One period binomial model considers asset prices $S(i)$ where $i=0,1$ where $S(0) = S$ and we have either $S(1) = Su$ or $S(1) = Sd$, where $0 < d< u$, and nominal interest rate per period is $r$. The condition for no arbitrage is $d < 1+ r < u$.
The proof of this depends on the no arbitrage formula $p_{1}r_{1}(1) + p_{2}r_{1}(2) = 0$. Here $r_{1}(j)$ is the return function with $j=1$ corresponding to case $S(1) = Su$ and $j=2$ corresponding to $S(1) = Sd$, and $(p_{1}, p_{2})$ is the probability vector.
I don't understand the way the return function is computed, in particular why the return at time $0$ is required (i.e. present value of gain if asset price moves up or down). So they say the discounted value of $S(1)$ at time $0$ is $\frac{S(1)}{1+r}$ and thus return at time $0$ is
$$\frac{S(1)}{1+r} - S,$$
thus $r_{1}(1) = \frac{Su}{1+r} - S$ and $r_{1}(2) = \frac{Sd}{1+r} - S$, after which $(p_{1}, p_{2})$ can be computed and hence no arbitrage condition.
What does this return at time $0$ even mean and why is needed? How is it possible when one buys asset at time $0$ and then sells at time $1$? Why don't they simply take return to be $r_{1}(1) = \frac{Su - S}{S}$ and $r_{1}(2) = \frac{Sd - S}{S}$ ?
Please help.
The no-arbitrage condition $$\tag{*}d \leqslant 1+r \leqslant u$$ can be proved without this return terminology. However, my interpretation where the aforementioned return arises is as follows.
In the binomial market model there are two securities, the stock and a riskless money market instrument with a return of $1+r$ over the period. An arbitrage portfolio has a value process satisfying $V(0) = 0$ and $V(1) > 0$ with probability $1$. The main result is
To prove the forward implication, assume first that the inequality $d \leqslant 1+r$ does not hold. Form a portfolio that is long $\frac{1}{1+r}$ shares of stock financed by borrowing $\frac{S}{1+r}$ in cash. The initial portfolio value is
$$V(0) = \frac{1}{1+r}\cdot S - \frac{S}{1+r} = 0,$$
and the ending value of the portfolio is
$$V(1) = \frac{1}{1+r}\cdot S(1)- \frac{S}{1+r} \cdot (1+r) = \frac{S(1)}{1+r} - S$$
Hence, the change ion portfolio value or "return" is
$$V(1) - V(0)= \frac{S(1)}{1+r} - S$$
The smallest possible return occurs when $S(1) = Sd$ and if $d > 1+r$ we have an arbitrage, since in all possible states,
$$V(1) - V(0)\geqslant \frac{Sd}{1+r}-S > 0$$
Consequently, if there is no arbitrage we must have $d \leqslant 1+r$.
Similarly, we can prove that the inequality $1+r \leqslant u$ must hold if there is no arbitrage. In this case, we form a portfolio that is short $\frac{1}{1+r}$ shares of stock with the proceeds of the short sale $\frac{S}{1+r}$ invested in the money market account. The initial portfolio value is again $V(0) = 0$ and the "return" is
$$V(1) - V(0) = S - \frac{S(1)}{1+r}$$
The smallest possible return occurs when $S(1) = Su$ and if $u < 1+r$ there is an arbitrage, since and in all possible states,
$$V(1) - V(0) \geqslant S - \frac{Su}{1+r} > 0$$
Consequently, if there is no arbitrage we must have $1+r \leqslant u$