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


1 Answer 1


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, 2016 at 16:23
  • $\begingroup$ "How do you price it?".. The way I answered. That's all there is to it. $\endgroup$
    – Quantuple
    Mar 17, 2016 at 20:09

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