Binomial Tree Option Pricing Requires Path-Count Weights

Summary

The document diagnoses an error in a currency option valuation using a six-step binomial tree. The calculation sums discounted payoffs across terminal nodes but weights each node only by the risk-neutral probability of one particular sequence of up and down moves. Since multiple sequences can reach the same terminal node, that probability alone understates the node's total probability.

The correction is to multiply each sequence probability by the binomial coefficient counting how many paths reach that node. For the six-step tree, the answer notes that the second and third terms require factors of six and fifteen, respectively. This addresses the missing probability weights in the stated payoff sum. The response does not recalculate the option price or independently verify the tree parameters, payoff setup, or currency conventions, so it resolves the identified combinatorial issue rather than validating the full valuation.

Key ideas

• A terminal node in a binomial tree can be reached by multiple sequences of up and down moves.
• The probability weight for a terminal payoff must include the number of paths reaching that node.
• Binomial coefficients provide the path counts used to aggregate risk-neutral probabilities.
• Correcting path weights does not by itself validate the other assumptions in an option valuation.

Tags

# Put-Call Parity on Currency and Binomial Trees


# Put-Call Parity on Currency and Binomial Trees












I tried solving the below problem without knowing the shortcut of thinking about this in terms of a put versus a call. I can't seem to arrive at the correct answer using my method and I'm wondering where I'm going wrong.

$u = e^{(0.03 - 0.05)/12 + 0.05\sqrt{1/12}} = 1.012848937$

$d = e^{(0.03 - 0.05)/12 - 0.05\sqrt{1/12}} = 0.984028496$

$p^* = \frac{1}{1 + e^{0.05\sqrt{1/12}}} = 0.496391623$

The call pays off at nodes: $u^6, u^5d, u^4d^2, u^3d^3, u^2d^4, u^1d^5$.

The desired price is then:

$\big[(0.8u^6 - 1.35^{-1}){p^*}^6(1 - p^*)^0 + (0.8u^5d^1 - 1.35^{-1}){p^*}^5(1 - p^*)^1 + (0.8u^4d^2 - 1.35^{-1}){p^*}^4(1 - p^*)^2 + (0.8u^3d^3 - 1.35^{-1}){p^*}^3(1 - p^*)^3 + (0.8u^2d^4 - 1.35^{-1}){p^*}^2(1 - p^*)^4 + (0.8u^1d^5 - 1.35^{-1}){p^*}^1(1 - p^*)^5\big]$

$= 0.005760287$, multiplied by the discount factor $e^{-0.03 \cdot 0.5}$ and $\$1,000,000$ to arrive at $5,760.28$ euros.

## Answer by spaceisdarkgreen (score 4, accepted)

https://quant.stackexchange.com/a/36114

You have forgotten the combinatorial factors for binomial probabilities on your terms. You need $$ {n\choose k} p^n(1-p)^{n-k},$$ not just $$ p^n(1-p)^{n-k}.$$ The second term should have a factor of $6$ and the third should have a factor of $15,$ etc.

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.