 2 πℓ 3 ds d ℓ 2 + √ , 2 πℓ 3 2 0 0 0 where p ( s , ℓ ) is the conditional density in s of the elapsed time T 2 given that the “last excursion” has length ℓ . 43 / 50

Probability Density Function 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 2.0 x

Concluding remarks 44 / 50