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I want to show if $W_t$ and $\tilde{W_t}$ are independent Brownian motions then $$X_t=pW_t+\sqrt{1-p^2}\tilde{W_t}$$ is also Brownian. I am stuck with independent increments. If $s<t\leq u<v$ then $W_v-W_u$ and $W_t-W_s$ are independent and $\tilde{W_v}-\tilde{W_u}$ and $\tilde{W_t}-\tilde{W_s}$ are independent. Can I then infer that $$X_v-X_u=p(W_v-W_u)+\sqrt{1-p^2}(\tilde{W_v}-\tilde{W_u})$$ and $$X_t-X_s=p(W_t-W_s)+\sqrt{1-p^2}(\tilde{W_t}-\tilde{W_s})$$ are independent?

user30523
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  • Yes you can, because the much stronger independence of $$(W_u,\tilde W_u){u\leqslant t}$$ and $$(W_v-W_t,\tilde W_v-\tilde W_t){u\leqslant t}$$ holds. – Did Nov 24 '17 at 14:23

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