Consider a two-period binomial model for a risky asset with each period equal to a year and take $S_0 = 1$, $u = 1.03$ and $l = 0.98$. How do you price a look-back option with payoff($\max_{t=0,1,2}S_t - 1)_{+}$? Hint: The price is path dependent and each path has a payoff.

I don't understand this question at all, any suggestions is greatly appreciated


For a standard European option (i.e. non path - dependent payoff):

$$V_0 = \frac {1}{1+R} E [ V (S_T) ] $$

Because, in a 2 period binomial tree, the terminal stock price $S_T$ can take 3 distinct values: $S_{uu}=S_0u^2$, $S_{ul}=S_0ul$ and $S_{ll}=S_0l^2$, you can write the expectation:

$$V_0 = \frac {1}{1+R} (q_{u}^2 V (S_{uu}) + {\color {red}{2}} q_u q_l V ( S_{ul} )+ q_l^2 V (S_{ll}))$$

With $q_u $ and $q_l $ figuring risk-neutral probabilities of an up/low move over a period:

$$q_u = \frac{(1+R) -l}{u-l}$$ $$q_l = 1-q_u $$

and the terminal option values $V (S_{uu})$, $V (S_{ul})$, $V (S_{ll})$ can be determined through the payoff function, for instance $$ V (S_T) = \max ( S_T - K , 0 ) $$ for a call option.

For a path dependent option it's almost the same, except $V (.) $ is now a function of the whole path $\{S_0,S_{T-1}, S_T\}$, not just $S_T $

$$V_0 = \frac {1}{1+R} E [ V (S_0,S_{T-1},S_T) ] $$

Because, in a 2 period binomial tree, you have 4 possible paths: up/up, up/low, low/up, low/low, you can write the expectation as:

$$V_0 = \frac {1}{1+R} (q_{u}^2 V (S_0, S_u, S_{uu}) + q_u q_l V ( S_0, S_u, S_{ul} ) + q_l q_u V ( S_0, S_l, S_{ul} ) + q_l^2 V (S_0, S_l, S_{ll}))$$

With, as given in the text: $$ V (S_0, S_{T-1}, S_T) = \max ( \max ( S_0, S_{T-1}, S_T) - K, 0) $$ for a call with lookback feature.

Note how in the non path - dependent case the paths up/low and low/up both finish with the same stock value, hence can be aggregated, hence the factor $ {\color {red}{2}}$.

  • $\begingroup$ The question asks "how would you price" not "price the payoff" hence we just need to show how one would do it correct? $\endgroup$ – Wolfy Mar 17 '16 at 16:23
  • $\begingroup$ "How do you price it?".. The way I answered. That's all there is to it. $\endgroup$ – Quantuple Mar 17 '16 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.