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Set $N_t={\int }_0^t{Z}_s^{-1}d{Z}_s$, the stochastic logarithm of $Z$. On each finite horizon, a strictly positive continuous path has its reciprocal bounded; stopping at levels of $Z$, $Z^{-1}$ and $[Z]$ rigorously localizes this stochastic integral. Thus $N$ is a zero-start continuous local martingale, withThe Itô formula for the ordinary logarithm givesAccordingly, with the exponential convention printed in this question,The initial constant in $M$ contributes no quadratic variation. Under the standard normalized Doléans-Dade exponential convention, which subtracts the driver's initial value in the exponent, the same formula is written $Z=Z_0\mathcal{E}(N)$.

For uniqueness, suppose another continuous driver $L$ gives the printed exponential. Evaluating at zero forces $L_0=\log Z_0$. Applying the Itô formula to its exponential gives $d{Z}_t=Z_{t}d{L}_t$. Since $Z_t>0$, associativity of stochastic integration yields $d{L}_t=Z_t^{-1}d{Z}_t$. Thus $L_t=L_0+N_t=M_t$, up to indistinguishability of stochastic processes.

There is an initial-integrability qualification if “local martingale” is used with the usual integrable-initial-value definition. The displayed $M$ is a continuous local martingale when $\log Z_0\in L^1$, in particular when $Z_0$ is a positive deterministic number. An arbitrary positive integrable $Z_0$ does not ensure this. For example, take an ${\mathcal{F}}_0$-measurable positive integer $W$ with $\mathbb{P}(W=n)=6/({\pi }^2{n}^2)$ and put $Z_t=e^{-W}$ for all $t$. This is a bounded strictly positive martingale, but any representing driver must have $M_0=-W$, which is not integrable. Thus the unqualified claim is false under that definition. For general $Z_0$, the always-valid local-martingale statement is the zero-start representation $Z=Z_0\mathcal{E}(N)$; the printed representation additionally requires the stated integrability or the convention that allows arbitrary finite initial constants.

Work on a common stochastic basis on which $B,B^{\prime }$ are independent Brownian motions and the solutions are adapted; independent weak solutions can be put on the product stochastic basis. Independence gives $[B,B^{\prime }]=0$. Put $S=Z+Z^{\prime }\ge 0$ and define the predictable weightsThey are bounded and satisfy $u_t^2+v_t^2=1$ everywhere. DefineThis is a zero-start continuous local martingale, and its quadratic variation is $[\beta {]}_t={\int }_0^t(u_s^2+v_s^2)ds=t$. By the Lévy characterization of Brownian motion, $\beta$ is a Brownian motion in the common filtration.

On $\{S_t>0\}$, $\sqrt{S_t}{u}_t=\sqrt{Z_t}$ and $\sqrt{S_t}{v}_t=\sqrt{Z_t^{\prime }}$. On $\{S_t=0\}$, nonnegativity forces both $Z_t,Z_t^{\prime }$ to be zero, so the same identities hold. Associativity of stochastic integration therefore givesThis proves the addition law for square-root branching diffusions: $Z+Z^{\prime }$ is a weak solution with initial value $z+z^{\prime }$. Specifying the weights at zero is necessary to keep $[\beta {]}_t=t$ after extinction; simply setting both weights to zero would fail to produce a Brownian driver.

Because $K$ is continuous, adapted, and strictly positive, $K^{-1/2}$ is predictable and locally integrable against $M$. DefineThe quadratic variation of a stochastic integral isThus $W$ starts at zero and is a continuous local martingale with bracket $t$. The Lévy characterization of Brownian motion makes it a Brownian motion in the original filtration.

The process $H$ is continuous and adapted, and ${\int }_0^t{H}_s^{2}ds={\int }_0^t{K}_{s}ds<\infty$ almost surely. The associativity of stochastic integration givesThis is the stochastic-integral representation from absolutely continuous quadratic variation.

The Tanaka equation is an example:It has weak solutions and uniqueness in law, but lacks pathwise uniqueness. The value $s(0)=1$ is deliberate: using the sign function with value zero at zero would admit the identically zero solution and would change the example.

For a quick verification, take a Brownian motion $W$ and put $B=\int s(W)dW$. Its quadratic variation is $t$, so it is a Brownian motion by the Lévy characterization of Brownian motion, and associativity of stochastic integration gives $W=\int s(W)dB$. Every weak solution likewise has quadratic variation $t$ and therefore the law of Brownian motion. On this same enlarged filtration, both $W$ and $-W$ solve the equation driven by $B$: $s(-W)=-s(W)$ away from zero, and the Brownian zero set has zero time measure, so the discrepancy at zero contributes nothing to the Itô integral. These solutions differ with positive probability at any positive time. This proves the claimed distinction without relying on a choice of zero convention left unstated.

Set $K_t=\sqrt{H_t}$ and define the stochastic integralStrict positivity and predictability of $H$ make the integrand locally admissible. The process $W$ is a continuous local martingale starting from zero, and the quadratic variation of a stochastic integral givesBy the Lévy characterization of Brownian motion, $W$ is a Brownian motion. The associativity of stochastic integration then yieldswhich is the required representation.

With $s(x)=1$ for $x\ge 0$ and $s(x)=-1$ for $x<0$, the equation starting at zero has uniqueness in law but not pathwise uniqueness. For existence, take a Brownian motion $W$ and set $B=\int s(W)dW$; the Lévy characterization of Brownian motion and associativity of stochastic integration give $W=\int s(W)dB$. Every solution has quadratic variation $t$ and hence the Brownian motion law. Both $W$ and $-W$ are solutions driven by the same $B$, since the time spent at zero is zero and $s(-W)=-s(W)$ away from zero. Using $s(0)=0$ instead would allow the constant solution and invalidate this uniqueness-in-law example.