Conditional Expectation of lognormal distributed process

Question

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

Answer 1

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.