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A continuous martingale $M$ is L2-bounded when ${\sup }_{t\ge 0}\mathbb{E}|M_t{|}^2<\infty$. The L2 martingale convergence theorem gives a terminal value $M_{\infty }$ with convergence in $L^2$ and almost sure convergence. The resulting conditional expectation identity is $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$. Boundedness on each separate finite horizon is a weaker condition. The subspace with $M_0=0$ is commonly denoted ${\mathcal{M}}_0^2$.

For a fixed environment, use two independent copies of the walk to express $Z_t^2=⟨{\kappa }_t^{(1)}{\kappa }_t^{(2)}⟩$. The product is finite and nonnegative, so we can interchange the two expectations. At each time $j$, if the walks occupy different sites, the two environment signs are independent and their factors have product expectation $1$. If they coincide, both factors use the same sign andDifferent time layers are independent even when a site is visited again. Therefore, with the simultaneous collision count $I_t={\sum }_{j=1}^t{1}_{\{{\xi }_j^{(1)}={\xi }_j^{(2)}\}}$,This is the collision-count second moment for independent path weights. It involves equal-time intersections, not arbitrary intersections of the two spatial ranges.

The difference walk $D_j={\xi }_j^{(1)}-{\xi }_j^{(2)}$ has independent bounded increments symmetric about zero and a full-dimensional step distribution. It may hold at zero for one step; such a hold counts as a positive-time return. The allowed transience fact givesMoreover $\rho >0$ since the two first steps can be equal. Restarting the independent increments after every return to zero shows that the total positive-time return count $I_{\infty }$ has the zero-based geometric distributionIndeed the probability of at least $k$ returns is ${\rho }^k$, and subtracting successive tails gives this law. Consequently, whenever $\rho (1+{\epsilon }^2)<1$,A sufficient small-noise condition is $0<\epsilon <\min (1,\sqrt{(1-\rho )/\rho })$.

The L2 martingale convergence theorem now gives $Z_t\to \zeta$ in $L^2$ as well as almost surely. One can see the mean preservation directly: for $t\ge s$, the martingale identity gives $\mathbb{E}(Z_t-Z_s{)}^2=\mathbb{E}{Z}_t^2-\mathbb{E}{Z}_s^2$. The second moments are increasing and bounded, so the sequence is Cauchy in $L^2$ and its limit is the already identified almost-sure limit. Thus

Work first with $M_0=0$. More generally, the assertion needs $M_0\in L^2$ and then applies to $M-M_0$. Without this assumption, an integrable but non-square-integrable initial random variable kept constant in time has zero quadratic variation but is not bounded in L2 space.

Choose increasing stopping timesThey tend to infinity almost surely, because the continuous path and its quadratic variation are finite on every compact time interval. The stopped process is a bounded martingale, and $M_{t\wedge {\tau }_n}^2-[M{]}_{t\wedge {\tau }_n}$ is bounded and a local martingale. Taking expectations givesSuppose $C=\mathbb{E}[M{]}_{\infty }<\infty$. These squared expectations are at most $C$. For each fixed $t$, the stopped values converge almost surely to $M_t$ and are uniformly integrable, since they are uniformly bounded in L2 space. They therefore converge in $L^1$. Passing in the stopped conditional-expectation identity, using contraction of conditional expectation in $L^1$, gives $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$ for $s\le t$. Thus $M$ is a true martingale. The Fatou lemma gives $\mathbb{E}{M}_t^2\le C$, uniformly in $t$, so it is an L2-bounded continuous martingale.

Conversely, suppose $M$ is a true martingale with ${\sup }_t\mathbb{E}{M}_t^2\le C$. For fixed $t$, bounded-time optional sampling theorem gives $M_{t\wedge {\tau }_n}=\mathbb{E}(M_t\mid {\mathcal{F}}_{t\wedge {\tau }_n})$. Conditional Jensen inequality then yieldsThe brackets increase to $[M{]}_t$. Applying the monotone convergence theorem first in $n$ and then in $t\uparrow \infty$ proves $\mathbb{E}[M{]}_{\infty }\le C<\infty$.

The exact identities and terminal limits can also be justified. On each fixed horizon $t$, the Doob L2 maximal inequality givesThis is an integrable dominating variable for $M_{t\wedge {\tau }_n}^2$, so the dominated convergence theorem and increasing convergence of the brackets remove the stopping:The L2 martingale convergence theorem supplies $M_{\infty }$ with $M_t\to M_{\infty }$ almost surely and in L2 space. Consequently the squares converge in $L^1$, while $[M{]}_t\uparrow [M{]}_{\infty }$. We obtain the integrable terminal bracket criterion with the precise terminal identityFor $M_0\in L^2$ rather than zero, the final identity becomes $\mathbb{E}{M}_{\infty }^2=\mathbb{E}{M}_0^2+\mathbb{E}[M{]}_{\infty }$.

