A few comments on a result of A. Novikov and Girsanov's theorem

TL;DR

The paper provides a simple proof linking a limit condition on a continuous local martingale to the expectation of an exponential involving the martingale and its quadratic variation, clarifying a result related to Girsanov's theorem.

Contribution

It offers a straightforward proof of a specific limit condition implying an exponential expectation equality for continuous local martingales.

Findings

Established a simple proof for the limit condition and exponential expectation equivalence.

Clarified the relationship between the limit infimum condition and Girsanov's theorem.

Contributed to the theoretical understanding of martingale exponential moments.

Abstract

We give a simple proof that for a continuous local martingale $M_{t}$ $$ \liminf_{\varepsilon\downarrow0}\varepsilon \log Ee^{(1-\varepsilon) \langle M\rangle_{\infty}/2}<\infty \Longrightarrow E\exp(M_{\infty}-\langle M\rangle_{\infty}/2)=1. $$

Equations

$$\underset{\epsilon \downarrow 0}{\underline{\lim }}\epsilon \log E{e}^{(1-\epsilon )⟨M{⟩}_{\infty }/2}<\infty ⟹E\exp (M_{\infty }-⟨M{⟩}_{\infty }/2)=1.$$

$$M=M_{\infty },\rho =\rho (M)=e^{M-⟨M⟩/2},{\rho }_t={\rho }_t(M)=e^{M_t-⟨M{⟩}_t/2}.$$

$$E{e}^{⟨M⟩/2}<\infty ⟹E\rho =1.$$

$$\underset{\epsilon \downarrow 0}{\underline{\lim }}\epsilon \log E{e}^{(1-\epsilon )⟨M⟩/2}<\infty ⟹E\rho =1.$$

$$E\rho \le 1;\exists \epsilon >0:E{e}^{(1+\epsilon )⟨M⟩/2}<\infty ⟹E\rho =1$$

$$E{e}^{(1+\epsilon {)}^2⟨(1-\epsilon )M⟩/2}=E{e}^{(1-{\epsilon }^2{)}^2⟨M⟩/2}<\infty ,$$

$$1=E\rho ((1-\epsilon )M)=E{e}^{(1-\epsilon )(M-⟨M⟩/2)}{e}^{(1-\epsilon )\epsilon ⟨M⟩/2}{I}_{⟨M⟩\le T}$$

$$+E{I}_{⟨M⟩>T}{e}^{(1-\epsilon )(M-⟨M⟩/2)}{e}^{(1-\epsilon )\epsilon ⟨M⟩/2}$$

$$\le (E\rho {)}^{1-\epsilon }(E{e}^{(1-\epsilon )⟨M⟩/2}{I}_{⟨M⟩\le T}{)}^{\epsilon }+(E\rho I_{⟨M⟩>T}{)}^{1-\epsilon }(E{e}^{(1-\epsilon )⟨M⟩/2}{)}^{\epsilon }.$$

$$1\le E\rho +\mathrm{const}E\rho I_{⟨M⟩>T},$$

$$\cos(\sqrt{2\lambda}\pi)E\exp(\lambda\tau)=1,\quad E\exp(\lambda\tau)=\big{[}\cos(\sqrt{2\lambda}\pi)\big{]}^{-1}$$

$$[Ee^{(1-\varepsilon)\langle M\rangle/2)}]^{\varepsilon}=[Ee^{(1-\varepsilon)\tau/8}]^{\varepsilon}=\big{[}\cos\frac{\sqrt{1-\varepsilon}}{2}\pi\big{]}^{-\varepsilon}\rightarrow 1,$$

$$E{e}^{⟨M⟩/2-c⟨M{⟩}^{1/2}}<\infty ⟹E\rho =1.$$

$$I:=E{e}^{⟨M⟩/2-c⟨M{⟩}^{1/2}}<\infty ⟹\underset{\epsilon \downarrow 0}{\underline{\lim }}\epsilon \log E{e}^{(1-\epsilon )⟨M⟩/2}\le \log I-c^2<\infty .$$

