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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.

Yes: different adapted volatilities can give perfectly correlated terminal prices. The counterexample in part ii already establishes this with one nonconstant volatility. Both can be nonconstant as well. Put $h(x)=1+\frac{1}{2}\tanh x$ and takeThe Doléans-Dade exponential has volatility $h(B_t)$, bounded between $1/2$ and $3/2$. The second stock has volatility $h(B_t)S_t/(1+S_t)$, also bounded and nonconstant. Their squared volatility difference integrates toThe Novikov condition and boundedness give true square-integrable martingales, and the Itô isometry gives positive variance for each $t>0$. Yet $S_t^{\prime }=(1+S_t)/2$ gives correlation one. The mechanism is positive affine dependence, not equality of the multiplicative volatility coefficients; it also explains the failure of the literal part-ii premise.

The Doléans-Dade exponentialis a positive continuous local martingale with $X_0=1$, and part c shows $X_t\to 0$. Apply part a with $a=e^y$:

Integrating the variance equation givesThe Doléans-Dade exponential solution of $d{\tilde{S}}_t=\rho {\tilde{S}}_t\sqrt{v_t}d{W}_t$ isSubstitution yields