[proofplan] We verify the Brownian motion property by computing the conditional characteristic function. For any $\theta \in \mathbb{R}^d$, we form the scalar process $Y_t = \theta \cdot X_t$ and use the covariation hypothesis to compute $\langle Y \rangle_t = |\theta|^2 t$. We then apply Ito's formula to the complex exponential $Z_t = \exp(iY_t + \frac{1}{2}|\theta|^2 t)$ and show that $Z$ is a continuous local martingale. Since $|Z_t| \leq 1$, the local martingale $Z$ is bounded on bounded intervals and therefore a true martingale. The martingale property $\mathbb{E}[Z_t \mid \mathcal{F}_s] = Z_s$ yields the conditional characteristic function identity $\mathbb{E}[e^{i\theta \cdot (X_t - X_s)} \mid \mathcal{F}_s] = e^{-\frac{1}{2}|\theta|^2(t-s)}$, which simultaneously establishes the Gaussian distribution of increments and their independence from $\mathcal{F}_s$. [/proofplan] [step:Reduce to verifying the conditional characteristic function identity] To show that $(X^1, \ldots, X^d)$ is a standard $d$-dimensional Brownian motion, it suffices to verify three properties: (i) $X^i_0 = 0$ for each $i$ (given by hypothesis), (ii) continuity of paths (given since each $X^i$ is a continuous local martingale), and (iii) for all $0 \leq s < t$, the increment $X_t - X_s$ is independent of $\mathcal{F}_s$ and distributed as $N(0, (t-s)I_d)$. By the Fourier Uniqueness Theorem, the joint distribution of $X_t - X_s$ is determined by its characteristic function. Moreover, $X_t - X_s$ is independent of $\mathcal{F}_s$ if and only if the conditional characteristic function $\mathbb{E}[e^{i\theta \cdot (X_t - X_s)} \mid \mathcal{F}_s]$ is deterministic for every $\theta \in \mathbb{R}^d$. It therefore suffices to show that for all $\theta \in \mathbb{R}^d$ and $0 \leq s \leq t$, \begin{align*} \mathbb{E}\!\left[e^{i\theta \cdot (X_t - X_s)} \;\middle|\; \mathcal{F}_s\right] = e^{-\frac{1}{2}|\theta|^2(t-s)}. \end{align*} The right-hand side is the characteristic function of $N(0, (t-s)I_d)$ evaluated at $\theta$, and it is deterministic (not depending on $\omega$), so both the distribution and the independence follow from this single identity. [guided] Why does a deterministic conditional characteristic function imply independence? If $\mathbb{E}[e^{i\theta \cdot (X_t - X_s)} \mid \mathcal{F}_s] = \phi(\theta)$ is deterministic for all $\theta$, then for any bounded $\mathcal{F}_s$-measurable random variable $Z$, \begin{align*} \mathbb{E}[Z \, e^{i\theta \cdot (X_t - X_s)}] = \mathbb{E}[Z \, \mathbb{E}[e^{i\theta \cdot (X_t - X_s)} \mid \mathcal{F}_s]] = \mathbb{E}[Z] \cdot \phi(\theta) = \mathbb{E}[Z] \cdot \mathbb{E}[e^{i\theta \cdot (X_t - X_s)}]. \end{align*} This factorization of joint characteristic functions characterizes independence. Moreover, $\phi(\theta) = e^{-\frac{1}{2}|\theta|^2(t-s)}$ is exactly the characteristic function of the multivariate Gaussian $N(0, (t-s)I_d)$, so the distribution is identified as well. [/guided] [/step] [step:Form the scalar process $Y_t = \theta \cdot X_t$ and compute its quadratic variation] Fix $\theta = (\theta_1, \ldots, \theta_d) \in \mathbb{R}^d$ and define \begin{align*} Y: \Omega \times [0, \infty) &\to \mathbb{R} \\ (\omega, t) &\mapsto \sum_{i=1}^d \theta_i X^i_t(\omega). \end{align*} Since each $X^i$ is a continuous local martingale and the sum is finite, $Y$ is a continuous local martingale with $Y_0 = \theta \cdot X_0 = 0$. Compute the quadratic variation of $Y$ using bilinearity of the covariation and the hypothesis $\langle X^i, X^j \rangle_t = \delta_{ij} t$: \begin{align*} \langle Y \rangle_t = \left\langle \sum_{i=1}^d \theta_i X^i, \sum_{j=1}^d \theta_j X^j \right\rangle_t = \sum_{i,j=1}^d \theta_i \theta_j \langle X^i, X^j \rangle_t = \sum_{i,j=1}^d \theta_i \theta_j \delta_{ij} t = |\theta|^2 t. \end{align*} [/step] [step:Define $Z_t = \exp(iY_t + \frac{1}{2}|\theta|^2 t)$ and apply Ito's formula to show $Z$ is a local martingale] Define the process \begin{align*} Z: \Omega \times [0, \infty) &\to \mathbb{C} \\ (\omega, t) &\mapsto \exp\!