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Two continuous local martingales are orthogonal when their quadratic covariation is zero. With zero initial values, the martingale product identity says that their product is a local martingale. For two Brownian motions in a common filtration, orthogonality and the Lévy characterization of multidimensional Brownian motion make the pair a two-dimensional Brownian motion and imply independence. Merely having the two marginal Brownian motion laws on the same space does not imply orthogonality.

The martingale product identity and the Itô isometry for cross terms giveSince $B_s\sim N(0,s)$, its moment-generating function gives $\mathbb{E}{e}^{B_s}=e^{s/2}$. Therefore the answer is

The martingale product identity says that $N_t{K}_t-[N,K{]}_t$ is a martingale. Passing to the terminal values of the square-integrable martingales and using $N_0=K_0=0$ givesThe quadratic covariation identity for a stochastic integral isApplying the same product identity to $H\cdot M$ and $K$ therefore gives