This is a question due to the answer of Did in this post Independent increments of $X_t:=\int_0^t\phi(s) dW_s$. Precisely, we assume that the dynamics of a stock prices follows
$$dS_t=S_t(\phi_tdW_t)$$
where $\phi_t$ is a deterministic function, and assume bounded. Hence $S_t=\mathcal{E}(\phi\bullet W)=S_0e^{\int_0^t\phi_sdW_s-\frac{1}{2}\int_0^t\phi_s^2ds}$. For simplicity, $S_0=1$ and, regarding the question above, $X_t:=\int_0^t\phi_sdW_s$. Let $K>0$ be a constant, and we want to find the arbitrage free price of an European call option with strike $K$ and maturity $T$ at time $t$, i.e.
$$\pi_t:=E[(S_T-K)^+|\mathcal{F}_t]$$
As usual, with $A=\{S_T>K\}$ write this as
$$\pi_t=E[S_T\mathbf1_A|\mathcal{F}_t]-KE[\mathbf1_{A}|\mathcal{F}_t]$$
The second term can be easily calculated, using the above question, i.e.
$$KE[\mathbf1_{A}|\mathcal{F}_t]=KP[\frac{S_T}{S_t}>\frac{K}{S_t}|\mathcal{F}_t]$$
since $\frac{S_T}{S_t}$ is independent of $\mathcal{F}_t$ and $\frac{K}{S_t}$ is $\mathcal{F}_t$ measurable, we get
$$KP[\frac{S_T}{S_t}>\frac{K}{S_t}|\mathcal{F}_t]=KP[Y>\frac{K}{x}]|_{x=S_t}$$
where $Y$ is lognormal distributed with mean $-\frac{1}{2}\int_t^T\phi_s^2 ds$ and variance $\int_t^T\phi_s^2ds$. To calculate the first conditional expectation, $E[S_T\mathbf1_A|\mathcal{F}_t]$, I would define a new measure (on $\mathcal{F}_T$ through
$$\frac{dQ}{dP}:=S_T$$
since $S_T$ is a $P$ Martingale and $E[S_0]=1$ (if $S_0\not=1$, just use the measure $\frac{dQ}{dP}:=\frac{S_T}{S_0}$). Hence the density process is given by $Z_t:=E[\frac{dQ}{dP}|\mathcal{F}_t]=S_t$. Using Bayse, we get
$$S_tE[\frac{S_T}{S_t}\mathbf1_A|\mathcal{F}_t]=S_tE_Q[\mathbf1_A|\mathcal{F}_t]$$.
If we know that under $Q$, $\frac{S_T}{S_t}$ is still independent of $\mathcal{F}_t$, we could conclude in the same way as above. If this is not the case, how would you then calculate the first conditional expectation?