\left(iY_t(\omega) + \tfrac{1}{2}|\theta|^2 t\right). \end{align*} Write $Z_t = \exp(U_t)$ where $U_t = iY_t + \frac{1}{2}|\theta|^2 t$. The process $U$ is a continuous semimartingale: its local martingale part is $iY_t$ and its finite variation part is $\frac{1}{2}|\theta|^2 t$. Apply [Ito's Formula](/theorems/2099) to the function $\exp: \mathbb{C} \to \mathbb{C}$ (viewed as $\exp: \mathbb{R}^2 \to \mathbb{R}^2$ applied to the real and imaginary parts of $U$). Since $\exp' = \exp'' = \exp$, the formula gives \begin{align*} dZ_t = Z_t \, dU_t + \frac{1}{2} Z_t \, d\langle U \rangle_t. \end{align*} The quadratic variation of $U$ comes solely from its local martingale part $iY$: \begin{align*} \langle U \rangle_t = \langle iY \rangle_t = i^2 \langle Y \rangle_t = -|\theta|^2 t, \end{align*} where we used $\langle cY \rangle_t = c^2 \langle Y \rangle_t$ for a constant $c \in \mathbb{C}$ and $\langle Y \rangle_t = |\theta|^2 t$ from the previous step. Substituting $dU_t = i \, dY_t + \frac{1}{2}|\theta|^2 \, dt$ and $d\langle U \rangle_t = -|\theta|^2 \, dt$: \begin{align*} dZ_t = Z_t \left(i \, dY_t + \frac{1}{2}|\theta|^2 \, dt\right) + \frac{1}{2} Z_t \left(-|\theta|^2 \, dt\right) = Z_t \left(i \, dY_t + \frac{1}{2}|\theta|^2 \, dt - \frac{1}{2}|\theta|^2 \, dt\right) = i Z_t \, dY_t. \end{align*} The finite variation terms $\frac{1}{2}|\theta|^2 \, dt$ and $-\frac{1}{2}|\theta|^2 \, dt$ cancel exactly, leaving \begin{align*} dZ_t = i Z_t \, dY_t. \end{align*} Equivalently, $Z_t = Z_0 + i \int_0^t Z_s \, dY_s = 1 + i \int_0^t Z_s \, dY_s$. Since $Y$ is a continuous local martingale and $Z$ is a continuous adapted process, the stochastic integral $\int_0^t Z_s \, dY_s$ is a continuous local martingale. Therefore $Z$ is a continuous local martingale. [guided] The cancellation of the $dt$ terms is the heart of the proof and explains the specific choice of $Z_t = \exp(iY_t + \frac{1}{2}|\theta|^2 t)$. The extra factor $\exp(\frac{1}{2}|\theta|^2 t)$ is chosen precisely to cancel the Ito correction. To see why: if we naively tried $\tilde{Z}_t = e^{iY_t}$, Ito's formula would give $d\tilde{Z}_t = i\tilde{Z}_t \, dY_t + \frac{1}{2}(i)^2 \tilde{Z}_t \, d\langle Y \rangle_t = i\tilde{Z}_t \, dY_t - \frac{1}{2}|\theta|^2 \tilde{Z}_t \, dt$. This has a non-zero drift $-\frac{1}{2}|\theta|^2 \tilde{Z}_t \, dt$, so $\tilde{Z}$ is not a local martingale. Multiplying by $\exp(\frac{1}{2}|\theta|^2 t)$ adds a compensating drift that exactly cancels the Ito correction, leaving a pure stochastic integral. More precisely, with $U_t = iY_t + \frac{1}{2}|\theta|^2 t$, we compute: - $dU_t = i \, dY_t + \frac{1}{2}|\theta|^2 \, dt$ (the finite variation contribution) - $d\langle U \rangle_t = (i)^2 d\langle Y \rangle_t = -|\theta|^2 \, dt$ (only the martingale part $iY$ contributes to quadratic variation) When we apply $\exp$ via Ito's formula, the drift in $Z$ is $Z_t(\frac{1}{2}|\theta|^2 - \frac{1}{2}|\theta|^2) \, dt = 0$. The cancellation is exact because we designed $U$ to include the compensating term $\frac{1}{2}|\theta|^2 t$. [/guided] [/step] [step:Show $Z$ is a true martingale using boundedness] We have $Z_t = \exp(iY_t + \frac{1}{2}|\theta|^2 t)$. Since $Y_t$ is real-valued, the modulus of $Z_t$ is \begin{align*} |Z_t| = \left|\exp\!