Lévy’s Characterization of Brownian Motion

Standard Brownian motion, {\{B_t\}_{t\ge 0}}, is defined to be a real-valued process satisfying the following properties.

  1. {B_0=0}.
  2. {B_t-B_s} is normally distributed with mean 0 and variance ts independently of {\{B_u\colon u\le s\}}, for any {t>s\ge 0}.
  3. B has continuous sample paths.

As always, it only really matters is that these properties hold almost surely. Now, to apply the techniques of stochastic calculus, it is assumed that there is an underlying filtered probability space {(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},{\mathbb P})}, which necessitates a further definition; a process B is a Brownian motion on a filtered probability space {(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},{\mathbb P})} if in addition to the above properties it is also adapted, so that {B_t} is {\mathcal{F}_t}-measurable, and {B_t-B_s} is independent of {\mathcal{F}_s} for each {t>s\ge 0}. Note that the above condition that {B_t-B_s} is independent of {\{B_u\colon u\le s\}} is not explicitly required, as it also follows from the independence from {\mathcal{F}_s}. According to these definitions, a process is a Brownian motion if and only if it is a Brownian motion with respect to its natural filtration.

The property that {B_t-B_s} has zero mean independently of {\mathcal{F}_s} means that Brownian motion is a martingale. Furthermore, we previously calculated its quadratic variation as {[B]_t=t}. An incredibly useful result is that the converse statement holds. That is, Brownian motion is the only local martingale with this quadratic variation. This is known as Lévy’s characterization, and shows that Brownian motion is a particularly general stochastic process, justifying its ubiquitous influence on the study of continuous-time stochastic processes.

Theorem 1 (Lévy’s Characterization of Brownian Motion) Let X be a local martingale with {X_0=0}. Then, the following are equivalent.

  1. X is standard Brownian motion on the underlying filtered probability space.
  2. X is continuous and {X^2_t-t} is a local martingale.
  3. X has quadratic variation {[X]_t=t}.

This result carries directly through to the d-dimensional situation. A d-dimensional Brownian motion {X=(X^1,\ldots,X^d)} on a filtered probability space {(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},{\mathbb P})} is a continuous adapted process with {X_0=0} such that, for any {t>s\ge 0}, {X_t-X_s} is independent of {\mathcal{F}_s} and multivariate normal with zero mean and covariance matrix {(t-s)I_d}.

Theorem 2 Let {X=(X^1,\ldots,X^d)} be a d-dimensional local martingale with {X_0=0}. Then, the following are equivalent.

  1. X is a Brownian motion on the underlying filtered probability space.
  2. X is continuous and {X^i_tX^j_t-\delta_{ij}t} is a local martingale for {1\le i,j\le d}.
  3. X has quadratic covariations {[X^i,X^j]_t=\delta_{ij}t} for {1\le i,j\le d}.

Here, {\delta_{ij}\equiv 1_{\{i=j\}}} is the Kronecker delta. Theorems 1 and 2 are both special cases of the following characterization of multidimensional centered Gaussian processes with independent increments. Note that, as processes are continuous whenever their quadratic variation is continuous, statement 3 of Theorems 1 and 2 automatically imply continuity of X.

Theorem 3 Let {X=(X^1,\ldots,X^d)} be a d-dimensional continuous local martingale with {X_0=0}. Also, let {\{\Sigma_t\}_{t\ge 0}} be symmetric {d\times d} real matrices such that {\Sigma_0=0} and {t\mapsto a^{\rm T}\Sigma_t a} is continuous and increasing for all {a\in{\mathbb R}^d}. Then, the following are equivalent.

  1. {X_t-X_s} is independent of {\mathcal{F}_s} and normally distributed with mean zero and covariance matrix {\Sigma_t-\Sigma_s}, for all {t>s\ge 0}.
  2. {X^i_tX^j_t-\Sigma^{ij}_t} is a local martingale for {1\le i,j\le d}.
  3. X has quadratic covariations {[X^i,X^j]_t=\Sigma^{ij}_t}.

