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S TOCHASTIC P ROCESSES II
PART I – M ARTINGALES
Conditional expectations
Measure theory
o In a probability space ( W, , ), a sigma field is a collection of events, each of which as a subset of W^.^ It^ satisfies^ (i)^ Æ Î^ ^ (ii)^ A^ Î^ ^ ^ Ac Î^ (iii) Ai Î È i ¥ (^) = 1 Ai Î. Notes: (i) and (ii) W Î
c^ c
i^ ¥ =^ Ai^ =^ i ¥= Ai , so also closed under infinite (and finite) intersection.
o A random variable maps X ( w ) : W . When we say X is measurable with
respect to and write X Î , we mean { w : X ( w ) £ x }Î " x.
Conditional expectations
o ( X | Y ) is a random variable. ( X | Y )( w ) = ( X | Y = Y ( w )). In other words,
the fact Y = Y ( w ) “reveals” a “region” of W in which we are located. We then find the expected value of X given we are in that “ region ”. In terms of the definition below, we can write ( X | Y ) =( X | s ( Y )), where s ( Y ) is the sigma-field generated by Y – in other words,
s ( Y ) = { { w : Y ( w ) £ x }: x Î } – every event can would be revealed by
Y.
o W = ( X | ) is a random variable. ( X | )( w )is a bit harder to understand – effectively, it takes the expectation of X over the smallest that contains w. In other words, let A be the smallest element of that contains w – then we restrict ourselves to some region of W and find the expectation over that region; ( X | )( w ) =( X A ). Formal properties:
W Î : information as to where we are in W only ever “reaches” us via knowledge of which part of we’re in, so this is obvious. (^) ( W A (^) ) = (^) ( X (^) A ) for all A Î : we are now restricting ourselves to a region of W that is -measurable. Provided A is the smallest element for which (^) w Î A , (^) W ( w ) = ( X A ), and the result follows trivially. (If it is not the smallest element, the result requires additional thought). o Some properties i. éêë X^ | ùúû if X Î ii. éêë^ éêë X^ | ù =úûùúû ( X ) iii. (^) ( XZ | (^) ) = Z ( X |) if Z Î iv. Tower: éêë^ ( X | (^) ) | ù úû=( X | (^) ) if Í : in this case, is “more descriptive” than , so the result makes sense. Proof : Use ( ( ( X | (^) )| ) A (^) ) = ( ( X | (^) ) A ) for (^) A Î . Then use the fact that A Î to show this is equal to (^) ( X A (^) ). v. Linearity vi. Jensen’s: for convex f , éêë^ f X ( ) | ùúû^ ³ f (^) ( éêë X^ |ùúû)
o Notes éêë X^ ùúû^ = éêë X^ | { Æ W, }ùúû (the RHS is a constant, because whatever w we choose, the only element of {Æ W , } that contains it will be W ). Thus, (ii) is a special case of (iv). Integrability of X implies integrability of éêë X^ | ùúû: ( )
( )
(vi) (ii) éêë^ ( X | ) ùúû^ £ éêë^ X |ùúû = X
o Example : Let W be countable. Let = (^) { 1 , 2 ,} be a partition of W , and
be the set of all subsets of . Then ( X | ) takes value å^^ w^^ Î^ ^ i ( X ( i ) w )(^ w ) with probability ( i ).
Modes of convergence, etc…
Modes of convergence
o Almost sure : (^) ( lim (^) n Xn ( w ) = X ( w )) = 1 , or ( Xn X i.o.) = 0 o In probability : lim (^) n ¥ ( X (^) n - X ³ e ) = 0 " e > 0 Can extract a subsequence (^) { X (^) nk } that tends to X almost surely. Chose (^) ( Xn (^) k (^) + 1 - X ³ (^) k^1 )£ 21 k. o L 1 : lim (^) n ¥ Xn - X = 0 (assuming Xn Î L 1 ) Implies lim n ¥ Xn = X , since (^) ( X (^) n - X (^) ) £ Xn - X (derive by writing a = a - b + b and using triangle inequality). Implies lim (^) n ¥ ( X (^) n ) =( X ), since ( X (^) n - X )£ Xn - X , by Jensen. Implies Xn p X by Markov’s.
