[proofplan] Both parts follow from the independence of the increment $B_t - B_s$ from $\mathcal{F}_s^+$ (established in [Independence from the Germ Field](/theorems/1177)) combined with the known moments of the Gaussian [distribution](/page/Distribution). For part (i) the conditional expectation of the increment vanishes. For part (ii) we expand $B_t^2$ using the decomposition $B_t = B_s + (B_t - B_s)$ and compute each conditional expectation term separately. [/proofplan] [step:Prove $(B_t)$ is a martingale] Fix $0 \leq s \leq t$. Write $B_t = B_s + (B_t - B_s)$. By [Independence from the Germ Field](/theorems/1177), the increment $B_t - B_s$ is independent of $\mathcal{F}_s^+$, and $B_s$ is $\mathcal{F}_s^+$-measurable. Therefore \begin{align*} \mathbb{E}[B_t \mid \mathcal{F}_s^+] = B_s + \mathbb{E}[B_t - B_s \mid \mathcal{F}_s^+] = B_s + \mathbb{E}[B_t - B_s] = B_s + 0 = B_s, \end{align*} where $\mathbb{E}[B_t - B_s] = 0$ since $B_t - B_s \sim \mathcal{N}(0, t-s)$. Integrability is immediate: $\mathbb{E}[|B_t|] \leq (\mathbb{E}[B_t^2])^{1/2} = \sqrt{t} < \infty$. [/step] [step:Prove $(B_t^2 - t)$ is a martingale] Fix $0 \leq s \leq t$. Write $B_t = B_s + (B_t - B_s)$ and expand \begin{align*} B_t^2 = B_s^2 + 2B_s(B_t - B_s) + (B_t - B_s)^2. \end{align*} Taking conditional expectations with respect to $\mathcal{F}_s^+$, using $\mathcal{F}_s^+$-measurability of $B_s$ and independence of $B_t - B_s$ from $\mathcal{F}_s^+$: \begin{align*} \mathbb{E}[B_t^2 \mid \mathcal{F}_s^+] &= B_s^2 + 2B_s \, \mathbb{E}[B_t - B_s] + \mathbb{E}[(B_t - B_s)^2] \\ &= B_s^2 + 2B_s \cdot 0 + (t - s) \\ &= B_s^2 + (t - s), \end{align*} where $\mathbb{E}[(B_t - B_s)^2] = \operatorname{Var}(B_t - B_s) = t - s$ since $B_t - B_s \sim \mathcal{N}(0, t-s)$. Therefore \begin{align*} \mathbb{E}[B_t^2 - t \mid \mathcal{F}_s^+] = B_s^2 + (t-s) - t = B_s^2 - s. \end{align*} Integrability: $\mathbb{E}[|B_t^2 - t|] \leq \mathbb{E}[B_t^2] + t = 2t < \infty$. [/step]