Generalization of Doob's Inequality and A Tighter Estimate on Look-back Option Price
aa r X i v : . [ q -f i n . M F ] J u l GENERALIZATION OF DOOB’S INEQUALITY AND ATIGHTER ESTIMATE ON LOOKBACK OPTIONPRICE
JIAN SUNFUDAN UNIVERSITY
Abstract.
In this short note, we will strengthen the classic Doob’s L p inequality for sub-martingale processes. Because this inequalityis of fundamental importance to the theory of stochastic process,we believe this generalization will find many interesting applica-tions. Introduction
Doob’s maximum inequality for the sub-martingale process has playedan important role in the stochastic process theory. It has become astandard result which appears in almost every introductory text inthis subject. Let { X t } be be a process defined on a probability spacewith a filtration F t . Its maximum is defined by M t = max ≤ s ≤ t X s By the definition we have M = X . If X t is a positive continuoussub-martingale, the Doob’s L p inequality states that(1) k M T k p ≤ pp − k X T k p . In particular, when p = 2, we have(2) k M T k ≤ k X T k . Even though the coefficient on the right hand side is not important forthe purpose of establishing the finiteness of the L integrability of M T , The author would like to thank Professor Steven Shreve at Carnegie MellonUniversity and Professor Jiangang Ying at Fudan University. The author also wouldlike to thank Peter Carr and Bruno Dupire for their inspirational discussions. it may become important for some other applications. For example,when X T is a martingale with X = 0, we infer from (2)(3) E ( M T ) ≤ E ( M T ) ≤ E ( X T ) . This provides an estimate on the expectation of the Maximum M T interms of the standard deviation of X T . In fact, we will see later thatwe can have a much tighter estimate(4) E ( M T ) ≤ E ( X T ) , for any continuous martingale with X = 0. When X = 0 we will have(5) E ( M T ) ≤ √ E ( X T ) . In either case, we obtain a better result. In Finance, people usuallyuse martingales to model the stock prices or any other tradable assets.Their maximum M T sometimes represent payoff of certain derivatives.The inequality above actually gives a good estimate of this derivativepayoff in terms of European type option prices. In this area, the mag-nitude of the coefficient matters a lot to the applications.In general when p >
2, Doob’s inequality is equivalent to(6) E ( M pT ) ≤ (cid:18) pp − (cid:19) p E ( X pT ) , and we are going to strengthen this inequality to(7) E ( M pT ) + pp − X p ≤ (cid:18) pp − (cid:19) p E ( X pT )by adding a term of initial position X .Our method is to first prove an identity which will involve X t and M t . From this identity we will use the standard methods to derive ourinequalities. Because our starting point is an identity rather than anestimates, we could prove a tighter inequality. Our methodology alsoprovides a totally different proof of Doob’s maximum inequality.2. Classic Results
For completeness and comparison, we state and prove the classicmaximal inequality in this section.
Theorem 1 (Doob’s Inequality ) . Let X t be an nonnegative martingaleprocess. For any real a > we have (8) aP ( M T ≥ a ) ≤ E ( X T M T ≥ a ) ENERALIZATION OF DOOB’S INEQUALITY AND A TIGHTER ESTIMATE ON LOOKBACK OPTION PRICE3
Proof.
We define the stopping time(9) τ = inf { t : X t ≥ a } . and we claim to have(10) 1 τ ≤ T + X T − X T ∧ τ a ≤ X T M T ≥ a a . It is obvious to check its validity. Take the expectation and use thefact that X T ∧ τ is also a martingale, we get the result. (cid:3) Theorem 2 (Maximal inequality) . For the nonnegative martingaleprocess, We have the following L p norm inequality: for any p > , (11) k M T k p ≤ pp − k X T k p Classical proof.
This is based on the Doob’s inequality. Use the stan-dard measure theory and the Doob’s inequality, E ( M pT ) = Z ∞ p x p − P ( M T > x ) dx ≤ Z ∞ p x p − E ( X T M T ≥ x ) dx = E (cid:18)Z ∞ p x p − X T M T ≥ x dx (cid:19) = E (cid:18)Z M T p x p − X T dx (cid:19) = pp − E (cid:0) M p − T X T (cid:1) ≤ pp − E ( M pT ) ( p − /p E ( X pT ) /p Further simplify this inequality we will get the result. Please note thatwe have used H¨older inequality(12) E ( M p − T X T ) ≤ E ( M pT ) p − p E ( X pT ) p in this proof. (cid:3) A New Inequality
We will try to strengthen the maximal inequality proved in the pre-vious section. First we prove an identity.
JIAN SUN FUDAN UNIVERSITY
Theorem 3.
Let X t be a nonnegative continuous martingale. For any p > , if (13) Z T M pt d [ X, X ] t < ∞ , we have the following identity: (14) E ( X T M pT ) = pp + 1 E ( M p +1 T ) + 1 p + 1 X p +10 . Proof.
