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Let $f:[0,t]\rightarrow \mathbb{R^+}$ be a deterministic and integrable and $(B_t)_{t\geq 0}$ is a standard Brownian motion. If $X_t=\int_o^tf(s)dB_s$, we know that $X_t$ has normal distribution with mean zero.

  1. Does the correlation between $X_t$ and $B_t$ equal to $1$?
  2. If 1. is true, can I consider $X_t$ as $g(t)B_t$ for some $g$?
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  1. Hint: Use Itô's isometry, i.e. $$\mathbb{E}\left[ \left( \int_0^t f(s) \, dB_s \right) \left( \int_0^t g(s) \, dB_s \right) \right] = \mathbb{E} \left( \int_0^t f(s) g(s) \, ds \right).$$
  2. No, in general we cannot expect this. Just consider e.g. $f(s) = s$, then, by Itô's formula, $$\int_0^t s \, dB_s = t B_t - \int_0^t B_s \, ds.$$ We cannot expect to write the right-hand side as a function of $B_t$ since the right-hand side does depend on the sample path of the Brownian motion up to time $t$.
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