Proof: Suppose that 1 holds and set {Y=a^{\rm T}X}, {Z=b^{\rm T}Y} for some {a,b\in{\mathbb R}^d}. Then, {Y_t-Y_s}, {Z_t-Z_s} are joint normal with zero mean and covariance {a^{\rm T}(\Sigma_t-\Sigma_s)b}, independently of {\mathcal{F}_s}. So

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rcl} \displaystyle{\mathbb E}\left[Y_tZ_t-Y_sZ_s\mid\mathcal{F}_s\right]&\displaystyle=&\displaystyle{\mathbb E}\left[(Y_t-Y_s)(Z_t-Z_s)\mid\mathcal{F}_s\right]\smallskip\\ &&\displaystyle+ {\mathbb E}\left[Y_s(Z_t-Z_s)+Z_s(Y_t-Y_s)\mid\mathcal{F}_s\right]\smallskip\\ &\displaystyle=&\displaystyle a^{\rm T}(\Sigma_t-\Sigma_s)b. \end{array}

Setting {M_t=Y_tZ_t-a^{\rm T}\Sigma_t b} gives {{\mathbb E}[M_t-M_s\mid\mathcal{F}_s]=0}, so M is a martingale and 2 holds.

Now suppose that 2 holds, choose any {a\in{\mathbb R}^d}, and set {Y=a^{\rm T}X}. Then, {Y^2_t-a^{\rm T}\Sigma_t a} is a local martingale, as is {Y^2-[Y]}. It follows that {V_t=[Y]_t-a^{\rm T}\Sigma_t a} is also a local martingale and, being the difference of continuous increasing processes, V is also an FV process. So, V is constant, giving {[Y]_t=a^{\rm T}\Sigma_t a}. By the polarization identity,

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle[a^{\rm T}X, b^{\rm T}X]_t&\displaystyle=([(a+b)^{\rm T}X]_t-[(a-b)^{\rm T}X]_t)/4\smallskip\\ &\displaystyle= \left((a+b)^{\rm T}\Sigma_t(a+b)-(a-b)^{\rm T}\Sigma_t(a-b)\right)/4\smallskip\\ &=a^{\rm T}\Sigma_t b \end{array}

giving property 3.

Finally, suppose that 3 holds, choose {a\in{\mathbb R}^d} and set {Y=a^{\rm T}X}, so that {[Y]_t=a^{\rm T}\Sigma_t a}. The process

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle M_t&\displaystyle=f(Y_t,[Y]_t)\equiv\exp\left(iY_t+\frac{1}{2}[Y]_t\right)\smallskip\\ &\displaystyle= \exp\left(ia^{\rm T}X_t+ \frac{1}{2}a^{\rm T}\Sigma_t a\right) \end{array}

is bounded by {\vert M_t\vert\le\exp(a^{\rm T}\Sigma_t a/2)}. Applying Ito’s lemma for continuous semimartingales to f gives

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle dM_t&\displaystyle=f_1(Y_t,[Y]_t)\,dY_t+f_2(Y_t,[Y]_t)\,d[Y]_t+\frac{1}{2}f_{11}(Y_t,[Y]_t)\,d[Y]_t\smallskip\\ &\displaystyle= iM_t\,dY_t. \end{array}

By preservation of the local martingale property, M is a local martingale. As it is also bounded over finite intervals, M is a proper martingale. So,

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}[\exp(ia^{\rm T}(X_t-X_s))\mid\mathcal{F}_s]&\displaystyle={\mathbb E}[M_t\exp(-ia^{\rm T}X_s-a^{\rm T}\Sigma_t a/2)\mid\mathcal{F}_s]\smallskip\\ &\displaystyle=M_s\exp(-ia^{\rm T}X_s-a^{\rm T}\Sigma_t a/2)\smallskip\\ &\displaystyle=\exp(a^{\rm T}(\Sigma_s-\Sigma_t)a/2). \end{array}

This is the characteristic function of the multivariate normal, independently of {\mathcal{F}_s}, with mean zero and covariance matrix {\Sigma_t-\Sigma_s}, as required. ⬜

Theorem 1 can also be used to characterize those processes which can be expressed as a stochastic integral with respect to a Brownian motion. Other than the local martingale property, it is necessary for the quadratic variation to have absolutely continuous sample paths. A function {f\colon[a,b]\rightarrow{\mathbb R}} is said to be absolutely continuous if it is continuous with finite variation and {\int\alpha\,df=0} for all bounded measurable functions {\alpha} with {\int_a^b\vert\alpha(s)\vert\,ds=0}. Equivalently, the signed measure {\int\cdot\,df} is absolutely continuous with respect to the Lebesgue measure which, by the Radon-Nikodym theorem, is equivalent to f being expressible as an integral {f(x)=f(a)+\int_a^xf^\prime(y)\,dy} for some Lebesgue-integrable function {f^\prime}. In the stochastic setting, FV processes with absolutely continuous sample paths can be expressed as an integral of a predictable process.

