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I encounter a problem to understand this:

The price of a forward option is : $C(K,t,T)=\mathbb{E}[((S_{T}/S_{t})-K)+]$ OK

The option should only depend on $T-t$ because the yield randomness (for a week) in 2 years should be the same as the yield randomness (for a week) in 3 years and that will be the same as the yield randomness (for a week) today if no known events are expected WHY ?

Then taking $X_{t}=ln(S_{t}/S_{0})$ we should have: for all $u$ and for all $a$, $X_{u+a}-X_{u}=X_{a}$ (equality in distribution) WHY ?

Thank you

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  • $\begingroup$ This appears a forward start option. For a forward option, you may refer to quant.stackexchange.com/questions/17153/…. $\endgroup$ – Gordon May 6 '15 at 13:24
  • $\begingroup$ I don't understand why the link you provided would help me... I search on google "forward start option" but i found nothing that would help me to understand the last 2 sentences of my first post. For the first one: is it related to the Black-Scholes model ? For the second one; how do we conclude with the stationary increments ??? Thank you $\endgroup$ – glork May 6 '15 at 13:42
  • $\begingroup$ below may answered partial of your question. $\endgroup$ – Gordon May 6 '15 at 13:59
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For the last question. We assume that \begin{align*} S_t = S_0 e^{(r-q-\frac{1}{2}\sigma^2)t + \sigma W_t}, \end{align*} where $W$ is a standard Brownian motion, $r$ is the interest rate, $q$ is the dividend yield, and $\sigma$ is the volatility. Then, \begin{align*} X_{u+a}-X_a &= (r-q-\frac{1}{2}\sigma^2)a + \sigma(W_{u+a}-W_u)\\ &\sim (r-q-\frac{1}{2}\sigma^2)a + \sigma W_a\\ &= X_a. \end{align*}

For the forward start option, note that \begin{align*} S_T/S_t &= e^{(r-q-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T- W_t)}\\ &= e^{(r-q-\frac{1}{2}\sigma^2)(T-t) + \sigma \sqrt{T-t}\xi}, \end{align*} where $\xi$ is a standard normal random variable. Then \begin{align*} C(K, t, T) &= e^{-rT} \mathbb{E}\big(S_T/S_t -K)^+ \big)\\ &= e^{-rT}\big[N(d_1) - KN(d_2) \big], \end{align*} where $N$ is the cumulative distribution function of a standard normal random variable, \begin{align*} d_{1} = \frac{\ln (1/K) + (r-q+ \frac{1}{2}\sigma^2 )(T-t)}{\sigma \sqrt{T-t}}, \end{align*} and \begin{align*} d_{2} = \frac{\ln (1/K) + (r-q- \frac{1}{2}\sigma^2 )(T-t)}{\sigma \sqrt{T-t}}. \end{align*}

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