1. Let $(B_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ be two standard Brownian motions and let $X_t := B_t W_t$. Is $(X_t)_{t \geq 0}$ a martingale?
The easiest way to proceed seems to be to apply Ito's Lemma, from which we get: \begin{align} dX_t = B_t dB_t + W_t dW_t + d<B, W>_t \end{align}
Hence, $X_t$ is a sum of two Ito integrals (which are martingales), a constant ($X_0$) and a last term, $d<B, W>_t$. $X$ should only be a martingale if the last term is null. In general, this term should be $d<B, W>_t = \rho dt$ and its magnitude depends on the correlation between the two Brownian motions. In general, the answer should thus be no, with the exception being two independent Brownian motions.
2. Let $Z_t \sim N(0,1)$ and $X_t = \sqrt{t} Z_t$. Is $X_t$ a standard Brownian motion?
We have:
\begin{align} E(X_{t+s} - X_t) = 0 \\ V(X_{t+s} - X_t) = E \left( (t+s)Z_{t+s}^2 + (t)Z_t^2 - \sqrt{t}\sqrt{t+s}Z_{t+s} Z_t \right) \\ \leftrightarrow V(X_{t+s} - X_t) = (t + s)(1) + t(1) + 0 = 2t + s \neq s. \end{align}
Hence, this isn't a standard Brownian motion.
3. Are disjoint increments of a martingale uncorrelated?
A martingale is a measurable and integrable process $(M_t)_{t \geq 0}$ with the following property $E(M_t | F_s) = M_s \; \forall t \geq s \geq 0$ from which we can get $E(M_t - M_s| F_s) = 0$. Let $0 < i < j < k < l$, then \begin{align} E((M_j - M_i)(M_l - M_k)) &= E\left( M_j (M_l - M_k) - M_i(M_l - M_k) \right) \\ &= E\left( M_j (M_l - M_k) \right) - E \left(M_i(M_l - M_k) \right) \\ &= E\left( M_j E(M_l - M_k | F_j) \right) - E \left(M_i E(M_l - M_k| F_i) \right) \\ &= 0 + 0 = 0 \end{align}
Hence, the answer is yes.
Am I correct or did I make a mistake somewhere?