Lemma 4 Let V be an FV process such that {\int_0^t\alpha\,dV=0} (almost surely) for each {t\ge 0} and bounded predictable process {\alpha} satisfying {\int_0^t\vert\alpha_s\vert\,ds=0} (almost surely).

Then, {V=V_0+\int\xi_s\,ds} for some predictable process {\xi} satisfying {\int_0^t\vert\xi_s\vert\,ds<\infty} (almost surely) for each {t\ge 0}.

Proof: Let us first suppose that V has integrable variation and let {\mathcal{P}} denote the predictable sigma-algebra. Define the following measures on {({\mathbb R}_+\times\Omega,\mathcal{P})}

\displaystyle  \lambda(\alpha)={\mathbb E}\left[\int_0^\infty\alpha_s\,ds\right],\ \mu(\alpha)={\mathbb E}\left[\int_0^\infty\alpha\,dV\right]

for all nonnegative bounded and predictable {\alpha}. Here {\lambda} is a (nonnegative) sigma-finite measure and {\mu} is a finite signed measure.

By the condition of the lemma, {\mu(\alpha)=0} whenever {\lambda(\vert\alpha\vert)=0} so, by the Radon-Nikodym theorem, there exists a predictable process {\alpha=d\mu/d\lambda} so that {\mu(\xi)=\lambda(\xi\alpha)} for all bounded predictable {\xi}. The process {W\equiv V-\int\alpha_s\,ds} is continuous, has integrable variation over all bounded time intervals and satisfies

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}\left[\int_0^t\xi\,dW\right]&\displaystyle={\mathbb E}\left[\int_0^t\xi\,dV\right]-{\mathbb E}\left[\int_0^t\xi_s\alpha_s\,ds\right]\smallskip\\ &\displaystyle=\mu\left(1_{(0,t]\times\Omega}\xi\right)-\lambda\left(1_{(0,t]\times\Omega}\xi\alpha\right)\smallskip\\ &\displaystyle=0 \end{array}

for bounded predictable {\xi}. So, W is a martingale as well as a continuous FV process and must be constant. This gives {W=V_0} as required.

Now, for general V satisfying the conditions of the lemma, define stopping times

\displaystyle  \tau_n=\inf\left\{t\ge 0\colon\int_0^t\,\vert dV\vert\ge n\right\}.

Then the stopped processes {V^{\tau_n}} have variation bounded by n and, by the above argument, there are predictable processes {\alpha^n} such that {V^{\tau_n}=V_0+\int\alpha^n_s\,ds}. The result now follows by taking {\alpha_t=\sum_n1_{\{\tau_{n-1}<t\le\tau_n\}}\alpha^n_t}. ⬜

Any process of the form {X=\int\xi\,dB} for a Brownian motion B has quadratic variation {[X]=\int\xi^2_s\,ds}, which is absolutely continuous. We prove that the converse statement also holds. That is, if X is a local martingale with absolutely continuous quadratic variation, then it can be expressed as a stochastic integral with respect to a Brownian motion. Before doing this, there is one technical issue to consider. On any probability space it is always possible to define the process {X_t=0}, which is identically zero everywhere. This trivially has zero quadratic variation and we would like to write {X=\int\xi\,dB} where {\xi=0}. However, this is only possible if there is at least one Brownian motion B defined on the underlying filtered probability space. This is not a major problem though, as it is always possible to enlarge the space to add a Brownian motion. For example, if {(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},{\mathbb P})} is the original filtered probability space and {(\Omega^\prime,\mathcal{F}^\prime,\{\mathcal{F}^\prime_t\}_{t\ge 0},{{\mathbb P}}^\prime)} is any other space with a Brownian motion defined on it, then

\displaystyle  (\Omega^{\prime\prime},\mathcal{F}^{\prime\prime},\{\mathcal{F}^{\prime\prime}_t\}_{t\ge 0},{{\mathbb P}}^{\prime\prime})\equiv(\Omega\times\Omega^\prime,\mathcal{F}\otimes\mathcal{F}^\prime,\{\mathcal{F}_t\otimes\mathcal{F}^{\prime}_t\}_{t\ge0},{\mathbb P}\otimes{{\mathbb P}}^\prime)

defines a new, enlarged, filtered probability space which contains a Brownian motion and to which all processes and random variables on our original space can be lifted. The statement of the representability of processes as stochastic integrals with respect to Brownian motion is as follows.

