# Independent increments of $X_t:=\int_0^t\phi(s) dW_s$

Motivated through the following question Can we prove directly that $M_t$ is a martingale, I want to ask this in a separate question. Suppose we have a deterministic function $\phi$ which belongs to $L^2[0,T]$. We then look at $$X_t:=\int_0^t\phi(s)dW_s$$ where $W$ is a Brownian Motion. We know that $X_t$ is normal distributed with mean $0$ and variance $\int_0^t\phi(s)^2ds$. Moreover it is a martingale. The question is, how can we prove that the increments are independent. Do we have to assume some further specification about the filtration? Obviously we assume that $W$ is a Brownian Motion w.r.t. to $(\mathcal{F}_t)$.

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The argument was already given in an answer to one of your previous questions: if $0\leqslant s\leqslant t$, $X_s$ is measurable with respect to the sigma-algebra $\mathcal G_s=\sigma(W_u,u\leqslant s)$ and $X_t-X_s$ is measurable with respect to the sigma-algebra $\mathcal H_s=\sigma(V_u,u\geqslant 0)$, where, for every $u\geqslant0$, $V_u=W_{u+s}-W_s$.
Now, $(V_u)_{u\geqslant 0}$ is independent of $(W_u)_{u\leqslant s}$ hence the sigma-algebras $\mathcal G_s$ and $\mathcal H_s$ are independent hence the random variables $X_s$ and $X_t-X_s$ are independent.
The same argument shows the stronger result that, for every $s$, the processes $(X_t-X_s)_{t\geqslant s}$ and $(X_t)_{t\leqslant s}$ are independent.
regarding your answer, I wonder about the following: since $Z_T:=\mathcal{E}(\phi\bullet W)$ is a martingale, you can define an equivalent measure $\frac{dQ}{dP}=Z_T$. For option pricing, you have often something like $E[\frac{Z_T}{Z_t}\mathbf1_A|\mathcal{F}_t]=E_Q[\mathbf1_A|\mathcal{F}_t]$, where $A=\{\frac{Z_T}{Z_t}>\frac{K}{Z_t}\}$ for a constant $K$. Trivially $\frac{K}{Z_t}$ is $\mathcal{F}_t$ measurable, but is $\frac{Z_T}{Z_t}$ again independent of $\mathcal{F}_t$ under $Q$? Then one could easily calculate this conditional expectation. And if so, why? – math Feb 10 '13 at 11:49