No-Arbitrage Bounds and Convex Combinations in the Binomial Model

# How does $1 + R = q_u · u + q_d · d $ follow from $d ≤ (1 + R) ≤u$ in the Binomial Pricing Model?


# How does $1 + R = q_u · u + q_d · d $ follow from $d ≤ (1 + R) ≤u$ in the Binomial Pricing Model?












I've been reading Tomas Bjork's 'Arbitrage theory' and it says:

> To say that $d ≤ (1 + R) ≤u$ holds is equivalent to saying that $1 + R$ is a convex combination of u and d, i.e. $1 + R = q_u · u + q_d · d $

I understand why the condition $d ≤ (1 + R) ≤u$ should hold for there not to be an arbitrage opportunity, and I also understand that $1 + R = q_u · u + q_d · d $ means that the expected return of the stock is equal to the risk-free return, but how does that inequality holding imply this equality? What's the proof to get to this convex combination?

## Answer by nbbo2 (score 4, accepted)

https://quant.stackexchange.com/a/58230

The equality $1 + R = q_u · u + q_d · d $ is not particularly significant or difficult to prove.

In fact, any number $b$ can be written as a linear combination of 2 other distinct arbitrary numbers $a,c$: $b=qa+(1−q)c$. (Easy: just set $q=\frac{b−c}{a−c}$). But in addition iff $a\le b\le c$ then it is a convex linear combination i.e. $0\le q \le 1$ and $0\le(1−q) \le 1$ and this I think is the essential point here, $q$ will be between 0 and 1. (Spoiler warning: later in the book Bjork will argue that since q is between 0 and 1 it can be interpreted as a probability).

You quoted a passage from Bjork, I don't have access to the book right now, but a more complete statement of what Bjork is trying to say would be:

> To say that $d ≤ (1 + R) ≤u$ holds is equivalent to saying that $1 + R$ is a convex combination of u and d, i.e. $1 + R = q_u · u + q_d · d $, where it is guaranteed that $0\le q_u,q_d \le 1$.

The final part (the inequalities for the two q's) is the most important. From the "no-arbitrage inequality" $d ≤ (1 + R) ≤u$ we deduce that $q_u$ is a "pseudo-probability" i.e. a number between 0 and 1.