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Cross posted from here.

Let $B$ be a $Q$-Brownian motion and $X^{s,x}$ given by

$$dX_t = X_t(\mu_t dt + \sigma_t dB_t),\quad X_s = x$$

for $\mu, \sigma$ deterministic. Let $\mu_{s,t}=\int_s^t \mu_u du$ and $\sigma^2_{s,t} = \int_s^t \sigma^2_u du$, i.e. $X^{s,x}_t = x\exp(\mu_{s,t} - \sigma^2_{s,t}/2 + \int_s^t\sigma(u)dB_u)$.

If $\mu_t \equiv \mu$, $\sigma_t\equiv \sigma$ are constant I know that

$$\begin{align} E_Q\left((X_t^{0,1} - K)^+\vert\mathcal{F}_s\right) & =E_Q\left((X_{t}^{s,x} - K)^+\right)\vert_{x = X_s^{0,1}}\\ & = x\exp(\mu_{s,t})\Phi(d_1(x,s,t)) - K\Phi(d_2(x,s,t))\vert_{x = X_s^{0,1}}\\ \end{align}$$

where $d_1(x,s,t) = \frac{log(x/K) + \mu_{s,t} - \sigma_{s,t}^2/2}{\sqrt{\sigma^2_{s,t}}}$ and $d_2(x,s,t) = d_1(x,s,t) - \sqrt{\sigma^2_{s,t}}$, using the Markov property and the standard computations from Black-Scholes pricing.

Question: The same formula should hold for $\mu$, $\sigma$ non-constant but how do I show that in a non-sketchy way? I tried to use markov kernels but...

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