We consider the following differential identity:(15) d ( X t M pt ) = M pt dX t + pM p − t X t dM t . Written in the integral term and make use the observation dM t = 0 ⇒ X t = M t , we have(16) X T M pT − X p +10 = Z T M pt dX t + Z T pM pt dM t . Take the expectation and use the fact that X t is a martingale, therefore(17) E (cid:18)Z T M pt dX t (cid:19) = 0 , we have E ( X T M pT ) = pp + 1 E ( M p +1 T ) + 1 p + 1 X p +10 . This finishes the proof. (cid:3)
Theorem 4.
Let X t be a continuous martingale, if (18) Z T M t d [ X, X ] t < ∞ , we have the following identity: (19) E ( X T M T ) = 12 E ( M T ) + 12 X . Proof.
The proof is the same as above since when p = 1, we don’trequire X t to be positive anymore. (cid:3) Theorem 5.
Let X t be a nonnegative continuous sub-martingale. Forany p > , if (20) Z T M pt d [ X, X ] t < ∞ , ENERALIZATION OF DOOB’S INEQUALITY AND A TIGHTER ESTIMATE ON LOOKBACK OPTION PRICE5 we have the following inequality: (21) E ( X T M pT ) ≥ pp + 1 E ( M p +1 T ) + 1 p + 1 X p +10 . Proof.
We basically follow the proof in Theorem (3). We notice thatwhen X t is a sub-martingale,(22) Z T M pt dX t is also a sub-martingale so(23) E (cid:18)Z T M pt dX t (cid:19) ≥ (cid:3) Theorem 6 (Generalization of Doob’s maximal inequality) . For anonnegative continuous sub-martingale process, if (24) Z T M pt d [ X, X ] t < ∞ , we then have (25) E ( M p +1 T ) + p + 1 p X p +10 ≤ (cid:18) p + 1 p (cid:19) p +1 E ( X p +1 T ) for any p > .Proof. We use the H¨older inequality:(26) E ( M pT X T ) ≤ E ( X p +1 T ) p +1 E ( M p +1 T ) pp +1 . For any 0 < ε <
1, we can write(27) E ( M pT X T ) ≤ (cid:0) ε − p E ( X p +1 T ) (cid:1) p +1 (cid:0) εE ( M p +1 T ) (cid:1) pp +1 . Again, we use the H¨older inequality(28) a p +1 b pp +1 ≤ p + 1 a + pp + 1 b to get(29) E ( M pT X T ) ≤ ε − p p + 1 E ( X p +1 T ) + p εp + 1 E ( M p +1 T ) . Now use the inequality (21), pp + 1 E ( M p +1 T ) + 1 p + 1 X p +10 ≤ ε − p p + 1 E ( X p +1 T ) + p εp + 1 E ( M p +1 T ) . JIAN SUN FUDAN UNIVERSITY
Rearranging the terms,(30) E ( M p +1 T ) + 1 p (1 − ε ) X p +10 ≤ p + 1 p ε p (1 − ε )( p + 1) E ( X p +1 T ) . Now we minimize the function(31) min <ε< ε p (1 − ε )( p + 1) = (cid:18) p + 1 p (cid:19) p , the equality takes place when ε = p/ ( p + 1). Put everything back into(30), we get E ( M p +1 T ) + p + 1 p X p +10 ≤ (cid:18) p + 1 p (cid:19) p +1 E ( X p +1 T ) . (cid:3) Some Implications
Proposition 1.
For X T is a continuous Martingale, then we have (32) E ( M T ) + 2 X ≤ E ( X T ) . Proof.
For the martingale X t , by Theorem 4, E ( M T ) + X = 2 E ( X T M T ) ≤ E ( M T ) + 2 E ( X T )and arranging terms will prove the inequality. (cid:3) It is interesting to compare with the the classical result which onlygives(33) E ( M T ) ≤ E ( X T ) . If we use again H¨older inequality, we will get(34) E ( M T ) ≤ E ( M T ) ≤ E ( X T ) . and consequently, we have(35) E ( M T ) ≤ E ( X T ) . We can in fact get stronger result by using the Identity 14.
Proposition 2.
For continuous martingale X T , we have (36) E ( M T ) ≤ √ E ( X T ) . ENERALIZATION OF DOOB’S INEQUALITY AND A TIGHTER ESTIMATE ON LOOKBACK OPTION PRICE7
Proof.
Let p = 1 in the identity 14, we have E ( M T ) + X = 2 E ( X T M T )which is equivalent to(37) E (( M T − X T ) ) = E ( X T ) − X . Use H¨older inequality, we have(38) E ( M T ) − E ( X T ) ≤ q E ( X T ) − X , hence,(39) E ( M T ) ≤ X + q E ( X T ) − X . Now use the inequality ( a + b ) ≤ a + b ), we have(40) E ( M T ) ≤ E ( X T ) . which is what we want. (cid:3) Proposition 3.
When X T is a martingale and X = 0 , then we have (41) E ( M T ) ≤ E ( X T ) . Proof.
In the Inequality 38, take X = 0. It is evident. (cid:3) References [1] Ioannis Karatzas, Steven Shreve
Brownian Motion and Stochastic Calculus ,Springer Verlag, 1991.[2] Steven Shreve
Stochastic Calculus Models for Finance
Springer Verlag, 2002.[3] Daniel Revuz, Marc Yor