Deriving a Replicating Portfolio Position by Algebra

Summary

This short explanation shows how to isolate the bond position in a one-period option replication model. Starting from the equation for the up-state payoff, it substitutes the stock position that was derived earlier, moves the stock term to the other side, and puts the terms over a common denominator. The resulting expression gives the bond holding in terms of the up- and down-state option payoffs and stock prices, divided by the difference between the state prices of the stock.

The material is an algebraic walkthrough rather than a full treatment of option pricing. It assumes the underlying two-state replication equations and the previously obtained stock holding are already known. It offers no numerical example or discussion of model assumptions, so readers need the surrounding derivation to understand why the equations represent a replicating portfolio.

Key ideas

• Substitute the previously derived stock holding into the up-state payoff equation.
• Rearrange the equation to place the bond position by itself on one side.
• Use a common denominator to combine payoff terms into a single fraction.
• The derivation relies on the two-state replication setup introduced elsewhere.

Tags

# How does this book (Financial Theory w/ Python) arrive at the solution at the bottom?


# How does this book (Financial Theory w/ Python) arrive at the solution at the bottom?












I am trying to work through understanding this but I do not know how they got to the solution at the bottom (b*). Any help?

## Answer by TickaJules (score 2)

https://quant.stackexchange.com/a/69251

Just algebra. Plug their $s^*$ into the first of the 2 equations $b\cdot B_1...$ then move things around so that $b$ is alone on the left hand side.

Like so: $$b \cdot B_1 + \Big( {{C^u_1-C^d_1}\over{S^u_1-S^d_1}} \Big) S^u_1 = C^u_1,$$ then $$b \cdot B_1 = C^u_1 - \Big( {{C^u_1-C^d_1}\over{S^u_1-S^d_1}} \Big) S^u_1,$$ then $$b \cdot B_1 = C^u_1 \Big( {{S^u_1-S^d_1}\over{S^u_1-S^d_1}} \Big) - \Big( {{C^u_1-C^d_1}\over{S^u_1-S^d_1}} \Big) S^u_1,$$ then $$b \cdot B_1 = {{C^d_1 S^u_1-C^u_1 S^d_1}\over{S^u_1-S^d_1}}$$ 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.