[proofplan] We apply [Itô's Formula](/theorems/2099) to the function $f(x) = e^x$ with the semimartingale $X_t = M_t - \frac{1}{2}\langle M \rangle_t$. The finite-variation part $-\frac{1}{2}\langle M \rangle_t$ contributes nothing to quadratic variation, so the Itô correction is $\frac{1}{2}d\langle M \rangle_t$, which cancels the $-\frac{1}{2}d\langle M \rangle_t$ from the drift. This yields $dZ_t = Z_t \, dM_t$, showing $Z$ is a stochastic integral against $M$ and hence a continuous local martingale. For the uniform integrability claim under the bounded quadratic variation hypothesis $\langle M \rangle_\infty \leq b$, we use a sub-Gaussian tail estimate for the running supremum of $M$ to show $\mathbb{E}[\sup_t Z_t] < \infty$, which yields uniform integrability by de la Vallée-Poussin. [/proofplan] [step:Apply Itô's formula to $f(x) = e^x$ with the semimartingale $M_t - \frac{1}{2}\langle M \rangle_t$] Define the semimartingale \begin{align*} X_t = M_t - \tfrac{1}{2}\langle M \rangle_t. \end{align*} The decomposition of $X$ into local martingale and finite-variation parts is $X_t = M_t + A_t$ with $A_t = -\frac{1}{2}\langle M \rangle_t$. Since $M$ is a continuous local martingale and $\langle M \rangle$ is a continuous adapted increasing process, $X$ is a continuous semimartingale. The quadratic variation of $X$ is $\langle X \rangle_t = \langle M \rangle_t$, because the finite-variation part $A$ contributes nothing to quadratic variation (by the [Continuous Finite Variation Local Martingale is Identically Zero](/theorems/2081) theorem, a continuous finite-variation local martingale is zero, so the quadratic variation of a finite-variation process is zero). Define \begin{align*} f : \mathbb{R} &\to (0,\infty) \\ x &\mapsto e^x. \end{align*} Since $f \in C^2(\mathbb{R})$ and $X$ is a continuous semimartingale, [Itô's Formula](/theorems/2099) applies and gives \begin{align*} f(X_t) = f(X_0) + \int_0^t f'(X_s) \, dX_s + \frac{1}{2}\int_0^t f''(X_s) \, d\langle X \rangle_s. \end{align*} Since $f' = f'' = f = e^{(\cdot)}$, we have $f'(X_s) = f''(X_s) = Z_s$, and $f(X_0) = e^{M_0 - \frac{1}{2}\langle M \rangle_0} = e^0 = 1$ because $M_0 = 0$ and $\langle M \rangle_0 = 0$. Thus \begin{align*} Z_t &= 1 + \int_0^t Z_s \, dX_s + \frac{1}{2}\int_0^t Z_s \, d\langle X \rangle_s \\ &= 1 + \int_0^t Z_s \, dM_s - \frac{1}{2}\int_0^t Z_s \, d\langle M \rangle_s + \frac{1}{2}\int_0^t Z_s \, d\langle M \rangle_s \\ &= 1 + \int_0^t Z_s \, dM_s. \end{align*} The cancellation of the $\pm \frac{1}{2} Z_s \, d\langle M \rangle_s$ terms is exact, and the SDE $dZ_t = Z_t \, dM_t$ with $Z_0 = 1$ is established. [guided] The stochastic exponential is defined as $Z_t = \mathcal{E}(M)_t = \exp\bigl(M_t - \frac{1}{2}\langle M \rangle_t\bigr)$. The exponent is the semimartingale $X_t = M_t - \frac{1}{2}\langle M \rangle_t$. Why is this specific combination in the exponent? The answer is precisely the Itô correction: the $-\frac{1}{2}\langle M \rangle_t$ is placed there to cancel the Itô correction term, so that $Z$ satisfies the drift-free equation $dZ = Z \, dM$. To see this, decompose $X_t = M_t + A_t$ where $A_t = -\frac{1}{2}\langle M \rangle_t$ is a continuous finite-variation process. The quadratic variation of $X$ equals $\langle M \rangle_t$ because the finite-variation part $A$ has zero quadratic variation. (Why? Because $A$ is a continuous finite-variation process, and the sum $\sum_i (A_{t_i} - A_{t_{i-1}})^2 \leq \max_i |A_{t_i} - A_{t_{i-1}}| \cdot \sum_i |A_{t_i} - A_{t_{i-1}}|$ tends to zero as the mesh goes to zero: the first factor vanishes by uniform continuity, while the second is bounded by the total variation.) Define $f(x) = e^x$. We verify the hypotheses of [Itô's Formula](/theorems/2099): $f \in C^2(\mathbb{R})$ and $X$ is a continuous (real-valued, so $p = 1$) semimartingale. Itô's formula gives \begin{align*} f(X_t) = f(X_0) + \int_0^t f'(X_s) \, dX_s + \frac{1}{2}\int_0^t f''(X_s) \, d\langle X \rangle_s. \end{align*} For the exponential, $f' = f'' = f$, so $f'(X_s) = f''(X_s) = e^{X_s} = Z_s$. The initial value is $f(X_0) = e^{M_0 - \frac{1}{2}\langle M \rangle_0} = e^0 = 1$ since $M_0 = 0$ implies $\langle M \rangle_0 = 0$. Now expand the stochastic integral $\int_0^t Z_s \, dX_s$. Since $dX_s = dM_s - \frac{1}{2}d\langle M \rangle_s$, we get \begin{align*} \int_0^t Z_s \, dX_s = \int_0^t Z_s \, dM_s - \frac{1}{2}\int_0^t Z_s \, d\langle M \rangle_s. \end{align*} The Itô correction term is $\frac{1}{2}\int_0^t Z_s \, d\langle X \rangle_s = \frac{1}{2}\int_0^t Z_s \, d\langle M \rangle_s$. Adding these together: \begin{align*} Z_t &= 1 + \underbrace{\int_0^t Z_s \, dM_s}_{\text{martingale part}} \underbrace{- \frac{1}{2}\int_0^t Z_s \, d\langle M \rangle_s + \frac{1}{2}\int_0^t Z_s \, d\langle M \rangle_s}_{\text{these cancel}} \\ &= 1 + \int_0^t Z_s \, dM_s. \end{align*} The cancellation is the entire raison d'être of the $-\frac{1}{2}\langle M \rangle$ term in the definition of $\mathcal{E}(M)$. Without it, we would have $de^{M_t} = e^{M_t}(dM_t + \frac{1}{2}d\langle M \rangle_t)$, which has a non-zero drift term $\frac{1}{2}e^{M_t} d\langle M \rangle_t$ and is not a local martingale. [/guided] [/step] [step:Conclude that $\mathcal{E}(M)$ is a continuous local martingale] The integral representation $Z_t = 1 + \int_0^t Z_s \, dM_s$ shows that $Z$ is a stochastic integral of the process $s \mapsto Z_s$ against the continuous local martingale $M$. Since $Z$ is continuous (as the exponential of a continuous function) and hence locally bounded, and $M$ is a continuous local martingale, the [Properties of the Semimartingale Integral](/theorems/2094) guarantee that $\int_0^t Z_s \, dM_s$ is a continuous local martingale. Adding the constant $1$ preserves the local martingale property (the reducing sequence is unchanged). Moreover, $Z_0 = 1$ by the computation above. [/step] [step:Show $\mathcal{E}(M)$ is a uniformly integrable martingale when $\langle M \rangle_\infty \leq b < \infty$ a.s.] Suppose $\langle M \rangle_\infty := \lim_{t \to \infty} \langle M \rangle_t \leq b$ a.s. for some constant $b > 0$. We show that $\mathbb{E}[\sup_{t \geq 0} Z_t] < \infty$, which implies $Z$ is a uniformly integrable martingale. Since $Z_t = \exp\bigl(M_t - \frac{1}{2}\langle M \rangle_t\bigr)$ and $\langle M \rangle_t \geq 0$, we have \begin{align*} \sup_{t \geq 0} Z_t \leq \exp\Bigl(\sup_{t \geq 0} M_t\Bigr). \end{align*} We bound $\mathbb{E}[\exp(\sup_{t \geq 0} M_t)]$ using the sub-Gaussian tail of the running supremum. Define the stopping time $\tau_a = \inf\{t \geq 0 : M_t \geq a\}$ for $a > 0$. Since $M$ is a continuous local martingale with $\langle M \rangle_\infty \leq b$, the process $N_t = \exp\bigl(M_t - \frac{1}{2}\langle M \rangle_t\bigr)$ is a non-negative local martingale by the first part of this proof, hence a supermartingale by the [Non-negative Local Martingale is a Supermartingale](/theorems/2077) theorem. In particular $\mathbb{E}[N_t] \leq N_0 = 1$ for all $t$. On the event $\{\sup_{t \geq 0} M_t \geq a\} = \{\tau_a < \infty\}$, we have $M_{\tau_a} = a$ by continuity. By the optional stopping theorem applied to the supermartingale $N$ at the stopping time $\tau_a$: \begin{align*} \mathbb{P}\!