Interchange arguments
o Concerned with whether X (^) n X a.s. lim (^) n ¥( X (^) n ) =( X ). o Holds if Bounded convergence : Xn £ c , c is constant. Proof : Define Bcn = An = (^) { X (^) n - X < e }. Write ( Xn - X ) as an integral and split over An and Bn. Use the fact that ( Bn ) 0. [Can replace with convergence in probability; X (^) n £ c still holds for subsequence]. Monotone Convergence : 0 £ Xn X almost surely Proof : Xn £ X ( Xn ) £ ( X ) lim sup (^) n ( Xn ) £( X ). Together with Fatou, gives our result. Dominated convergence : Xn £ Y , Y integrable. Proof : Condition implies Y Xn ³ 0. Apply Fatou to both, subtract ( Y )< ¥from both sides. o Fatou : If X (^) n > 0, (^) ( lim inf n (^) X (^) n )£lim inf n (^) ( X (^) n )
Proof : Let Ym = inf n > m Xn , and note that inf n (^) > m ( X (^) n ) ³( Y (^) m ). Replace Y (^) m with min[ Y (^) m , k ] – bounded because Xn ’s positive. Use bounded convergence and let k ¥.
Uniform integrability
o Definition : (^) { Xn (^) } is uniformly integrable if
0 0
0, s.t.
l sup ,
im (^) n n
n (^) X a n (^) X a
a n X e a X e n a a
> >
¥ éêë^ ù úû= " > $ éê^ ùú£ " ³ ë û
o Lemma : (^) { X (^) n u.i.} sup n Xn £ K Proof : Write Xn = (^) ( Xn (^) Xn £ a ) + (^) ( Xn (^) X (^) n > a )£ a + e. o Lemma : éêësup^ n X (^) n ù úû< ¥ { Xn }u.i. Proof : Let Y = sup n Xn. [ Y ] = [ Y Y (^) £ a ] +[ Y Y (^) ³ a ]. By monotone convergence, first term approaches ( Y ), second gets arbitrary close to 0, and X n £ Y , so u.i. follows. o Lemma : $ d^^ >^ 0 s.t.^ Xn^^1 +^ d £^ K^ < ¥ { X (^) n } u.i. Proof : (^) ( Xn (^) X (^) n a ) Xn Xan Xn a (^) a^1 Xn^1 Xn a aK
d (^) d d d
> > > £ æç^ é^ ù^ ö÷÷^ = æç^ ö÷£ (^) çççè (^) êë úû (^) ÷÷ø (^) çè (^) ÷÷ø. By splitting
into intervals Xn £ 1 and Xn > 1 , can also show that for any 0 < d ¢£ d , also integrable. o Theorem : (^) { Xn (^) } u.i. & X (^) n p X Xn L 1 X Proof : First, show X is integrable by writing éêë^ X ùúû^ = éêëlim inf^ n Xn ùúû, using Fatou’s, writing lim inf < sup and using the definition of u.i [use subsequences for convergence in p ]. Then, set Yn = Xn - X £ Xn + X – it is u.i. because X integrable. Split ( Y (^) n ) into Yn > a and Yn £ a. First one can be made arbitrarily small by u.i. Bounded convergence applies to the second one, and
since Yn 0 , it can also be made arbitrarily small by letting n ¥.
o Theorem : Xn p (^) X & Xn X < ¥ { Xn } u.i.