The Novikov condition holds on every horizon, sinceTherefore $Z=\mathcal{E}(M)$ is a true martingale and the Girsanov theorem applies through each finite $T$.

The deterministic bound also gives a genuine infinite-horizon density. For any $p>1$, the stochastic exponential identity isA nonnegative local martingale is a supermartingale, soThis $L^p$ bound gives uniform integrability. Also $M$ is a L2-bounded continuous martingale, since localization and the Itô isometry give $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t\le C$. The L2 martingale convergence theorem gives a finite limit $M_{\infty }$; hence $Z_{\infty }=\exp (M_{\infty }-[M{]}_{\infty }/2)>0$ and $\mathbb{E}{Z}_{\infty }=1$. Weighting by $Z_{\infty }$ defines an equivalent probability measure on ${\mathcal{F}}_{\infty }$. This is the bounded-bracket criterion for a stochastic exponential.

Work on a filtered probability space with the usual conditions for a filtration. A continuous martingale belongs to the class of L2-bounded continuous martingales whenThis means a uniform bound over the entire time interval, not merely being square-integrable at each individual time. The L2 martingale convergence theorem gives a terminal variable $M_{\infty }$ and almost sure convergence and convergence in the Lebesgue space $L^2$; moreover $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$ by conditional expectation.

The predictable sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is the smallest sigma-algebra making every left-continuous adapted process measurable. Equivalently, it is generated bywhere the sample coordinate comes first. A previsible process is precisely a process measurable for this predictable sigma-algebra. Values at time zero matter for its definition, although they make no contribution to integration against a continuous quadratic variation starting at zero.

Write $N_2=\Vert M_{\infty }{\Vert }_2$ and $N_{\ast }=\Vert {\sup }_{t\ge 0}|M_t|{\Vert }_2$. The L2 martingale convergence theorem ensures the existence of $M_{\infty }$ and yields $|M_{\infty }|\le {\sup }_t|M_t|$ almost surely. Hence $N_2\le N_{\ast }$.

For a finite horizon $T$, the Doob L2 maximal inequality for the submartingale $|M|$ givesThe inequality does not require $M_0=0$. Send $T$ to infinity: the left side increases to $N_{\ast }^2$ by the monotone convergence theorem, and the right side tends to $4N_2^2$ by convergence in the Lebesgue space $L^2$. Thus the equivalent terminal and maximal norms for L2-bounded continuous martingales satisfyBoth are genuine norms on L2-bounded continuous martingales modulo indistinguishability of stochastic processes: a zero terminal norm gives $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]=0$, first at rational times and then at all times by continuity. This proves that the two norms are equivalent norms.

Set $M_n={\sum }_{j=0}^n{X}_j$ and ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$. The independence and zero expected values give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$, so $(M_n)$ is a martingale. The variance additivity for independent random variables givesThe Cauchy-Schwarz inequality implies ${\sup }_n\mathbb{E}|M_n|\le \sqrt{V}$. Applying the martingale convergence theorem from (b),In fact the L2 martingale convergence theorem also gives convergence in L2. Independently, for $n>m$ the same variance additivity for independent random variables yields $\mathbb{E}|M_n-M_m{|}^2={\sum }_{j=m+1}^n\mathbb{E}[X_j^2]\to 0$, confirming that the partial sums form a Cauchy sequence in $L^2$.

The L2 martingale convergence theorem gives $M_t\to M_{\infty }$ in $L^2$, soConversely, apply the Doob L2 maximal inequality on $[0,T]$:Monotone convergence as $T\to \infty$ givesThus the two norms are equivalent.

For the simple symmetric random walk, $S_n$ is a martingale and the square-minus-time martingale of a simple symmetric random walk isIndeed, conditioning on ${\mathcal{F}}_n$ and using $\mathbb{E}[X_{n+1}]=0$ and $X_{n+1}^2=1$ gives $\mathbb{E}[S_{n+1}^2\mid {\mathcal{F}}_n]=S_n^2+1$.

Apply the optional sampling theorem for a supermartingale to the bounded stopping time $T\wedge n$:Thus the stopped martingale $(S_{T\wedge n})$ is bounded in $L^2$. The L2 martingale convergence theorem gives convergence in $L^2$, and because $T<\infty$ almost surely its limit is $S_T$. Meanwhile the monotone convergence theorem gives $\mathbb{E}[T\wedge n]\to \mathbb{E}[T]$. Therefore

Finite mean is essential. Let $T=\inf \{n\ge 1:S_n=0\}$ be the first return to zero. The one-dimensional simple symmetric random walk is recurrent, so $T<\infty$ almost surely, but its first-return time has infinite mean. Since $S_T=0$,