63.1. HOW TO RECOGNIZE A MARTINGALE 2117

Proof: First of all, the sum converges because eventually τk∧t = t. Therefore, for largeenough k, M (τk+1∧ t)−M (τk ∧ t)≡ ∆Mk = 0. Consider first the finite sum, k ≤ q.

E

( q

∑k=0

(ξ k,∆Mk)

)2 (63.1.2)

When the sum is multiplied out, you get mixed terms. Consider one of these mixedterms, j < k

E((ξ k,∆Mk)

(ξ j,∆M j

))Using Corollary 63.1.3 and Doob’s optional sampling theorem, Theorem 60.5.4, this equals

E(

E((ξ k,∆Mk)

(ξ j,∆M j

)|Fτk

))= E

((ξ j,∆M j

)E((ξ k,∆Mk) |Fτk

))= E

((ξ j,∆M j

)(ξ k,E

(M (τk+1∧ t)−M (τk ∧ t) |Fτk

)))= E

((ξ j,∆M j

)(ξ k,0)

)= 0

Note that in using the optional sampling theorem, the stopping time τk+1∧ t is bounded.Therefore, the only terms which survive in 63.1.2 are the non mixed terms and so this

expression reduces to

q

∑k=0

E (ξ k,∆Mk)2 ≤C2

q

∑k=0

E(||∆Mk||2

)

=C2q

∑k=0

E(||M (τk+1∧ t)−M (τk ∧ t)||2

)

= C2q

∑k=0

E(||M (τk+1∧ t)||2

)+E

(||M (τk ∧ t)||2

)−2E ((M (τk ∧ t) ,M (τk+1∧ t))) (63.1.3)

Consider the term E ((M (τk ∧ t) ,M (τk+1∧ t))) . By Doob’s optional sampling theorem formartingales and Corollary 63.1.3 again, this equals

E(E((M (τk ∧ t) ,M (τk+1∧ t)) |Fτk

))= E

((M (τk ∧ t) ,E

(M (τk+1∧ t) |Fτk

)))= E ((M (τk ∧ t) ,M (τk+1∧ t ∧ τk)))

= E(||M (τk ∧ t)||2

)It follows 63.1.3 equals

C2q

∑k=0

E(||M (τk+1∧ t)||2

)−E

(||M (τk ∧ t)||2

)≤C2E

(||M (t)||2

).

63.1. HOW TO RECOGNIZE A MARTINGALE 2117Proof: First of all, the sum converges because eventually t; \t =t. Therefore, for largeenough k, M (T41 At) -M(t, At) = AM; = 0. Consider first the finite sum, k < q.q 2E (x am) (63.1.2)k=0When the sum is multiplied out, you get mixed terms. Consider one of these mixedterms, j <kE ((E,,AMe) (€;,4M;) )Using Corollary 63.1.3 and Doob’s optional sampling theorem, Theorem 60.5.4, this equalsE(E ((&,4My) (€;,4M;) |Fa)) =E ((E)-AM;) E (Ex AMe) Fr.)=E ((€),AM;) (EE (M (tes At) —M (te At) | Fe) =E (EAM) (E,,0)) =0Note that in using the optional sampling theorem, the stopping time T;+ At is bounded.Therefore, the only terms which survive in 63.1.2 are the non mixed terms and so thisexpression reduces toqYe Ey,aM)* <C* YE (|iaMill)k=0= CY (\|M (tus A) —M (ADI?)k=0= CLE (\IM Ce ANP) +8 (iM (eA0IP)k=0~2E (M(t, At) M (tey1 At))) (63.1.3)Consider the term E ((M (t% At) ,M (Tx41 At))). By Doob’s optional sampling theorem formartingales and Corollary 63.1.3 again, this equalsE(E ((M(t At) ,M (t41 At)) | Fx))= E((M(t,At),E (M(th1 At) |Fe,)))E((M(t, At) ,M (Tey1 At A T)))= E (||M( AZ|)It follows 63.1.3 equalsChe ((l (ress Aa)|) —E (LM (te A0)|?) < CPE (lM @)IP).