So, I've just started looking into financial mathematics and the following question keeps bugging me. From what I understood, if the market is arbitrage-free and a given contingent claim of value $h$ is attainable, then there is a measure $Q$ such that the discounted asset prices $(\tilde{S}_t)$ form a martingale. Consequently, the discounted values of the portfolio $(\tilde{V}_t)$ also form a martingale. As such, it makes sense to write:

\begin{align} V_t = e^{-r(T-t)} E^Q[V_T|\mathcal{F}_t] = e^{-r(T-t)} E^Q[h|\mathcal{F}_t] \end{align}

And then we consider the fair value of the option at time $t$ to be $V_t$.

Now, I was reading another author and he writes: "The time-$t$ price of a European call on a non-dividend paying stock with spot price $S_t$, when the strike is $K$ and the time to maturity is $\tau = T − t$, is the discounted expected value of the payoff under the risk-neutral measure $Q$." Hence,

\begin{align} C_t = e^{-r\tau}E^Q[h] = e^{-r(T-t)}E^Q[\max(S_T-K,0)] \end{align}

My question is: Why can we take the "usual" expectation when computing $C_t$, instead of using the conditional expectation to $\mathcal{F}_t$? Are they the same? If so, I'm not seeing why... Any help is appreciated. Thanks in advance.

  • 3
    $\begingroup$ People sometimes omit the conditioning $\mathcal{F}_t$ out from the formula but still mean the same thing. Importantly when calculating the time $t$ price of the option one conditions on the stock price being at some $S_t$. $\endgroup$
    – fes
    Jul 3 at 21:55
  • $\begingroup$ Hmm... But he continues that derivation and ends up with an expression of the form $C_t = S_t P_1 - Ke^{-r(T-t)} P_2$, where $P_1$ and $P_2$ are probabilities. The author writes that $P_2 = E^Q(1_{S_T>K})$, for instance. This makes sense if we're talking about the "usual" expectation, since $E^Q(1_{S_T>K}) = Q(\{S_T>K\})$. But it doesn't make sense for conditional expectation, right? At least, I see $E^Q[1_{S_T>K}|\mathcal{F}_t]$ as a random variable... $\endgroup$
    – Oscar
    Jul 4 at 13:54
  • $\begingroup$ That formula is standard; the expectations are conditional. From time $t$ perspective the conditional expectation is not a random variable. $\endgroup$
    – fes
    Jul 4 at 14:08
  • $\begingroup$ I'm sorry @fes, but I'm not understanding what you mean by "From time 𝑡 perspective the conditional expectation is not a random variable." Why is $E^Q[1_{S_T > K}| \mathcal{F}_t]$ a probability? Let's assume that $\mathcal{F}_t = \sigma(S_t,v_t)$, where $v_t$ is the volatility random variable. (I'm studying this in the context of stochastic volatility models btw) $\endgroup$
    – Oscar
    Jul 4 at 14:14
  • 2
    $\begingroup$ Oh my bad, I simply didn't realise that, generally speaking, we have that $E[1_A | \mathcal{F}] = P(A|\mathcal{F})$. Thank you! $\endgroup$
    – Oscar
    Jul 4 at 14:46


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