Optional stopping
Definition (stopping time) : If T is an integer valued random variable, we say it is a stopping time with respect to a filtration n if (^) { T = n } Î n for all n (or (^) { T £ t } Î t for all t , in continuous time). Remark : If T 1 and T 2 are stopping times, so are T 1 (^) + T T 2 , 1 (^) T T 2 , 1 (^) T 2. Theorem : If (^) { X (^) n (^) } is a (sub)martingale and T is a stopping time, (^) { XT (^) n } is also a (sub)martingale. If (^) { Xn^ }is u.i., so it (^) { XT^ n }. Proof : Write XT (^) n = (^) å^ nk (^) - =^10 X (^) k { (^) T = k } + Xn { (^) T > - n 1}. Clearly, this is n measurable, and 1 0 T n n k n X (^) k X X - £ (^) å (^) = + so it is integrable. Conditioning follows. u i.. : to show u.i., first note that (^) { XT + n } is also a submartingale, and so ( X (^) T + ^ n )£ X (^) n + £ Xn. Taking limits, sup (^) n ( XT + n ) £ sup n Xn < ¥ [the last inequality follows by u.i. of (^) { X (^) n }]. By
{ Xn } u.i.
X n L 1 X
X n X < ¥
( X (^) n ) ( X )
X (^) n X a.s.
X n pX
0 £ X (^) n X X^ n £ Y
sup n Xn £ K
éêë^ sup n Xn ù úû< ¥
1 X n K +^ d £
X n £ c
Y < ¥
These three results require X (^) n p X
{ Xn } martingale
the theorem in the martingale convergence section, XT + (^) n n (^) (^) ¥ XT with XT < ¥. Finally, consider éêë^ XT (^) n (^) XT (^) n > a ùúû. Simply split it over T (^) £ n and T (^) > n , drop these indicators and use integrability of X (^) T and u.i. of (^) { X (^) n }. Example : Consider a gambler’s ruin with wealth S (^) t at time t with S 0 = i and with probably ½ of going each direction at each time step. Let ( ) ( )
Probability we hit , at which point we stop 1 Probability we hit 0, at which point we're ruined
N i p
p = >
And let T = inf (^) { n : Sn = N or Sn = (^0) } We can now use the OST o On St OST says that (^) ( ST ) = ( S 0 )= i. Logic says that (^) ( S (^) T ) = pN + 0. Together, we obtain p = i / N. o On St^2 - n OST says that ( ST^2 - T ) = ( S 0^2 )= i^2 , and so ( T ) = ( ST^2 )- i^2. Logic says that ( S (^) T^2 ) = pN^2 + 0. Together, (^ T^^ )=^ i N (^ - i ).
Counterexample : Consider the example above, but with T ¢ = inf (^) { n : Sn = N > i }. This is well defined, in that ( T ¢ < ¥ =) 1 (because the random walk is an irreducible Markov chain, which means every state will eventually be visited) but blindly applying the OST gives ( S (^) T (^) ¢ )= i , which implies that N = i. Clearly, something has gone awry. We need to develop conditions under which the OST works. One such condition is… … Theorem : If T £ n 0 a.s. then [ XT ] =[ Xn 0 ]. Proof : [ XT ] = [ XT (^) n 0 ] =[ X 0 ] Remark : This condition is not satisfied in the counterexample above because even though ( T ¢ < ¥ =) 1 , it is not bounded – following the MC analogy, ( T ¢)^ = ¥. More generally, there are three key “ingredients” to optional stopping (1)^ (2)^ (3) [ X (^) 0 ] = lim (^) n [ XT (^) n ] = [lim (^) n XT (^) n ] = [ XT ] Notes:
- This equality holds for every n (since XT (^) n is also a martingale) and so it still holds after we take t he limit.
- This equality requires an interchange argument – uniform integrability of (^) { XT (^) n } will achieve this, for example.
It is important to remember that when we invoke the martingale convergence theorem so say that (^) XT (^) n n (^) ¥ XT , we are implicitly implying that ( T < ¥ =) 1.
