Using Put-Call Parity to Scale a Linear Option Value Equation

Summary

The post asks why a combination of put and call prices at two strike prices can be valued at two thirds of a stated total. It is a question about interpreting an algebraic result obtained from put-call parity, rather than about an options trading strategy.

The accepted answer explains that the first equation has coefficients 165 and 3, while the target expression has corresponding coefficients 110 and 2. Each target coefficient is two thirds of its counterpart, so the entire left-hand expression is two thirds of the first equation's left-hand side. By equality, its value is also two thirds of the first equation's right-hand side, yielding the stated result. The reasoning relies on both terms scaling by the same factor; it does not require adding two equations out of three. The excerpt illustrates proportionality in a linear equation but does not provide broader option-pricing discussion.

Key ideas

• The target expression uses coefficients that are each two thirds of those in the given equation.
• Scaling every term on one side by the same factor scales the equation's value by that factor.
• The two-thirds result follows from proportional coefficients, not from selecting two of three equations.
• The example explains algebraic reasoning in a put-call parity setting.

Tags

# Problem solving using the put-call parity


# Problem solving using the put-call parity












I am self-studying for an actuarial exam on financial economics. I encountered this problem, and I am having difficulty seeing why the statement underlined is true:

How do we know that $P(60) - C(60) + P(50) - C(50) = 110e^{-rT} - 2Se^{-\delta T} = (2/3)(15)$?

The part with (2/3) confuses me. We are using 2 out of the 3 equations, but I don't see why that would necessarily imply that the value of adding 2 of the 3 equations is (2/3) the value of adding the three equations together.

## Answer by Alex C (score 2, accepted)

https://quant.stackexchange.com/a/21629

What a difficult problem. The first line gave $165 e^{-rt} -3 S e^{-dt} = 15$ [since 50+55+60 = 165]. In the second line we want to evaluate $110 e^{-rt} -2 S e^{-dt} $. We notice that this is exactly two thirds of the left side of the above, because 110 is two thirds of 165 and 2 is two thirds of 3. So we take two thirds of the right hand side of the first eqn. i.e. (2/3)*15, which is 10.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.