# 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?

• Thanks for the answer ! Should you submit it as an answer so I can accept it? – Metrician Sep 25 '20 at 9:51

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).
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.