Background Information:
This question follows from here
It is tempting to write $$V_0(X) = \beta\left[\left(\frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}\right)X(u) + \left(\frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}\right)X(d)\right]$$ as
$$V_0(X) = E_Q[\beta X]$$ where the expectation is taken with respect to the new purely formal probability measure $Q$ defined by $$Q(u) = \frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}$$ and $$Q(d) = \frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}$$
Note that $Q(u) + Q(d) = 1$; $Q$ will be a probability measure provided these values are non-negative.
Question:
Show why $ \ Q(u),Q(d) \geq 0$
My reasoning:
Suppose $$Q(u) = \frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)} < 0$$ then either $\beta^{-1}S_0 - S_1(d) > 0$ and $S_1(u) - S_1(d) < 0$ or $\beta^{-1}S_0 - S_1(d) < 0$ and $S_1(u) - S_1(d) > 0$. In either case there would be an arbitrage opportunity.
Consider the first case where $$\beta^{-1}S_0 - S_1(d) > 0 \ \ \text{and} \ \ S_1(u) - S_1(d) < 0$$ Then I believe we would short $\beta^{-1}S_0$ and use the proceeds to go long in a bond.
I am not sure if this is sort of the correct reasoning I need to fulfill the question. Any suggestions are greatly appreciated.