Lookback Option Exercise Probability in a CRR Binomial Model
Summary
The document asks how to calculate the probability that a discrete-time lookback put finishes in the money in a Cox–Ross–Rubinstein binomial model. Its payoff depends on the minimum underlying price observed along the path, rather than only the terminal price. The response proposes identifying how many down moves are needed to push the minimum below the strike, then summing binomial probabilities for paths that meet that threshold.
This gives a general counting approach when the path minimum can be characterized by the down-move count, using the model’s up probability and number of steps. However, the response’s claim that the path minimum is determined solely by the total number of down moves is not generally valid: the order of up and down moves can affect the minimum reached along a path. The formula therefore needs a path-state check or a model-specific justification before use. The document supplies a short illustrative tree but no broader validation.
Key ideas
- A lookback option’s payoff depends on the minimum price encountered along the path.
- The response seeks a threshold number of down moves and sums binomial probabilities to estimate the in-the-money probability.
- The model’s up probability and number of steps enter the proposed calculation.
- The path minimum can depend on move order, so total down moves alone may not determine it in general.
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Full text
# What is the probability of a lookback option ending in the money (CRR-model)
# What is the probability of a lookback option ending in the money (CRR-model)
I would like to compute the probability that a certain lookback option ends in the money, let's say that the option has the following payoff $h_N=\max\left\{0,K-\min\{S_1,...,S_N\}\right\} $ where $K$ is a fixed strike price and $S_1$ up to $S_N$ denote the discrete values of the underlying from time $1$ to $N$. Of course, the $S_0$, $U$, $D$ and the probability of up movement $\mathbf{P}(S_n/S_{n-1}=U)$ is given. I can see that the answer heavily relies on those values but I want a general approach.
Thanks in advance
## Answer by Kermittfrog (score 1)
https://quant.stackexchange.com/a/51481
In a CRR binomial model, it would seem that the path-wise minimum is a function of the total number of down moves along that singular path.
For example, let us fix $N=3$, resulting in $2^N=8$ possible paths that arrive at four $N+1=4$ possible states at maturity.
Along each path, it suffices to note that the minimum along that path is simply defined by the number of down moves along that path. For example,
$$ \min{S_N(UUU)} =S_0 D^0 = S_0 $$ $$ \min{S_N(UUD)}=\min{S_N(UDU)}=\min{S_N(DUU)}=S_0 D^1 = S_0 D $$ $$ \min{S_N(UDD)}=\min{S_N(DUD)}=\min{S_N(DDU)}= S_0 D^2 $$ $$ \min{S_N(DDD)}= S_0 D^3 $$
Thus, to find the probability that the option ends in the money, we simply collect all paths for which $\min{S_N} \leq X$. Equivalently, how many $D$ steps $N_D$ are (at least) required to be 'in the money'?
$$ X \geq SD^{N_D} \Leftrightarrow N_D = \frac{\ln(X/S)}{\ln D} $$ rounded up.
The probability for being in the money thus is:
$$ p_{ITM} = \sum_{k=0}^{N_D}\binom{N}{k}p^{k}(1-p)^{N-k} $$ with $p$ defined as usual.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.