\left(\sup_{t \geq 0} M_t \geq a\right) &= \mathbb{P}(\tau_a < \infty) \leq e^{-a} \, \mathbb{E}\!\left[e^{M_{\tau_a} - \frac{1}{2}\langle M \rangle_{\tau_a}} \, \mathbb{1}_{\{\tau_a < \infty\}}\right] \cdot e^{a} \cdot \sup_{\omega} e^{-\frac{1}{2}\langle M \rangle_{\tau_a}} \\ &\leq e^{-a} \cdot 1 \cdot e^{\frac{1}{2}b} \cdot e^{a} \cdot e^{-a} \end{align*} More precisely, we use the exponential martingale inequality directly. For any $\lambda > 0$, the process $\exp(\lambda M_t - \frac{\lambda^2}{2}\langle M \rangle_t)$ is a non-negative supermartingale with initial value $1$. By the maximal inequality for non-negative supermartingales, \begin{align*} \mathbb{P}\!\left(\sup_{t \geq 0} M_t \geq a\right) \leq \mathbb{P}\!\left(\sup_{t \geq 0} \exp\!\left(\lambda M_t - \frac{\lambda^2}{2}\langle M \rangle_t\right) \geq \exp\!\left(\lambda a - \frac{\lambda^2}{2}b\right)\right) \leq \exp\!\left(-\lambda a + \frac{\lambda^2}{2}b\right), \end{align*} where the last inequality uses the supermartingale maximal inequality $\mathbb{P}(\sup_t Y_t \geq c) \leq \mathbb{E}[Y_0]/c$ for a non-negative supermartingale $Y$. Optimizing over $\lambda$ by setting $\lambda = a/b$ gives \begin{align*} \mathbb{P}\!\left(\sup_{t \geq 0} M_t \geq a\right) \leq \exp\!\left(-\frac{a^2}{2b}\right). \end{align*} Now compute: \begin{align*} \mathbb{E}\!\left[\exp\!\left(\sup_{t \geq 0} M_t\right)\right] &= \int_0^\infty \mathbb{P}\!\left(\exp\!\left(\sup_{t \geq 0} M_t\right) \geq \lambda\right) \, d\mathcal{L}^1(\lambda) \\ &= 1 + \int_1^\infty \mathbb{P}\!\left(\sup_{t \geq 0} M_t \geq \log \lambda\right) \, d\mathcal{L}^1(\lambda) \\ &\leq 1 + \int_1^\infty \exp\!\left(-\frac{(\log \lambda)^2}{2b}\right) \, d\mathcal{L}^1(\lambda). \end{align*} The substitution $\lambda = e^u$, $d\mathcal{L}^1(\lambda) = e^u \, d\mathcal{L}^1(u)$, transforms the integral to \begin{align*} \int_0^\infty \exp\!\left(u - \frac{u^2}{2b}\right) \, d\mathcal{L}^1(u) = \int_0^\infty \exp\!\left(-\frac{1}{2b}(u - b)^2 + \frac{b}{2}\right) \, d\mathcal{L}^1(u) \leq e^{b/2} \sqrt{2\pi b} < \infty. \end{align*} Therefore $\mathbb{E}[\sup_{t \geq 0} Z_t] \leq \mathbb{E}[\exp(\sup_{t \geq 0} M_t)] < \infty$. Since $Z$ is a non-negative continuous local martingale with $\mathbb{E}[\sup_{t \geq 0} Z_t] < \infty$, the family $\{Z_T : T \text{ a stopping time}\}$ is dominated by the integrable random variable $\sup_{t \geq 0} Z_t$ and is therefore uniformly integrable. By the [Characterization of Martingales Among Local Martingales](/theorems/2078), $Z$ is a (true) martingale. The uniform integrability of $\{Z_t : t \geq 0\}$ follows from the domination $Z_t \leq \sup_{s \geq 0} Z_s \in L^1$. [guided] We want to promote the local martingale $Z$ to a uniformly integrable martingale under the hypothesis $\langle M \rangle_\infty \leq b$. The strategy is to find an integrable dominating random variable for $\sup_t Z_t$. Since $Z_t = \exp(M_t - \frac{1}{2}\langle M \rangle_t)$ and $\langle M \rangle_t \geq 0$, we have the pointwise bound $Z_t \leq e^{M_t}$, so $\sup_t Z_t \leq \exp(\sup_t M_t)$. The question reduces to showing $\mathbb{E}[\exp(\sup_t M_t)] < \infty$. The key tool is the **exponential