Martingale Property of Ito Integral & Girsanov Thm

Advanced Stochastic Processes. LECTURE 15 Martingale property of Ito integral and Girsanov theorem Lecture outline • Continuity of Ito integral • Martingale property of Ito integral. Martingale representation theorem. • Girsanov’s theorem. 15.1. Continuity and martingale properties of Ito process �t �t Ito process is a process of the form X(t) = X(0) + 0 U (s)ds + 0 V (s)dB(s). The first part of �t this expression, 0 U (s)ds is a continuous function of t (recall from real analysis that Reimann integral is continuous in t). What about the Ito integral part? It turns out that there always exists a continuous version of the Ito integral. Theorem 15.1. Given� V ∈ H 2 and T > 0 there exists an a.s. continuous stochastic process t Y (t) such that Y (t) = 0 V (s)dB(s) a.s. for all 0 ≤ t ≤ T . Namely, there exists a continuous �t version of 0 V (s)dB(s) on every finite interval. �t Note that the theorem does not imply that every version of 0 V (s)dB(s) is continuous. Can you think about two process which are a.s. equal to each other but one is a.s. continuous while the other is discontinuous everywhere? �t Proof. Recall that 0 V (s)dB(s) is the L2 limit along any sequence of simple processes Vn satisfying � t (15.2) E[ (Vn (t) − V (t))2 dt] → 0. 0 Fix any such sequence Vn and consider their corresponding Ito integral It (Xn ). �t Lemma 15.3. It (Vn ) = 0 Vn (t)dB(t) is a martingale. 1 2 Proof. Let Πn : 0 = t0 < t1 < · · · < tn = t be the partition corresponding to Vn . Fix s < t. Increase the partition by an extra point tk = s. Observe that It (Xn ) with respect to the new partition is the same as the original one. Also Fs = Ftk . We have � E[It (Vn )|Fs ] = E[ Vn (tj )(B(tj+1 ) − B(tj ))|Fs ] j = E[ � Vn (tj )(B(tj+1 ) − B(tj ))|Fs ] + E[ j≥k � Vn (tj )(B(tj+1 ) − B(tj ))|Fs ] j≤k−1 � � = E[ Vn (tj )E[B(tj+1 ) − B(tj )|Ftj ]|Fs ] + Vn (tj )(B(tj+1 ) − B(tj )) j≥k j≤k−1 = 0 + Is (Vn ), where the last equality holds since indeed a martingale. � j≤k−1 Vn (tj )(B(tj+1 ) − B(tj )) ∈ Fs . Therefore It (Vn ) is � We fix � > 0. From (15.2) it follows that there exists n0 such that for all m, n ≥ n0 � T E[ (Vn (t) − Vm (t))2 dt < �3 . 0 Fix any such pair m, n. From Lemma 15.3 we have It (Vn ) − It (Vm ) = It (Vn − Vm ) is also martingale. Applying the Doob­Kolmogorov inequality, we obtain E[(IT (Vn ) − IT (Vm ))2 ] P( sup |It (Vn ) − It (Vm )| ≥ �) ≤ �2 0≤t≤T E[(IT (Vn − Vm ))2 ] �2 �T E[ 0 (Vn (t) − Vm (t))2 dt] = �2 < �. = This means that we can construct a subsequence nk along which 1 1 P( sup |It (Vnk+1 ) − It (Vnk )| ≥ k ) ≤ k . 2 2 0≤t≤T In particular, the sum of these probabilities is finite. Applying the Borel­Cantelli Lemma, for almost all samples ω, there exists k(ω) such that for all k > k(ω) 1 sup |It (Vnk+1 (ω) ) − It (Vnk (ω))| < k . 2 0≤t≤T Now we need a short digression into real analysis. We define a sequence of functions ak (t), t ∈ [0, T ] to be uniformly Cauchy if for every � > 0 there exists k0 such that for all k, m > k0 we have sup0≤t≤T |ak (t) − am (t)| < �. The following is the result from real analysis. Proposition 1. Suppose ak is a uniform Cauchy sequence of continuous functions on [0, T ]. Then ak (t) converges everywhere on [0, T ] to some continuous function a(t). Applying this proposition to our case we conclude that for almost all ω the point­wise limit of It (Vnk (ω)) exists and is continuous on [0, T ]. We call this limit Y (t, ω) and now we show that Y (t) = It (V ) a.s. From the definition of Ito integral, we have that the subsequence It (Vnk ) LECTURE 15. ITO PROCESSES 3 converges to It (V ) in L2 . This implies convergence in probability. On the other hand a.s. convergence also implies convergence in probability. Therefore Yt and It (V ) are two limits in probability of It (Vnk ). Therefore (check that this is indeed the case) they are a.s. equal to each other. � We have created a version Yt of It (V ) which is a.s. continuous. From now on assume that we are