$$E{e}^{⟨M⟩/2-g(⟨M{⟩}^{1/2})}<\infty ⟹E\rho =1,$$

$$g(\tau )=\sqrt{2\tau \log \log \tau }$$

$$P\big{(}\int_{0}^{1}b^{2}(\xi_{\cdot},t)\,dt<\infty\big{)}=1.$$

$$Ef({\xi }_{\cdot })=Ef(w_{\cdot }^x)e^{\varphi },\varphi ={\int }_0^{1}b(w_{\cdot }^x,t)d{w}_t-(1/2){\int }_0^1{b}^2(w_{\cdot }^x,t)dt.$$

$$E\rho_{t\wedge\tau}^{p}\leq\bigg{(}E\rho_{t\wedge\tau}(prM)\bigg{)}^{1/r}\bigg{(}E\exp(\frac{rp-1}{2}p\frac{r}{r-1}\langle M\rangle_{t\wedge\tau})\bigg{)}^{(r-1)/r}.$$

Full text

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1. Main Result

Let $(\Omega,\mathcal{F},P)$ be a complete probability space and let $M_{t}$ be a continuous local martingale on $(\Omega,\mathcal{F},P)$, provided with an appropriate filtration of sub $\sigma$-fields of $\mathcal{F}$, such that $\langle M\rangle=\langle M\rangle_{\infty}<\infty$ (a.s.). Define

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The process $\rho_{t}$ is called an exponential martingale, although it is not necessarily a martingale. We will be discussing generalizations of the following celebrated result of A. Novikov (1972), which gives a sufficient condition for $\rho_{t}$ to be a martingale:

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This result is quite important in many applications related to absolute continuous change of probability measure and, in particular, makes available Girsanov’s theorem.

The original proof in [8] is based on knowing the distribution of the first exit time of the Wiener process with constant drift from a shifted positive half-axis. Some of other known proofs are even more involved (see, for instance, Section 8.1 in [1]). The latest easier proofs and the history revolving around Novikov’s condition are found and thoroughly discussed in [2] and [3]. Here we present a completely elementary proof of a result that is somewhat stronger than (1.1).

Theorem 1.1**.**

We have

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Proof. We start with two known facts (proved at the end of the paper):

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and show a solution of part of Problem 4.3.13 of [5] following the hint to that problem. Observe that for small enough $\varepsilon>0$

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which by (1.3) implies that $E\rho((1-\varepsilon)M)=1$.

Then use Hölder’s inequality and write that for small enough $\varepsilon>0$ and any constant $T\in(0,\infty)$

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As $\varepsilon\downarrow 0$, we get

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which gives $1\leq E\rho$ after letting $T\rightarrow\infty$. This together with the first relation in (1.3) implies our statement (1.2). The theorem is proved.

Assertion (1.2) is stronger than (1.1).

Example 1.2*.*

Take a one-dimensional Wiener process $w_{t}$ and let $\tau$ be the first exit time of $w_{t}$ from $(-\pi,\pi)$. Take $\lambda\in(0,1/8)$ and observe that $u_{\lambda}(x)=\cos(\sqrt{2\lambda}x)$ satisfies $(1/2)u^{\prime\prime}+\lambda u=0$ on $[-\pi,\pi]$. It follows by Itô’s formula that $m^{\lambda}_{t}:=u_{\lambda}(w_{t\wedge\tau})\exp(\lambda(t\wedge\tau))$ is a martingale and since $u_{\lambda}$ is bounded away from zero ($\lambda<1/8$), there is a $\varepsilon>0$ such that, for any $t<\infty$, $\varepsilon E\exp(\lambda(t\wedge\tau))\leq Em^{\lambda}_{t}=1$. As $t\to\infty$, we get $\varepsilon E\exp\lambda\tau\leq 1$, which by the dominated convergence theorem allows us to send $t\to\infty$ in $Em^{\lambda}_{t}=1$ and obtain that

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if $\lambda<1/8$ and then $E\exp(\lambda\tau)=\infty$ if $\lambda=1/8$. Therefore, for the martingale $M_{t}=w_{t\wedge\tau}/4$ one easily finds that as $\varepsilon\downarrow 0$

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so that the assumption in (1.2) is satisfied, whereas $E\exp(\langle M\rangle/2)=\infty$ and Novikov’s criterion is not applicable.

