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


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.

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  • $\begingroup$ Thanks for answering, but if i am not mistaken the number of downs is not sufficient to find the minimum since the "place" the downs appear affect the min, see this example lets say we have one down movement and two ups then we can have that $S_N=(110,121,108.9)$ for U=1.1 D=0.9 , and in this case the min is 108.9 or say the first move is down i.e $S_N=(90,99,108.9)$ in this case you see that the min is different. Did i mis-understood your answer? $\endgroup$ – Stelios Kounis Mar 5 at 9:53
  • $\begingroup$ To my understanding, in the CRR model we have U=e^{\sigma\sqrt{\Delta t}}=1/D , thus the timing is irrelevant. $\endgroup$ – Kermittfrog Mar 5 at 19:37
  • $\begingroup$ The timing is clearly not irrelevant if you look at the example above, where we indeed have U=1/D. Also, would your calculations above not be for the risk neutral probability of ending up in the money? I imagine that is all we can hope to calculate, but might be worth noting the distinction. $\endgroup$ – Oscar Apr 3 at 22:32
  • $\begingroup$ Comment removed $\endgroup$ – Kermittfrog Apr 4 at 5:22

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