dealing only with continuous versions of It (V ). Another important property of It (V ) is that it is a martingale. Theorem 15.4. For every V ∈ H 2 , It (V ) is a martingale. Proof. We first need to check E[|I�t (V )|] < ∞. This will be true, provided that E[It2 (V )] < ∞. t But by Ito isometry E[It2 (V )] = E[ 0 V 2 (s)ds] < ∞ since V ∈ H 2 . Fix s < t. Take any sequence of simple processes Vn satisfying (15.2). Let Ys = E[It (V )|Fs ]. Consider E[(Ys − Is (V ))2 ] = E[(Ys − Is (Vn ) + Is (Vn ) − Is (V ))2 ] ≤ 2E[(Ys − Is (Vn ))2 ] + 2E[(Is (Vn ) − Is (V ))2 ] By definition, we have limn E[(Is (Vn ) − Is (V ))2 ] = 0. On the other hand, since by Lemma 15.3 we have Is (Vn ) = E[It (Vn )|Fs ], then E[(Ys − Is (Vn ))2 ] = E[(E[It (V )|Fs ] − E[It (Vn )|Fs ])2 ] = E[(E[It (V ) − It (Vn )|Fs ])2 ] ≤ E[E[(It (V ) − It (Vn ))2 |Fs ]] = E[(It (V ) − It (Vn ))2 ], where we use conditional Jensen’s inequality in the second step. But again limn E[(It (V ) − It (Vn ))2 ] = 0 by definition. Combining, we conclude that E[(Ys − Is (V ))2 ] = 0 or Ys = E[It (V )|Fs ] = Is (V ), or It (V ) is indeed a martingale. � 15.2. Martingale Representation Theorem and Girsanov’ theorem We established in the previous section that Ito integral is a martingale. It turns out a converse statement is true. The proof of this fact is complicated, it is based on some complex analytic techniques and we skip it. Theorem 15.5 (Martingale Representation Theorem). Suppose M (t) ∈ Ft is a martingale with a.s. continuous sample paths. Then there exists a unique process X(t) ∈ H 2 such that � t M (t) = E[M (0)] + X(s)dB(s). 0 The martingale representation theorem is based on a related Dudley’s Theorem and is an important tool in finance. Another related tool is Girsanov’s theorem, which we now discuss. 4 First let us review the change of measure technique. Given a probability measure P and a non­negative random variable ψ such that E[ψ] < ∞, we may define a new probability measure E[ψ1{A}] . E[ψ] The following identity was established during the recitation. Given two fields F ⊃ G and a random variable X ∈ F such that EP [|X |], EP2 [|X |] < ∞ P2 (A) = EP [ψX|G] EP [ψ|G] Girsanov’s theorem is an important statement which provides a translation between different probability measures on the same space (Ω, F ) and filtration {Ft }. (15.6) EP2 [X|G] = �t �t Thus suppose we have an Ito process X(t) = X(0) + 0 U (s)ds + 0 V (s)dB(s). Recall that both X(t), B(t) ∈ Ft and correspond to the same probability space (Ω, F , P). We will derive Girsanov’s theorem first for the case V (s) = 1 and � t then state without proof the general case. For now, consider the process X(t) = X(0) + 0 U (s)ds + dB(s) or dX = U dt + dB. Consider the following process � t � 1 t 2 M (t) = exp(− U (s)dB(s) − U (s)ds). 2 0 0 This process is adapted to Ft , but does it belong to H 2 ? Not necessarily. So we only consider the case when it does: Definition 15.7. A process U is defined to satisfy Novikov’s condition if for every t 1 Rt E[e 2 0 U 2 (s)ds ] < ∞. It turns out (we do not prove this) that this suffices to ensure M ∈ H 2 . There is a reason we used the mnemonic M for the process – it gives rise to a martingale. Proposition 2. Consider an Ito process dX = U dt + dB. The processes M (t) and Y (t) = M (t)X(t) are martingales. Proof. It turns out (we skip the proof) that the Novikov condition implies E[M (t)] < ∞. This implies E[Y (t)] < ∞. By Ito’s lemma Y (t) is an Ito process since M, X are Ito processes. Using the Ito formula we have 1 1 dM = M (−U dB − U 2 dt) + M U 2 dt = −M U dB. 2 2 Since Ito integral is a martingale, M (t) is a martingale. We use the multidimensional Ito formula applied to function g(x, y) = xy: dY = XdM + M dX + (dM dX). Therefore dY = X(−M U dB) + M (U dt + dB) + (−M U dB)(U dt + dB) = (−XM U + M )dB where we use (−M U dB)(U dt+dB) = −M U U dBdt−M U (dB)2 = −M U dt and this cancels with M U dt. We conclude that Y is an Ito integral (it does not have a non­Brownian component). We