See the solution to Exercise 7.6 here.

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The solution calculates $E^Q (S(T_1)/S(T_0))$ and then just plugs that into the risk neutral valuation formula. But why? The risk neutral valuation formula holds for claims that depend on $T$ alone. Why can it be used also for this case where the claim depends on one more date?

My solution was similar, but I still don't trust it. How can I use a theorem which solves a different problem than this one, where the claim depends on multiple dates?


2 Answers 2


The valuation formula for a contingent claim delivering a payoff at $T$, as seen of today $t$ knowing that the underlying is currently worth $s$ reads $$ \Pi(t,s) = e^{-r(T-t)} \Bbb{E}^\Bbb{Q} \left[ f(S_T) \mid \mathcal{F}_t \right] $$ where $f(S_T)$ is the $\mathcal{F}_T$-measureable payoff of your contingent claim.

In the exercise you mention, we have $t < T_0 < T_1=T$ along with $f(S_{T_1}) = \frac{S_{T_1}}{S_{T_0}}$.

Because $f(S_{T_1})$ is $\mathcal{F}_{T_1}$ measurable (at $T_1$ you "known" both $S_{T_0}$ and $S_{T_1}$) then you can seamlessly use the standard valuation formula to write:

$$ \Pi(t,s) = e^{-r(T_1-t)} \Bbb{E}^\Bbb{Q} \left[ \frac{S_{T_1}}{S_{T_0}}\mid \mathcal{F}_t \right] $$


The spot price process is driven by a constant coefficient geometric Brownian motion. Thus, the ratio $S \left( T_1 \right) / S \left( T_0 \right)$ is

  1. independent of $\mathcal{F} \left( T_0 \right)$ and
  2. its distribution only depends on the length of the time interval $T_1 - T_0$.

It follows that

\begin{equation} S \left( T_1 \right) / S \left( T_0 \right) \sim S \left( T_1 - T_0 \right) / S(0), \end{equation}

where $\sim$ denotes equality in distribution. I.e. the solution is the same as the one that only involves the single date $T_1 - T_0$. Note that in the general case where the payoff function is given by $f \left( S \left( T_0 \right), S \left( T_1 \right) \right)$, this might no longer hold.


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