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:

  1. Level 1 or a price of $M_{4}=8$, I would have $ n \cdot \left [ 8-4 \right ]$ payoffs for this level.
  2. Level 2 or a price of $M_{4}=16$, I would have $ o \cdot \left [ 16-4 \right ]$ payoffs for this level.
  3. Level 3 or a price of $M_{4}=32$, I would have $ p \cdot \left [ 32-4 \right ]$ payoffs for this level.
  4. 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:

  1. P(equaling level 1 at T=4) = 0, there are no paths that end at 1 after 4 trials.
  2. 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.

enter image description here

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?

  • $\begingroup$ @Glen_b could you please move to quant.SE, it seems the question will have a better chance of being answered there. Thanks. $\endgroup$ May 19 '17 at 0:23
  • $\begingroup$ To solve that I believe you need the mirror principle described for example in 4.1. to 4.4 of www2.math.uu.se/~sea/kurser/stokprocmn1/slumpvandring_eng.pdf $\endgroup$
    – Ami44
    May 20 '17 at 10:57
  • $\begingroup$ @Ami44 that's a handy link, thanks for that. I have solved this problem (not sure if it's the most efficient way), I'll post the solution in a few days. Thanks again for the link. $\endgroup$ May 22 '17 at 12:45

The general formula to answer this question can be found on page 105-106 of Introduction to Mathematical Finance by Pliska.

In general: $\bar{\mathbb{P}}\left ( M_{4} \geq 4\left ( 2^{i} \right ) \right )$ (or the probability the maximum to date price was 4, 8, 16 etc) is equal to:

  • the probability of the stock price finishing at $M_{4}$; plus
  • the probability the stock price exceeding $M_{4}$; plus
  • the probability the stock price reached $M_{4}$ at some point but finished below $M_{4}$.

The formula to evaluate this can be found on the pages of Pliska's book mentioned above.

However to answer this question, what needs to be calculated is $\bar{\mathbb{P}}\left ( M_{4} = 4\left ( 2^{i} \right ) \right )$ for each level $i$. The only levels of $i$ that contribute value to the option are for $1 \leq i \leq 4$. The probability of the maximum being exactly $i$ can be calculated via:

  • $\bar{\mathbb{P}}\left ( M_{4} \geq 4\left ( 2^{i} \right ) \right ) - \bar{\mathbb{P}}\left ( M_{4} \geq 4\left ( 2^{i+1} \right ) \right )$.

Once again, using the formulas in Pliska's book, $\bar{\mathbb{P}}\left ( M_{4} = 8\right )$ is given by:

  • $\bar{\mathbb{P}}\left ( M_{4} \geq 8\right ) - \bar{\mathbb{P}}\left ( M_{4} \geq 16 ) \right )$ = 25.92%
  • $\bar{\mathbb{P}}\left ( M_{4} \geq 8\right )$ = 49.6%
  • $\bar{\mathbb{P}}\left ( M_{4} \geq 16\right )$ = 23.68%

In this example I have assumed r =10% so the risk neutral probability of an up move is 40%.

Therefore there is a 25.92% chance of receving $(8-4)=4$

The same calculations need to be done for $M_{4} = 16,32$ and $64$ to determine the price of the option.


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