Say an option with five years left before maturity has a value of $x$ today. Theoretically, under the B/S framework, what is its expected value in five years (upon maturity)? Do we assume it will simply grow in line with the five-year risk free rate? Can you please also provide any reference I can cite for this?


or $E(v)_{t=5}=xe^{5R_f}$ using a continuously compounded $R_f$

Is this right? If so, is it only correct under a risk-neutral framework? And can we say $R_f$ is the drift in this case (is it proper semantics)?

Thank you for your help!

Edit: on second thought, this seems somewhat flawed if you're deep out-of-the-money... Really not sure how to proceed. Basically my question is what is the expected value of the option at maturity when you're at time t?


OP is absolutely right in his approach and this is the underlying idea behind risk neutral valuation or even BS model. If Black-Scholes model assumptions hold, then a derivative payoff can always be replicated in such a way it would never provide return more than risk free interest rate, otherwise it will lead to arbitrage opportunities. But assumption never holds in reality and we find deviation in actual price from BS price.

But theoretically, you are right.

EDIT : The OP has asked about the expected change in option price not the actual change. Under the risk neutral measure$(\mathbb{Q})$, the price of option $v_t$ at time $t$ is given by $$v_t= e^{-r(T-t)}\mathbb{E}_{\mathbb{Q}}\big[(S_T-K)^+\big]$$ where, $\mathbb{E}_{\mathbb{Q}}\big[(S_T-K)^+\big]$ is expected payoff at maturity and can be written as $$\mathbb{E}_{\mathbb{Q}}\big[(S_T-K)^+\big]=v_te^{r(T-t)}$$ which show that option price is expected to increase at risk free rate as pointed out in OP.

Formal Derivation $$v_t= e^{-r(T-t)}\mathbb{E}_{\mathbb{Q}}\big[(S_T-K)^+|\mathscr{F_t}\big]\quad \tag{1}$$

Let's suppose $t_1 \in (t,T]$, so option price at $t_1$ $$v_{t_1}= e^{-r(T-t_1)}\mathbb{E}_{\mathbb{Q}}\big[(S_T-K)^+|\mathscr{F_{t_1}}\big]\quad \tag{2}$$

But we don't have history of an process upto time $t_1$ and we are still at time $t$, so $v_{t_1}$ at time $t$ is $$v_{t_1}|\mathscr{F}_t= e^{-r(T-t_1)}\mathbb{E}_{\mathbb{Q}}\big[(S_T-K)^+|\mathscr{F_t}\big] \quad \tag{3}$$

Dividing equation 3 from equation 1, we get $$\frac{v_{t_1}}{v_t}=\frac{e^{-r(T-t_1)}}{e^{-r(T-t)}}$$ $$v_{t_1}=v_te^{r(t_1-t)}$$ where $v_{t_1}$ is expected option price at time $t_1$ given at time $t$.


The expected value of the option at maturity is simply $$\mathbb{E}[(S_T-K)^+]$$ Note that this is under the real world measure. In a B-S framework this value is given by $$e^{rT}C(\alpha;S_0, K, \sigma, T)$$ Where $C(r; S_0, K, \sigma, T)$ is the B-S call option price. Hence the expected growth rate (using a simple return) is $$\frac{e^{rT}C(\alpha;S_0, K, \sigma, T)-C(r; S_0, K, \sigma, T)}{C(r; S_0, K, \sigma, T)T}$$ However the expected continuously compounded return cannot be so easily solved since the option price is non-linear in the growth rate and Jensen's inequality comes into play.


The "best" approximation comes from the one of the greeks: $\Theta$. This is the derivative of the option value with respect to time-to-maturity. For a call option it is given by:

$$\Theta(\tau) = -\frac{\sigma}{2\sqrt{\tau}}S\phi(d_+) - r K e^{-r\tau} \Phi(d_+)$$

where $\tau = T - t$ is the time to maturity, $t$ is the current time, $T$ is the maturity and you are probably familiar with the other symbols (if not check the wiki).

The approximate change in value of the call option is then

$$ C(t+\delta t) \approx C(t) + \delta t \frac{\partial C}{\partial t} = C(t) - \delta t \Theta(\tau)$$

This is the first order approximation and it neglects the change of the underlying spot price. So if we assume that the stock price does not change, then this is a good approximation to the change in option value.

In reality we cannot actually ignore the underlying. The stock price will fluctuate up and down, and this will be reflected by time dependency of the option value.


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