Theorem 5 Let X be a local martingale such that [X] has absolutely continuous sample paths. Also, suppose that there exists at least one Brownian motion defined on the filtered probability space.

Then, {X=X_0+\int\xi\,dB} for a Brownian motion B and B-integrable process {\xi}.

Proof: Applying Lemma 4, we can write {[X]=\int\alpha_s\,ds}. Then, as quadratic variations are increasing, {\alpha} will be almost-everywhere nonnegative. Set {\xi=1_{\{\alpha\ge 0\}}\sqrt{\alpha}}, so {[X]=\int\xi^2_s\,ds}.

The integral

\displaystyle  \int_0^t1_{\{\xi_s\not=0\}}\xi^{-2}_s\,d[X]_s=\int_0^t1_{\{\xi_s\not=0\}}\,ds<\infty

shows that {1_{\{\xi_s\not=0\}}\xi^{-1}_s} is X-integrable. So, picking any Brownian motion W defined on the filtered probability space, we can define

\displaystyle  B=\int1_{\{\xi\not=0\}}\xi^{-1}\,dX + \int1_{\{\xi_s=0\}}\,dW_s

By preservation of the local martingale property, B is a local martingale. Its quadratic variation satisfies

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle d[B]_t&\displaystyle=1_{\{\xi_t\not=0\}}\xi^{-2}_t\,d[X]_t+1_{\{\xi_t=0\}}\,d[W]_t\smallskip\\ &\displaystyle=1_{\{\xi_t\not=0\}}\xi^{-2}_t(\xi^2_t\,dt) + 1_{\{\xi_t=0\}}\,dt\smallskip\\ &\displaystyle=dt \end{array}

so, by Lévy’s characterization, B is a Brownian motion. As {\int\xi^2\,d[B]=[X]<\infty}, {\xi} is B-integrable and,

\displaystyle  \int\xi\,dB=\int 1_{\{\xi\not=0\}}\,dX=X-X_0-\int1_{\{\xi=0\}}\,dX. (1)

However, {Y\equiv\int1_{\{\xi=0\}}\,dX} is a local martingale with quadratic variation {\int1_{\{\xi=0\}}\,d[X]=\int1_{\{\xi_s=0\}}\xi^2_s\,ds=0}. So, {{\mathbb E}[Y_t^2]=0} and (1) gives the result. ⬜

8 thoughts on “Lévy’s Characterization of Brownian Motion

  1. Hi, I have a little question about the proof of theorem 3.

    You claim there : “Then,Y_t- Y_s,Z_t- Z_s are joint normal with zero mean and covariance a^T(\Sigma_t-\Sigma_s)b, independently of \mathcal{F}_s.”.

    Well it is not obvious to me why they should be normal independently of \mathcal{F}_s ?

    I’m not sure about that but maybe this should be added to the proof or added as an hypothesis in 1 (in this last case this additional property should be proven in the last implication of 3 implies 1).

    Best regards

    1. Yes, I think you’re right. This should have been included among the hypotheses of statement (1). I updated it, thanks.

      Also, the proof of (3) ⇒ (1) is ok as it is. As the characteristic function of Xt – Xs conditioned on \mathcal{F}_s is not random, it is automatically independent of \mathcal{F}_s.

  2. Still in the proof of theorem 3, didn’t you mean Y=a’X and Z=b’X (I write b’ for the transposition of b)?
    Because you are mentioning the covariance of Y and Z with Z defined as b’Y but then have not the same dimension (Y is in R and Z has 1 line and d columns).
    Let me know what you think!


  3. Hi George,
    Thanks for the insightful writings.
    I have a question on Thm 3, that there seems to be a missing detail. You proved that M satisfies the martingale property, but haven’t shown it to be integrable, a necessary condition to be a martingale. Maybe this is obvious, but not to me at the moment.

    1. There is no need to assume continuity of X in Theorem 1, condition 3. This is because the condition already implies that X is continuous. In fact, the jumps of X are given by (\Delta X)^2=\Delta[X]=\Delta t=0.

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