Levy exponential local martingale

We will review a sufficient condition for Levy stochastic exponential to be a local martingale.

Let Levy-type stochastic integral {Y} be given by

\displaystyle  d Y(t) = G(t) dt + F(t) dW(t) + H(t,x) \tilde N(dt, dx) + K(t,x) N(dt, dx). \ \ \ \ \ (1)

Our first goal is to find necessary and sufficient conditions for a Levy-type stochastic integral {Y} to be a martingale. First, we impose some conditions on {K} and {G} with the notion of {\mathcal{P}_2(T,E)} and {\mathcal{H}_2(T)} (see definitions here).

  1. (LM1) {K\in \mathcal{P}_2(t, B_1^c(0))} for each {t>0}.
  2. (LM2) {\mathbb{E} [\int_0^t \int_{|x|\ge 1} |K(s,x)| \nu(dx) ds ] <\infty}.
  3. (LM3) {G^{1/2} \in \mathcal{H}_2(t)} for each {t>0}.

It follows from (LM2) that {\int_0^t \int_{|x|\ge 1} |K(s,x)| \nu(dx) ds <\infty} a.s. and we can write

\displaystyle  \int_0^t \int_{|x|\ge 1} K(s,x) N(ds, dx) = \int_0^t \int_{|x|\ge 1} K(s,x) \tilde N(ds, dx) + \int_0^t \int_{|x|\ge 1} K(s,x) \nu(dx) ds,

for each {t\ge 0} with the compensated integral being a local martingale.

Theorem 1 If {Y} is a Levy-type stochastic integral of the form (1) and the assumptions (LM1)–(LM3) are satisfied, then {Y} is a local martingale if and only if

\displaystyle  G(t) + \int_{|x|\ge 1} K(t,x) \nu(dx) = 0 \hbox{ a.s.}

for Lebesgue almost all {t\ge 0}.

Directly applying Ito’s formula on {e^{Y(t)}}, we have

Corollary 2 (Exponential martingale) With the same condition as Theorem 1, {e^Y} is a local martingale if and only if

\displaystyle  G(s) + \frac 1 2 F^2(s) + \int_{|x|<1} (e^{H(s,x)} -1 - H(s,x)) \nu(dx) + \int_{|x|\ge 1} (e^{K(s,x)} -1) \nu(dx) = 0,

almost surely and for almost all {s\ge 0}.

So, {e^Y} is a local martingale if and only if for {t\ge 0},

\displaystyle  \begin{array}{ll} e^{Y(t)} & = 1 + \int_0^t e^{Y(s-)} F(s) d W(s) + \int_0^t \int_{|x|<1} e^{Y(s-)} (e^{H(s,x)} -1) \tilde N(ds, dx) \\ & \hspace{2in} + \int_0^t \int_{|x|\ge 1} e^{Y(s-)} (e^{K(s,x)} -1) \tilde N(ds, dx). \end{array} \ \ \ \ \ (2)

The process {e^Y} given by equation (2) is called an exponential local martingale. Two important examples are:

  1. The Brownian case: If {Y} is a Brownian integral of

    \displaystyle  Y(t) = \int_0^t F(s) dW(s) + \int_0^t G(s) ds,

    where {G(t) = - \frac 1 2 F^2(s)}. Then, {e^Y} is exponential local martingale. A sufficient condition to be a martingale is the well-known Novikov condiion, i.e.

    \displaystyle \mathbb{E} \Big[ \exp\Big\{\int_0^\infty \frac 1 2 F^2(s) ds \Big\}\Big] <\infty.

  2. The Poisson case: If {Y} is a Poisson integral driven by a Poisson process {N} of intensity {\lambda} and has the form of

    \displaystyle  Y(t) = \int_0^t K(s) d N(s) + \int_0^t G(s) ds,

    If {G(t) = -\lambda (e^{K(t)} -1) } almost surely in {t}, then we have {e^Y} as exponential local martingale.

A general exponential local martingale of the form {\exp \{M(t) - \frac 1 2 \langle M, M\rangle(t)\}, t\ge 0)} to be a martingale, some additional conditions are needed, see [RY99], pp307–309. In particualr, One may expect exponential martingale in Theorem 1 under following (M1)-(M2) in place of (LM1)-(LM3),

  1. (M1) {\mathbb{E} [\int_0^t \int_{|x|\ge 1} \max \{ |K(s,x)|, |K(s,x)|^2 \} \nu(dx) ds ] <\infty}.
  2. (M2) {\int_0^t \mathbb{E} [|G(s)| ] ds < \infty} for each {t>0}.


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