Martingale Inequalities
Doob’s Inequality/The Maximal Inequality
o Motivation : Markov’s Inequality states that a (^) ( X > a (^) )£ X. o Theorem : If (^) { Xn (^) } is a submartingale and A = (^) { max (^0) £ £ k n Xk ³ a }, then a ( A ) £ [ Xn A (^) ] £[ Xn +] Proof : Define T = inf (^) { k : Xk ³ a or k ³ n }. Clearly, it’s a stopping time, and since T < n , [ XT ] =[ Xn ]. Write both sides of equation by splitting over A and A^ c , and note XT = Xn over Ac. Finally, note a ( ) a £[ XT A ]. Finally, note that X (^) n £ Xn +^ [ Xn (^) A ] £[ Xn +]. o If we let S 0 (^) = 0 and Sn = (^) å^ ni = 1 Xn were the Xn are IID with [ Xn ] = 0, ar[ Xn ]= s^2 , then (^) { Sn (^) } is a martingale, and (^) { K (^) n (^) } = (^) { Sn^2 } is a submartingale. As such, we get Kolmogorov’s Inequality : ( ) ( ) 2 2 2 (^0 0 ) max max [^ n ] k n k k n k S x X x S^ n x (^) x
s £ £ ³^ =^ £ £ ³^ £^ =
^
Azuma’s Inequality
o Theorem : (^) ( x -^1 - x -^3 ) j ( ) x £ F( ) x £ x -^1 j ( ) x for all x > 0 (where j is the normal density function). Proof : Notice that F( ) x μò x ¥^ e - y^2 /2d y. For the upper bound, multiply the integrand by (^) ( 1 + y -^2 ). For the lower bound, multiply the integral by (^) ( 1 - 3 y -^4 ). Integrating gives the required result. o Motivation : Let S 0 (^) = 0 and Sn = X 1 + + Xn , with Xi IID m s ,^2 < ¥. Define Yn = Sn^ n - m. We then have, by the Central Limit Theorem (^) ( Yn > e ) (^) » ( Z > (^) sn e ) (^) = 2 F( (^) sne )
Applying the result above gives an exponential approximation. (It is, by the way, summable, so we automatically recover the SLLN). o Theorem (Azuma’s Inequality) : Let (^) { Z (^) n (^) } be a zero-mean martingale with
bounded MG differences (ie: - a £ Z (^) i - Z (^) i - 1 £ b for a b ,^ ³^0 ). Then
( )
2 2 2 exp 2 n m^ n ( ) Z n e me a b
¥
> £ æçç^ - ö÷÷ ççè (^) + ÷÷÷ø
This bound is not as tight as the CLT’s, but it requires less. o Example : S (^) n = number of heads in n flips, where (Heads) = p. Zn = Sn - np is a martingale with - p £ Z (^) i - Z (^) i - 1 £ 1 - p. As such, we can use Azuma’s inequality and obtain (^) ( n^ ¥ = m Snn - p > e ) (^) £ 2 exp (^) ( - 2 me^2 ).
o Definition (Doob Martingale) : Let X be a random variable in L 1 and n be a set of filtrations. Then Xn = (^) ( X | n (^) )is a martingale. Proof : (^) ( Xn (^) + 1 | n (^) ) = éêë^ ( X | m (^) + 1 ) | n (^) ù úû= (^) ( X | n )= Xn.
o Let X =( X 1 , , Xn ), where the Xi are independent and with CDF Fi. Define i = s ( X 1 , , Xi ). Finally, let h : n such that, if x differs from y in only one component, h ( ) x - h ( ) y £ , for some ³ 0. Then Si = éêë h^ ( X ) | i ùúû is a Doob martingale. Provided we can prove Si - Si (^) - 1 £ , we can apply Azuma’s Inequality with a + b = to S (^) n = h ( X )
( )
2 ( ) ( ) 2 exp (^22) h - é h^ ù^ > n e £ æçç^ -^ ne ö÷÷÷ X êë X (^) úû (^) ççè ÷÷ø
Proof : To prove Si - Si (^) - 1 £ , note that ( ) ( )
1 1 1 1 1 1
1 1 1 1 , ,
, , , d ( )
) | , , , , , d ( ) d ( ) d ( )d ( )
i i n
i i x xn i i n n n i i i (^) x x i i n n n i i i i
S
S h X x x F x
h h X X x x x F x F
F x x
= éêë^ ù =ûú = (^) ò ò
X (^) ò (^) ò
As such, remembering that densities integrate to 1 1 d 1 d
S (^) i S (^) i - £ £
ò ò ò ò
As required.
