In a 4 period binomial model, I have a lookback put option that pays $\left [M_{4}-4 \right ]^{+}$, where $M_{4}$ is the maximum price reached during the sequence of 4 trials.
Lets say the starting price = 4, the number of trials = 4. The up-factor is 2 and down factor 1/2.
My idea to find the expected value of the option was to first find the number of paths that only reached (and did not exceed) each level, so:
- Level 1 or a price of $M_{4}=8$, I would have $ n \cdot \left [ 8-4 \right ]$ payoffs for this level.
- Level 2 or a price of $M_{4}=16$, I would have $ o \cdot \left [ 16-4 \right ]$ payoffs for this level.
- Level 3 or a price of $M_{4}=32$, I would have $ p \cdot \left [ 32-4 \right ]$ payoffs for this level.
- Level 4 or a price of $M_{4}=64$, I would have $ 1 \cdot \left [ 64-4 \right ]$ payoffs for this level.
Here's the problem
Take for example Level 1:
- P(equaling level 1 at T=4) = 0, there are no paths that end at 1 after 4 trials.
- P(reaching level 1, but ending at or below level 1) = $\frac{4!}{(4-3)!3!} + \frac{4!}{(4-4)!4!}=5$
However one of those 5 paths exceeds level 1, even though the path ends at level 1.
Hopefully the graph below illustrates the problem. Both paths end at level 0 (or 4), but the pink path breached level 1 and its payoff would be (16-4) instead of (8-4) for the green path.
Naturally if the number of periods grow, there could be many more than 1 path that does this.
Said another way, how do I calculate the number of red paths in an N period model?