martingale inequality**: for any $\lambda > 0$, the process $\exp(\lambda M_t - \frac{\lambda^2}{2}\langle M \rangle_t)$ is a stochastic exponential $\mathcal{E}(\lambda M)_t$, hence a non-negative local martingale by the first part of this proof. By the [Non-negative Local Martingale is a Supermartingale](/theorems/2077) theorem, it is a supermartingale with initial value $1$. For non-negative supermartingales, the maximal inequality states: if $Y$ is a non-negative supermartingale, then $\mathbb{P}(\sup_{t \geq 0} Y_t \geq c) \leq \mathbb{E}[Y_0] / c$ for any $c > 0$. Applying this to $Y_t = \exp(\lambda M_t - \frac{\lambda^2}{2}\langle M \rangle_t)$ with $c = \exp(\lambda a - \frac{\lambda^2}{2}b)$: On the event $\{\sup_t M_t \geq a\}$ and using $\langle M \rangle_t \leq b$, we have $\sup_t Y_t \geq \exp(\lambda a - \frac{\lambda^2}{2}b) = c$. Therefore \begin{align*} \mathbb{P}\!\left(\sup_{t \geq 0} M_t \geq a\right) \leq \frac{\mathbb{E}[Y_0]}{c} = \exp\!\left(-\lambda a + \frac{\lambda^2}{2}b\right). \end{align*} This holds for all $\lambda > 0$. The right-hand side is minimized by choosing $\lambda = a/b$ (differentiate $-\lambda a + \frac{\lambda^2}{2}b$ and set to zero), giving \begin{align*} \mathbb{P}\!\left(\sup_{t \geq 0} M_t \geq a\right) \leq \exp\!\left(-\frac{a^2}{2b}\right). \end{align*} This is a sub-Gaussian tail bound: the running supremum of $M$ has tail decay as fast as a Gaussian with variance $b$. To convert this tail bound into an $L^1$ bound on $\exp(\sup_t M_t)$, use the layer-cake representation: \begin{align*} \mathbb{E}\!\left[\exp\!\left(\sup_{t \geq 0} M_t\right)\right] = \int_0^\infty \mathbb{P}\!\left(\exp\!\left(\sup_t M_t\right) \geq \lambda\right) \, d\mathcal{L}^1(\lambda). \end{align*} Since $\exp(\sup_t M_t) \geq 1$ always (as $\sup_t M_t \geq M_0 = 0$), the integral from $0$ to $1$ contributes exactly $1$. For $\lambda \geq 1$, $\{\exp(\sup_t M_t) \geq \lambda\} = \{\sup_t M_t \geq \log \lambda\}$, so \begin{align*} \mathbb{E}\!\left[\exp\!\left(\sup_t M_t\right)\right] \leq 1 + \int_1^\infty \exp\!\left(-\frac{(\log \lambda)^2}{2b}\right) \, d\mathcal{L}^1(\lambda). \end{align*} Substituting $u = \log \lambda$, $d\mathcal{L}^1(\lambda) = e^u \, d\mathcal{L}^1(u)$: \begin{align*} \int_1^\infty \exp\!\left(-\frac{(\log \lambda)^2}{2b}\right) \, d\mathcal{L}^1(\lambda) = \int_0^\infty \exp\!\left(u - \frac{u^2}{2b}\right) \, d\mathcal{L}^1(u). \end{align*} Completing the square in the exponent: $u - \frac{u^2}{2b} = -\frac{1}{2b}(u - b)^2 + \frac{b}{2}$. Therefore \begin{align*} \int_0^\infty \exp\!\left(u - \frac{u^2}{2b}\right) \, d\mathcal{L}^1(u) = e^{b/2} \int_0^\infty \exp\!\left(-\frac{(u-b)^2}{2b}\right) \, d\mathcal{L}^1(u) \leq e^{b/2} \int_{-\infty}^\infty \exp\!\left(-\frac{(u-b)^2}{2b}\right) \, d\mathcal{L}^1(u) = e^{b/2}\sqrt{2\pi b}. \end{align*} The integral converges because the integrand decays as $e^{-u^2/(2b)}$ for large $u$, which is faster than any polynomial. Thus $\mathbb{E}[\sup_t Z_t] \leq \mathbb{E}[\exp(\sup_t M_t)] < \infty$. With $\sup_t Z_t \in L^1$, the family $\{Z_t\}_{t \geq 0}$ is dominated by this integrable random variable, so it is uniformly integrable. The [Characterization of Martingales Among Local Martingales](/theorems/2078) (which states that a local martingale whose stopped family is uniformly integrable is a true martingale) now promotes $Z$ from a local martingale to a genuine uniformly integrable martingale. [/guided] [/step]