{ } { }
2 1 2 2 2 2 1
inf : inf :
k k m k k m
T m T X a T m T X b
Odd stopping times mark the first time the martingale downcrosses a since its last upcrossing of b. Even stopping times are the other way round. These are clearly stopping times. o Consider a gambling strategy that sets Cm = 1 if T 2 (^) k - 1 < m £ T 2 k and Cm = 0 otherwise. We effectively “buy” only if we’re below a and sell otherwise. Every time there is an upcrossing, a profit of at least b – a is realized. o Theorem : Let U (^) n = sup (^) { k : T 2 (^) k £ n } – this is the number of upcrossings up to time n – and assume Xn is a submartingale. Then ( b - a ) éêë U^ (^) n ù £úû éêë(^ Xn - a ) +^ ùúû^ - éêë(^ X 0 - a )+ùúû (This inequality would also apply for a martingale). Remark : This is hardly very encouraging. The RHS is the amount that would have been earned by buying at the start and getting out at n. Proof : WLOG, shift the upcrossing range to [0, b - a ], and define Yn = ( X (^) n - a )+
- this is also a submartingale and has the same number of upcrossings. Define C (^) m as above, and since we earn at least b – a at every upcrossing, ( b - a U ) (^) n £ ( C Y ) n ( b - a )[ U (^) n ] £[( C Y ) n ]. Letting C (^) n = 1 - Cn , we get ( ) ( ) 1 ( 1 ) 0 n C Y (^) n + C Y (^) n = (^) å k = Yk - Y (^) k (^) - = Yn - Y. Since ( C Y ) n is also a martingale, éêë(^ C Y ) n ùúû^ ³ éêë(^ C Y ) 0 ùúû= 0. Thus, [( C Y ) ] n £ [ Y (^) n - Y 0 ].
Martingale Convergence
Theorem : If (^) { X (^) n (^) } is a submartingale and sup (^) n Xn < ¥ (this is a weaker condition than u.i.), then Xn X a.s.and ^ X^ < ¥. Remark : sup (^) n Xn < ¥ is equivalent to sup (^) n ( Xn +^ )< ¥ because x +^ £ x = 2 x +- x , and so, for example, ^ Xn^ =^ 2 [^ Xn^ +^ ]^ -^ [^ Xn^ ]^ £^ 2 [^ Xn^ +]^ - [^ X 0 ]. Proof : Note that ( X (^) n - a )+^ £ Xn ++ a and write the upcrossing inequality as ( X (^) n ) a U (^) n b a K
- (^) + éêë ù £úû (^) - = < ¥ [using the fact | X | dominates X +^ and the condition in the
theorem). However, U (^) n is increasing by definition and therefore tends to some U
(possibly ¥), but by monotone convergence, ( U n ) ( U )< ¥. Thus, the number of
up-crossings must be finite, and so (lim inf^ X^ n <^ a^ <^ b^ <^ lim sup Xn ) = 0. This is true for any a and b , and so (lim inf X (^) n = lim sup Xn ) = 1. Integrability of the limit follows by Fatou. Corollary : If (^) { Xn (^) } is a supermartingale and Xn > 0, then Xn X and [ X ] £[ X 0 ]. Proof : Let Yn = - X (^) n – this is a submartingale with [ Y (^) n +^ ] = 0. The condition of the theorem above (see the remark) is therefore satisfied. Note, however, that almost sure converge does not imply convergence in the means or variances, as the next two examples illustrate.. Example : Assume X (^) 0 = i > 0 , and Xn | X (^) n - 1 ~ Po( Xn - 1 ). Clearly, this is a martingale, and once we hit 0, we stay there. Let T = inf (^) { n : X (^) n = 0 or Xn ³ b }. By optional stopping, ( X (^) T ) = b (1 - p ), where b ^ ³ b and p = ( XT =0). But ( XT )= i. As such, 1 - p (^) bi 0 as b ¥. Thus, p 1. How do we know stopping time is finite? Note, however, that [ Xn^2 ] = [ ( X (^) n^2^ | Xn (^) - 1 )] = ( X (^) n^2 - 1 )+ i. As such, ar( Xn )= ni ; the variable itself tends to 0, but the variance blows up. Example : Let X 1 (^) ~ U [0,1], and X (^) n | Xn (^) - 1 ~ U [0, Xn - 1 ]. Let Yn = 2 nXn. We can write Xn = U 1 U (^) n with each U IID U [0,1]. This is a martingale, and, by the SLLN, n^1 log^ Yn^ =^ log 2^ +^1 n å^^ ni = 1 log^ U^ i ^ log^2 +( o^ l g^ U^1 )< ¥ a.s. So^ Yn^ ^0. Note, however, that ar( X (^) n ) = (^) ( )^43^ n - 1. Again, the variance blows up. Coupled with uniform integrability, however… … Theorem : When (^) { Xn (^) } is a martingale, the following are equivalent i. (^) { Xn^ } is u.i. (and therefore converges almost surely) ii. X n L 1 X iii. Xn can be written as a Doob martingale; Xn^ =^ ^ éêë X^ |^ n^ úûù, with ^ X^ < ¥. For submartingales, only (i) and (ii) are equivalent. Proof : ( i ) ( ii ) follows from the fact u.i. implies the boundedness condition in the convergence theorem. ( ii ) ( iii ): Let X be the L 1 limit of (^) { X (^) n (^) }; clearly, éêë X^ | n ùúû is a Doob martingale. To show this is equal to Xn , all we need to show is that [ Xn (^) A ] =[ X A ] for any A Î n. Two steps. (1) Use the martingale property to write,
o Covariance : Consider s < t. Then ov éêë B s^ ( ), B t ( ) ùúû^ = ov éêë B s^ ( ), B t ( ) - B s ( ) + B s ( ) ùúû = 0 + ar éêë B s^ ( ) ùúû= s = s t
o Joint density : Let t 1 (^) < t 2. Then
1 2 1
1 1 2 2 1 1 2 1 2 1 1 1 2 1 2 1 1 2 1
t (^ )^ t t (^ )
B t x B t x B t x B t B t x x B t x B t t x x f x f (^) - x x
éêë (^) = = ùúû (^) = éêë (^) = - = - ùúû = éêë^ = ùúû^ éêë^ - = - ùúû = -
o Conditional density : Let s < t. Then
( ) (^) ( ) ( )
2 2 2
1 2 1 (2(^ )) 2 2 ( ) 21 2
( ) | ( ) ( )^ ,^ ( )
exp exp ex
p
s t s t x b x x a^ t s t^ s t bt
B s x B t b B s^ x B f
t
x f b x f b
b B t b
p p p
éê (^) = = ùú= éêë^ é =^ ù= ùúû ë û ê = = -
=
úû
ë
By multiplying out and then completing the square, we find that ( B s ( ) |^^ B t ( )^^ =^ b^ ) ~^ N b (^ st^ ,^ s (^1 - st )) The mean is analogous to what we obtained in a Poisson process. The variance is less intuitive. o Hitting time : Let Ta = inf (^) { t : B t ( ) = a }, a > 0 (the Brownian path is
symmetric, so the result should be identical for – a ). Now consider (^) ( Bt ³ a (^) ) = (^) ( B (^) t ³ a | Ta £ t (^) ) (^) ( T £ t (^) ) + (^) ( Bt ³ a | Ta > t (^) ) ( T > t ) = 12 ( T £ t ) The last line follows from the so-called reflecting principle ; once Bt has hit a , it is equally likely to be above and below a at a later time, by symmetry. Now (^) ( T (^) a £ t (^) ) = 2 ( B (^) t ³ a ) = 2 F( at ) Now, as t ¥ , we would expect this probability to tend to 1, since the Brownian motion should eventually hit a. Indeed, (^) ( T (^) a £ t ) 2 (0)F = 1. However, Brownian motion is a null-recurrent Markov chain – the expected value of any hitting time is infinity:
(^2) /2 2 /2^2 / 0 1 1 ( ) 1 1 d d 2 2 2
a t a t^ a a T a^ e a^ e t ae^ t p (^) t p (^) t p t
¥ (^) - ¥ (^) -^ - ¥ = (^) ò ³ (^) ò ³ (^) ò = ¥ Note also that if Mt = max s £ tBs (a quantity always increasing in t ), we have (^) ( M (^) t ³ a (^) ) = ( T (^) a £ t ) = 2 F( at ) Similarly, note that min (^) s £ t Bs = - max (^) s £ t (- B (^) s ) =d - max s £ tBs.