2. Refined Novikov’s conditions

In [9] Novikov (1979) relaxes the conditions in (1.1) and shows that, for any constant $c\geq 0$,

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This condition is applicable in Example 1.2, although it is not very easy to see that. We need to know the tail of the distribution of $\tau$. On the other hand, the elementary inequality: $c\langle M\rangle^{1/2}\leq(\varepsilon/2)\langle M\rangle+c^{2}\varepsilon^{-1}$ implies that

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In the same article [9] Novikov gives a more elaborated condition

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where $g$ belongs to the lower Kolmogorov class, for instance,

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for large $\tau$. This condition is, of course, much weaker than ours, but checking it, generally, requires much more knowledge about the distribution of $\langle M\rangle$.

On the one hand, (2.2) provides the most general condition in terms of the distribution of $\langle M\rangle$. But on the other hand, this distribution, generally, has little to do with the equality $E\rho=1$. Indeed, if $\tau$ is the first time the Wiener process $w_{t}$ hits point 1 and $M_{t}=w_{t\wedge\tau}$, then by Kazamaki’s criterion (see [4]) we have $E\rho=1$ and at the same time even $E\sqrt{\langle M\rangle}=E\sqrt{\tau}=\infty$ not to mention any exponential moments. By the way, everything said about (1.2) has its natural counterpart for Kazamaki’s criterion (see http://arxiv.org/abs/math/0207013).

The conditions described above usually are interesting not exclusively in their own rights but in connection with the problem of absolute continuity of the distribution of a stochastic process with respect to the Wiener measure. For instance, let $\xi_{t}$ be a (adapted) solution of a stochastic equation $d\xi_{t}=dw_{t}+b(\xi_{\cdot},t)\,dt$, $\xi_{0}=0$, with nonanticipating $b(\xi_{\cdot},t)$ and we are interested to know when its distribution $\mu_{\xi_{\cdot}}$ on $C(0,1)$ is absolutely continuous with respect to the distribution $\mu_{w_{\cdot}}$ on $C(0,1)$ of the Wiener process $w_{t}$. According to Theorem 6 of [6], $\mu_{\xi_{\cdot}}\ll\mu_{w_{\cdot}}$ iff

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Under this condition for any nonnegative measurable function $f(x_{\cdot})$ on $C(0,1)$

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In particular, $Ee^{\phi}=1$. Thus (2.3) implies that $Ee^{\phi}=1$. More general multidimensional equations are treated in [7].

In conclusion, for the sake of completeness we prove (1.3) following the proof of Lemma 3 in [6]. This shows that unlike [8] and [9] no specific information about the Wiener process is needed to get our main result.

That $E\rho\leq 1$ follows from the fact that $\rho_{t}$ is a local martingale (Itô’s formula) and the fact that $\rho_{t}\geq 0$, so that it is a supermartingale, has the limit as $t\to\infty$ and Fatou applies.

Next, for any $p,r>1$, stopping time $\tau$ such that $\rho_{t\wedge\tau}$ is a martingale, and $t\in[0,\infty)$, we easily obtain by the Hölder inequality

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By 4, the first factor on the right is at most $1$. The second factor can be made bounded uniformly with respect to $t$, since $\langle M\rangle_{t}\leq\langle M\rangle$ and for $p=1+\delta,\,r=1+\sqrt{\delta}$ and sufficiently small $\delta>0$, it is not difficult to see that the coefficient of $\langle M\rangle_{t\wedge\tau}$ can be made smaller than $(1+\varepsilon)/2$. Fix $\delta,p$, with these properties. Then by Doob’s inequality $E\sup_{t\leq\tau}\rho_{t}^{p}\leq N$, where the constant $N$ is independent of $\tau$. This yields that $E\sup_{t<\infty}\rho_{t}^{p}\leq N$, the local martingale $\rho_{t}$ is bounded by a summable function independent of $t$ and $E\rho=\lim_{n\to\infty}E\rho_{\tau_{n}}=1$, where $\tau_{n}\to\infty$ is any localizing sequence for $\rho_{t}$.

Acknowledgment. The author is sincerely grateful to the referee for the comments and suggestions which certainly helped improve the presentation.