I have read the pricing procedure of a Forward-start option in a Black-Scholes world in Musiela-Rutkowski, but I don't find their proof clear (pp. 195-6). Let me summarize their argument:

Consider two dates $T_0 < T$. A forward-start call option allows the holder to receive, at time $T_0$ and with no additional cost, a call option expirying at $T$, with strike set equal to $S(T_0)K$, for some $K>0$. So, the option life starts at $T_0$, but the holder pays at time $0$ the price of the option.

So, let us see how to price such a contract. First, introduce the terminal payoff $$ FS(T)\colon = (S(T) - KS(T_0))^+ $$ and to find its price at time $0$, let us start by considering its value at time $T_0$. This is easily found to be $$ FS(T_0) = c(S(T_0), T-T_0, KS(T_0)). $$

At this point we see that, after some easy algebraic manipulation, we have $$ c(S(T_0), T-T_0, KS(T_0)) = S(T_0)\cdot c(1, T-T_0, K) $$

Here my problems begin. What exactly is it meant by the last symbol $c(1, T-T_0, K)$??

It is true that algebraically the above relation makes sense, $1$ stands for the value at time $T_0$ and the strike is $K$, but, what exactly means? The Black-Scholes formula for the price of an option on... what exactly? An asset whose value at $T_0$ is 1, and what is this asset??

Next, the proof proceeds like that: since $c(1, T-T_0, K)$ is nonrandom, the option's value at time $0$ equals $$ FS(0) = S(0)\cdot c(1, T-T_0, K) = c(S(0), T-T_0, KS(0)). $$ Why? I argued like that: the price at time $0$ of the contract should be, in the risk-neutral measure, the value $$ FS(0) = \tilde{\mathbf E}[D(T_0)\cdot c(S(T_0), T-T_0, KS(T_0))]=\tilde{\mathbf E} [D(T_0)S(T_0)\cdot c(1, T-T_0, K)], $$ where $D(T_0)$ is the discount factor at time $T_0$ (with constant interest rate). Since $c(1, T-T_0, K)$ is a constant (so I have guessed), can take it out of the Expectation symbol and obtain $$ \tilde{\mathbf E} [D(T_0)S(T_0)c(1, T-T_0, K)]= c(1, T-T_0, K)\tilde{\mathbf E} [D(T_0)S(T_0)] = c(1, T-T_0, K)S(0) $$ the last relation because $D(T_0)S(T_0)$ is a martingale. Is it this correct? It is my understanding that the value $S(T_0)$ is not known at time $t=0$, so we are treating it as a random variable, hence, it makes sense to take its discounted expectation back to time $0$, while the expression $c(1, T-T_0, K)$ (whose meaning it is unclear to me as I wrote before), it is just a number, since it is calculated at time $T_0$ when all the quantities which appear in $c(1, T-T_0, K)$ are determined.

I would appreciate if in your opinion this proof is ok, and what is your answer to the question I wrote in bold. Thanks in advance.


1 Answer 1


The proof is fine. For example, $D(t)S(t)$ is a martingale and then \begin{align*} E\big(D(t)S(t)\big) = S(0). \end{align*} Regarding the function $C(1, T-T_0, K)$, it is the value, at time $T_0$, of the option payoff \begin{align*} \left(\frac{S(T)}{S(T_0)} - K \right)^+. \end{align*} Here, you can treat $\frac{S(T)}{S(T_0)}$ as the normalized value or return of the asset. The identity \begin{align*} c(S(T_0), T-T_0, KS(T_0)) = S(T_0)\cdot c(1, T-T_0, K) \end{align*} can be seen directly from the Black-Scholes' formula or from the payoff equation \begin{align*} \big(S(T) - KS(T_0)\big)^+ = S(T_0) \left(\frac{S(T)}{S(T_0)} - K \right)^+. \end{align*}

  • $\begingroup$ Excellent answer and very clear and helpful, thanks. $\endgroup$
    – RandomGuy
    Commented Jul 18, 2016 at 18:06

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