o Arcsine law : Let X ( t ) be a Brownian Motion starting at a point other than 0. Note that ( ) ( ) ( ) ( )
0 0 0
min ( ) 0 | (0) max ( ) 0 | (0) max ( ) | (0) 0
u t u t u t a
X u X a X u X a X u a X T t
£ £ £ £ £ £
As such (^) ( At least one 0 in [ t 0 (^) , t 1 (^) ] | X t ( 0 ) = a (^) )= P a ( ) = ( T (^) a £ t ) Now, assume we know X (0) = 0, but do not know X ( t 0 ). The probability a of there being at least one 0 in [ t 0 , t 1 ]. We can find this by conditioning on the value of X ( t 0 ), and we get
( 0 1 ) 0 0 1 1
At least one 0 in ( t , t ) | X (0) 0 1 2 arcsin^ t^^2 arccos t = = - (^) p t = p t
o Consider A xt ( , y ) = ( X t ( ) > y , min (^) 0 £ £ u t X ( ) u > 0 | X ( 0 )= x ) Note that (^) ( X t ( ) > y | X (0) = x (^) ) = A x yt ( , ) + ( X t ( ) > y , min (^) 0 £ £ u t £ 0 | X (0) = x ) Consider the last term; by reflecting at the first time the process hits 0, we find that every sample path satisfying that term has a corresponding path that falls below 0 and with X ( t ) < – y (and vice versa). As such ( ) ( ) ( ) ( ) ( )
[ , ]
t t t t
A x y X t y X x X t y X x B y x B y x B y x y x
= Î - +
o Consider M (^) t = max 0 £ £ u tBt and Yt = Mt - Xt. Consider (^) ( M (^) t ³ m B , (^) t £ x (^) ) = ( Bt ³ 2 m - x )
This second argument is incorrect , because in going from the first to the second line, we condition Bt on the past while ignoring that we are at the same time condition Bt on the future (because we are conditioning on B 1 = 0).
Martingales Associated with Brownian Motion
o Recall that in continuous time, the martingale property reads, for all t > s , (^) ( X t ( ) | s (^) )= X s ( ). Defining a stopping time is more tricky; if T satisfies { T^ £^ t } Î^ t^ ,^ then^ since^ ^ is^ an^ increasing^ family, { T^ <^ t }^ =^ n ¥= 1 { T £^ t - n^1 }Î t^. However, if^ T^ only satisfies^ { T^ <^ t } Î^ t^ , then { T^ £^ t }^ =^ n { T^ <^ t +^1 n }Ï t^ (again,^ since^ the^ family^ is^ increasing).^ To conclude this, we need to assume right-continuity of the filtration. o Theorem : The following three processes are martingales: (^) { B (^) t (^) } (the “mean martingale”) (^) { B (^) t^2 - t } (the “variance martingale”) (^) { exp (^) ( qB (^) t - 12 q^2 t )}, where q is a deterministic parameter (the “exponential martingale”). Proof : The first two parts are trivial. For the last, recall that ( e qN^ (0,1)^ ) = eq^2 / and (^) ( e qB^ t^ -^12 q^2^^ t^ | s (^) ) = (^) ( e qB^ s^ -^12 q^2^^ t^ ) (^) ( e q (^^ Bt^ - B^ s^ )^ | s (^) ) =( e qB^ s^ -^12 q^2 t ) ( e q [^ N^ (0,1)] t^ - s ). o The exponential martingale can be used to generate many other martingales. Let 122 ( ; , ) (^0)! ( , ) x t^ n f q t x = e q^ - q^ = (^) å n ¥= nq Hn t x Where Hn is the n th^ Hermite polynomial, Hn ( , t x ) = f (^ n )(0; , t x ). Feeding this into the martingale property and exchanging summation and expectation, we can use the fact that this holds for any q including q = 0 , and conclude that for each n , { H (^) n^ ( , t B^ t^ )} is also a martingale. o We can apply the Optional Stopping Theorem to these martingales to get some interesting results. o Example : Define T = inf (^) { t : Bt = - a or b }. Using the mean martingale, we can find pb = a / ( a + b ). Using the variance martingale, we can find ( T )= ab.
