Reflection Counting for Barrier-Constrained Binomial Paths
Summary
The document considers a binary option paying only when a binomial stock path ends at a specified price without ever reaching an upper barrier. Because the up and down multipliers are reciprocals, the multiplicative stock process maps to an additive walk in up and down steps. The endpoint determines the net number of up moves, so unrestricted paths can be counted with a binomial coefficient.
To remove paths that cross the barrier and later return to the target, the answer uses reflection: each forbidden path corresponds to a reflected path ending at a higher level. Subtracting the count of those reflected paths gives the number of admissible paths. The answer flags an apparent mismatch in the question’s first binomial coefficient and gives a recurrence-based hint for counting bounded walks. It does not complete the discounted option valuation; the risk-free rate affects that valuation but not the path count.
Key ideas
- With reciprocal up and down factors, the stock path can be represented as an additive walk.
- The endpoint fixes the net number of up moves and determines the unrestricted path count.
- Reflection maps barrier-crossing paths ending at the target to paths ending at a higher level.
- Subtracting reflected paths gives the count that stays below the barrier.
- The answer identifies a likely typo in the proposed binomial expression and does not calculate the option price.
Tags
Full text
# Counting random paths
# Counting random paths
Assume the path of a certain stock can be modeled using a binomial tree. The initial price of the stock at time $t=0$ is 1024. The upstage factor of the stock price is $x=1.25$ and downstage factor of the stock price is $y=0.8$. Assume that risk is simply compounded with $r=3\%$. Determine the price of the binary option that pays \$1 if the stock price after 50 steps is \$2500 and it never touches the barrier of \$3125.
Using the random path approach, please explain why the number of paths is ${50\choose24}-{50\choose28}$
## Answer by Borun Chowdhury (score 2)
https://quant.stackexchange.com/a/25283
As I mentioned above, I am not sure what the variable $r$ is. If we ignore that, or assume the questioner wanted to say its the risk free interest rate, then it has no effect on the number of paths.
Then it is clear that after 50 steps going from \$1024 to \$2500 requires a net of 4 up movements with the given $x=y^{-1}=1.25$. Thus the number of steps without the barrier constraint is $\phantom{a}^{50} C_{23}$.
We need to subtract from this the number of paths that cross \$3125 and yet end up on \$2500. For each such path, there is one that passes through \$3125 at the same place but then is reflected across the line \$3125 to end up on \$3906.25. In other words the reflected path has a net 6 up movements. An example of this method is shown below. The numbers used are for a different problem.
So the number of such paths is $\phantom{a}^{50} C_{28}$. However, recall these are the paths we need to exclude as they crossed the barrier.
Thus the number of paths consistent with ending up on \$2500 having never crossed \$3125 is
$$ \phantom{a}^{50} C_{23}- \phantom{a}^{50} C_{28} $$
This is not the same as that asked in the question because of the difference in the first term but I suspect that is a typo.
## Answer by M. Jeunesse (score 1)
https://quant.stackexchange.com/a/25260
Two hints :
The number of paths never going up to $3125$ when starting from $1024$ and stepping up by a multiplicative factor of $5/4$ and down by a multiplicative factor $4/5$
is
the same as the number of paths starting from $0$ and and stepping up by an additive factor $+1$ and stepping down by an additive factor of $-1$ and never going up to $5$
Let $E(n,m)$ be the number of paths of length $n$ such that starting from $0$ and stepping up $+1$ and stepping down $-1$, then :
$E(n,m) = E(n-1,m-1) + E(n-1,m+1)$
it is then easy to think about Pascal triangle and to look for a solution like $E(n,m)=C(n,a \times n + b \times m)$
you can verify that $E(n,m) = C(n,\frac{n+m}{2})$ will satisfy the relation shipShown 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.