Expanding the Leisen–Reimer Tree Probability for Convergence Analysis

Summary

The document asks about an asymptotic expansion used in a paper on convergence of European option prices under binomial trees. It presents an expression for the inverse function used in the Leisen–Reimer construction, followed by a claimed large-step expansion in powers of the step count. It then substitutes the odd-step relation between the tree size and a second index, and asks why its own exponential expansion does not reproduce the displayed coefficients.

The material is a derivation question rather than a complete explanation: it does not include the questioner’s intermediate algebra, a resolution, or numerical checks. The expressions provide a concrete starting point for checking the expansion, including the square-root term and the conversion from the original step variable to the odd-step parameter. Readers should verify the asymptotic substitutions and coefficient matching independently; the document does not establish which line contains an error or whether the displayed expansion is correct.

Key ideas

• The question concerns an asymptotic expansion of the inverse mapping used in a Leisen–Reimer binomial tree.
• The displayed expansion expresses the probability in powers of the tree step count.
• The question then substitutes the odd-step relation to rewrite the expression using a different index.
• No derivation or resolution is supplied, so the coefficients and substitutions remain to be checked.

Tags

# Second order convergence for the Leisen-Reimer tree


# Second order convergence for the Leisen-Reimer tree












I have a question about this paper "Achieving higher order convergence for the prices of European options in binomial trees" by Mark Joshi, (Link: https://papers.ssrn.com/sol3/papers.cfm?abstract_id=976561 ). In section 9 he proves that the Leisen-Reimer tree has second order convergence. In the following, an exponential expansion has been used:

$h^{-1}(z) = 0.5 + 0.5 \left[ 1-\exp \left( - \left( \frac{z}{n + \frac{1}{3} + \frac{0.1}{n+1}} \right)^2 \left( n+ \frac{1}{6} \right) \right) \right]^{\frac{1}{2}} \\ = \frac{1}{2} + \frac{1}{2} \frac{z}{\sqrt{n}} - \frac{z}{8n^{3/2}} - \frac{z^3}{8n^{3/2}} + \mathcal{O}(n^{-5/2})$

Furthermore, we have that $n=2k+1$. So we get

$h^{-1}(z) = \frac{1}{2} + \frac{z}{2\sqrt{2}k^{1/2}} - \left( \frac{3}{16 \sqrt{2}} + \frac{1}{16 \sqrt{2}}z^3 \right) \frac{1}{k^{3/2}} + $

When I use exponential expansion, I am not able to get to the same result. Are there anyone who has an idea?

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