\left(iY_t + \tfrac{1}{2}|\theta|^2 t\right)\right| = \exp\!\left(\operatorname{Re}\!\left(iY_t + \tfrac{1}{2}|\theta|^2 t\right)\right) = \exp\!\left(\tfrac{1}{2}|\theta|^2 t\right). \end{align*} Wait — this shows $|Z_t|$ grows in $t$, so $Z$ is not bounded globally. However, for any fixed $T > 0$, we have $|Z_t| \leq \exp(\frac{1}{2}|\theta|^2 T)$ for all $t \in [0, T]$. A continuous local martingale that is bounded on $[0, T]$ is a true martingale on $[0, T]$: boundedness implies uniform integrability, and a uniformly integrable continuous local martingale is a martingale. Since $|Z_t| \leq \exp(\frac{1}{2}|\theta|^2 T)$ for $t \leq T$, the process $(Z_t)_{0 \leq t \leq T}$ is a martingale for every $T > 0$. [guided] We need to be careful about the distinction between local martingales and true martingales. A continuous local martingale $Z$ is a true martingale on $[0, T]$ if the family $\{Z_\tau : \tau \text{ a stopping time with } \tau \leq T\}$ is uniformly integrable. Boundedness $|Z_t| \leq C$ for all $t \leq T$ immediately gives uniform integrability (bounded families are uniformly integrable). Why does the modulus computation give $|Z_t| = \exp(\frac{1}{2}|\theta|^2 t)$? Since $Y_t$ is real, $iY_t$ is purely imaginary, so $\operatorname{Re}(iY_t) = 0$. Therefore $\operatorname{Re}(iY_t + \frac{1}{2}|\theta|^2 t) = \frac{1}{2}|\theta|^2 t$, and $|e^w| = e^{\operatorname{Re}(w)}$ for any $w \in \mathbb{C}$. The bound $|Z_t| \leq \exp(\frac{1}{2}|\theta|^2 T)$ for $t \leq T$ is deterministic (no randomness), so the uniform integrability is immediate. This is why the argument works without any moment conditions on $X$. [/guided] [/step] [step:Extract the conditional characteristic function identity from the martingale property] Since $(Z_t)_{0 \leq t \leq T}$ is a martingale for every $T > 0$, for $0 \leq s \leq t$: \begin{align*} \mathbb{E}[Z_t \mid \mathcal{F}_s] = Z_s. \end{align*} Substituting $Z_t = \exp(i\theta \cdot X_t + \frac{1}{2}|\theta|^2 t)$ and $Z_s = \exp(i\theta \cdot X_s + \frac{1}{2}|\theta|^2 s)$: \begin{align*} \mathbb{E}\!\left[\exp\!\left(i\theta \cdot X_t + \tfrac{1}{2}|\theta|^2 t\right) \;\middle|\; \mathcal{F}_s\right] = \exp\!\left(i\theta \cdot X_s + \tfrac{1}{2}|\theta|^2 s\right). \end{align*} The factor $\exp(i\theta \cdot X_s + \frac{1}{2}|\theta|^2 s)$ is $\mathcal{F}_s$-measurable (since $X_s$ is $\mathcal{F}_s$-measurable and $s$ is deterministic), so we may divide both sides by $Z_s$ (which has modulus $\exp(\frac{1}{2}|\theta|^2 s) > 0$, hence is never zero): \begin{align*} \mathbb{E}\!\left[\exp\!\left(i\theta \cdot (X_t - X_s) + \tfrac{1}{2}|\theta|^2(t - s)\right) \;\middle|\; \mathcal{F}_s\right] = 1. \end{align*} Since $\exp(\frac{1}{2}|\theta|^2(t-s))$ is a deterministic constant, we may pull it out of the conditional expectation: \begin{align*} \exp\!\left(\tfrac{1}{2}|\theta|^2(t-s)\right) \cdot \mathbb{E}\!\left[e^{i\theta \cdot (X_t - X_s)} \;\middle|\; \mathcal{F}_s\right] = 1. \end{align*} Dividing by $\exp(\frac{1}{2}|\theta|^2(t-s))$: \begin{align*} \mathbb{E}\!\left[e^{i\theta \cdot (X_t - X_s)} \;\middle|\; \mathcal{F}_s\right] = e^{-\frac{1}{2}|\theta|^2(t-s)}. \end{align*} This is the conditional characteristic function of $N(0, (t-s)I_d)$, and it is deterministic. As established in the first step, this implies that $X_t - X_s \sim N(0, (t-s)I_d)$ and $X_t - X_s$ is independent of $\mathcal{F}_s$. Since this holds for all $0 \leq s \leq t$ and $\theta \in \mathbb{R}^d$, the process $(X^1, \ldots, X^d)$ is a standard $d$-dimensional Brownian motion. [/step]