Conditional Expectation of lognormal distributed process

Suppose you have a process $X=\{X_t\}$ for $0\le t\le T$. And each $X_t$ is lognormal distributed. How do you calculate these two expressions:

$$E[(X_T-K)^+|\mathcal{F}_t]$$ and $$E[(X_T-K)^+]$$

Where $\{\mathcal{F}_t\}$ for $0\le t\le T$ is a filtration, such that $X_t$ is $\mathcal{F}_t$ measurable. Is there a general way / formula to compute this.

Motivation: If $X$ is a price process then this is the arbitrage free price (at $t$) of a European call option with maturity $T$ and Strike $K$. I can calculate this for special models, like Black-Scholes. However, I wondered if one could also obtain a formula for a general lognormal distributed process.

hulik

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General lognormal distributed process has general interdependence structure between its random variables, e.g. you can come up with the process such that $X_T$ is independent of $\mathcal F_t$, thus the conditional expectation would coincide with the unconditioned one. The latter one is the same for all processes which $X_T$ are lognormal - so you can use BS formula. –  S.D. Nov 29 '12 at 15:15
@S.D. How exactly do I see, that $X_T$ is independent of $\mathcal{F}_t$? Why can I just use the BS formula? Could you turn your comment into an answer, please? Then I would see how the details work out and also accept it. Thanks for your help. –  hulik Nov 29 '12 at 16:03
Let $(X_t)_{t\geq 0}$ be a stochastic process with values on $\mathbb R$ and say cadlag paths. Let us also require that for each $t\in \mathbb R_+$ the single random variable $X_t$ follows the lognormal distribution with some parameters. Then $X_T$ follows this distribution as well. To compute $$\mathsf E[(X_T - K)^+] \tag{1}$$ you only need to use the distribution of $X_T$, and it does not matter what are the distributions of $X_t$ for $0\leq t<T$. Thus, if you know how to compute $(1)$ for the case of the geometric Brownian motion as per BS model, the same formula also holds for all other processes.
Now, I guess that was also in your premise that there are many processes $X$ as defined above. Why is it so? The reason is that to define a stochastic process uniquely, you have to define all its possible finite dimensional distribution (fdd), i.e. the joint distributions of $(X_{t_1},\dots,X_{t_n})$. In our case, such distributions are not defined, as we talked before only about the distributions of single random variables $X_t$. Due to this reason, $\mathsf E[(X_T - K)^+|\mathcal F_t]$ depends on the choice of the fdd.
For example, let all $X_t$ be iid random variables, then $X_T$ is independent of $\mathcal F_t$ for all $0\leq t< T$ and thus $$\mathsf E[(X_T - K)^+|\mathcal F_t] = \mathsf E[(X_T - K)^+]$$ which is not the case e.g. in the BS model.
Thank for your answer. But my problem is, in the BS case, i.e. geometric Brownian motion, there we use explicitly that Brownian Motion has independent increments and that there are normally distributed. For a general $X_T$ lognormal distributed, we don't have a Brownian motion. So my question reduces to: Why "General lognormal distributed process has general interdependence structure between its random variables". Can you give me a reference for this statement? –  hulik Nov 29 '12 at 16:53
@hulik: look, you know that $X_T$ has a lognormal distribution and you are able to compute an expectation of a function of it. Does it matter, which method you use for the computation? –  S.D. Nov 29 '12 at 17:02