have established earlier that Ito integral is a martingale. � LECTURE 15. ITO PROCESSES 5 Now let us consider a change of measure implied by the transformation X → M X meaning the following. First fix T > 0. Consider the random variable M (T ) and a new measure PT = M (T )P introduced by M (T ): for every A ⊂ Ω, A ∈ F we let PT (A) = E[M (T )1{A}] = E[M (T )1{A}], E[M (T )] where the second equality follows since M is a martingale and therefore E[M (T )] = E[M (0)] = 1. This change of measure is called Girsanov’s transformation. Intuitively every sample ω ∈ Ω is reweighted from P(ω) to P(ω)MT (ω). We know that this transformation does introduce a new probability measure. The distribution of X(T ) under the new measure PT is then the distribution of X(T )M (T ) under the old measure P. This begs the question: how dependent is this transformation on the choice of T ? What if we chose a different T ? The wonderful thing about this transformation is that it ”works across” times as the following lemma establishes. Lemma 15.8. For every t ≤ T , the measures Pt = M (t)P and PT = E[M (T )]P are identical on Ft . In particular the distributions of X(t)M (t) and X(t)M (T ) are identical. Proof. Fix A ∈ Ft . We have using the martingale property of M from (2) PT (A) = E[M (T )1{A}] = E[E[M (T )1{A}|Ft ]] = E[1{A}E[M (T )|Ft ]] = E[1{A}M (t)] = Pt (A). � We turn to the first version of Girsanov’s theorem. It basically states that the Girsanov change of measure ”creates” a Brownian motion out of the process X: Theorem 15.9 (Girsanov’s Theorem I). For every T > 0, the process X(t), 0 ≤ t ≤ T is a Brownian motion with respect to the probability measure PT . Proof. The proof relies the the following so called Levi characterization of a Brownian motion Proposition 3. A continuous process B(t) is a standard Brownian motion if and only if it is a martingale and B 2 (t) − t is a martingale. We use this�characterization in order to prove Girsanov’s theorem. The non­Brownian part of t X(t), which is 0 U (s)ds is a Riemann integral and is therefore a continuous function of t (a fact from real analysis). The Brownian part is an Ito integral. Earlier we showed that Ito integral is a continuous process (always has a continuous representation). Thus X(t) is continuous. Now we show that X(t) is a martingale w.r.t. PT . Fix s < t. Using (15.6) EPT [X(t)|Fs ] = M (s)X(s) EP [M (t)X(t)|Fs ] = = X(s) M (s) EP [M (t)|Fs ] where we have used Proposition (2) and Lemma (15.8). It remains to show that X 2 (t) − t is a martingale w.r.t. PT . We use the Ito formula and the computations of Proposition 2 to compute d(M X 2 ): 1 1 d(M X 2 ) = X 2 dM + 2M XdX + 2X(dM )(dX) + 2M (dX)2 2 2 2 = X (−M U dB) + 2M XU dt + 2M XdB − 2XM U dt + M dt = X 2 (−M U dB) + 2M XdB + M dt 6 On the other hand again using Ito formula d(M t) = tdM + M dt = −tU M dB + M dt Therefore d(M (X 2 − t)) = d(M X 2 − M t) = X 2 (−M U dB) + 2M XdB − tU M dB Since this expression is an Ito integral, the process M (X 2 (t) − t) is a martingale. Again using (15.6) we show that under PT , X 2 (t) − t is martingale. We showed that Levi condition for X is satisfied. Thus X is a Brownian motion w.r.t. PT . � Let us turn to a more general version of Girsanov’s Theorem. It corresponds to the case when the multiplicative component of the Brownian part in X is non­unit. We will state it without the proof. The proof details are similar. Theorem 15.10 (Girsanov’s Theorem II). Consider an Ito process dX = U dt + V dB. Suppose there exists a process η ∈ H 2 satisfying η(t)V (t) = U (t) a.s. and suppose the Novikov condition holds for η. Let � t � 1 t 2 M (t) = exp(− η(s)dB(s) − η (s)ds). 2 0 0 � ˆ = t η(s)ds + B(t) is a Fix T > 0 and consider the change of measures PT = η(T )P. Then B(t) 0 Brownian motion w.r.t. PT and dX(t) = V (t)dB̂(t). ˆ and we recover the first version of In the special case V = 1 we obtain η = U and X = B Girsanov’s theorem. 15.3. Additional reading materials • Øksendal [1], Chapters III,IV, VIII