o Example : Let Xt = mt - sBt , with m s , > 0 ; this could be seen as the “net demand up to time t ”, where sB (^) t is a production process. We are interested in T = inf (^) { t : Xt = b > (^0) } (the first time stock depletes). ( T ) = bm is easily found using the mean martingale on BT. Use the variance martingale for ar( T ) = sm^23 b o Example : Xt = mt + sBt , T = inf (^) { t : Xt Ï -( a b , )}. Use 12 2 122 BT T XT^ T T e e q - q q (^) s^ - m - q =. Choose q = - 2 m / s , and use the OST (stopped martingale is bounded); ( e sq^ XT ) = 1. Directly gives p (^) b and p (^) a. Use OST on BT = ( XT - mT ) / s to find ( T ). Can let m < 0 and a -¥ and find (^) ( sup (^) t Xt ³ b ) = e -^2 b^ m s /^2. o Example : Following from the above example and letting T = inf (^) { t : Xt = b }, suppose we want ( e - gT ). Write - g T = qB (^) T - 12 q^2 T - bb. Use the OST on exp ( qB (^) T (^) t - 12 q^2 [ T t ] ) £exp( qBT t )[bounded.] Substitute b = mT + sBT and equate coefficients of T and BT.
Ito’s Formula
o For a deterministic xt , d x (^) t = x d t and d ( f x (^) t ) = f ¢(^ x (^) t )d xt + 12 f ¢¢( x (^) t )(d xt )^2 + , but (d x (^) t )^2 = x ^2 (d ) t^2 which vanishes. In BM, (d Bt )^2 ~ d t , and so (^12)
0 0
d ( ) ( ) d ( ) d ( ) (0) ( ) d 1 ( ) d 2
t (^) t t t tt t s s s
f B f B B f B t f B f f B B f B s
= ¢^ + ¢¢
= + (^) ò ¢^ + ò ¢¢ This is Ito’s Formula. The first integral above is called Ito’s Integral and can be approximated as
0 d^1 i 1 i i 1
n t t
t ò Xs^ Bs^ »^ å i = X^ - éêë B^^ - Bt - ùúû This is a martingale transformation, and is therefore a martingale under boundedness and predictability ( left-continuity) of X (^) t. Furthermore, as a martingale, the mean of the integral is 0. To find its variance, consider
( 1 1 ) 1
(^2 2 2 ) 0 d^1^ i^ i^ i^^1 (^ i )(^1 )^0 d
t (^) n n t Xs Bs (^) i = Xt (^) - Bt Bt (^) - i = Xt (^) - ti ti (^) - Xs s æçç ö÷÷ (^) » é (^) - ù = - » æçç ö÷÷ (^) çè ò (^) ÷ø (^) å (^) êë úû å çè ò ÷ø Where we have used orthogonality of martingale differences, and conditioned on t (^) i (^) - 1. This is knows as Ito’s Isometry. More generally, for a bivariate f ( t , x )
