Background: In an exercise about option pricing using binomial trees, the portfolio is riskless if
$$ S_{0u} \Delta - f_u = S_{0d} \Delta - f_d $$
So
$$ \Delta = \frac{f_u - f_d}{S_0(u-d)} $$
represents the amount of underlying I have to buy (if $\Delta >0$) or to sell (if $\Delta <0$).
Let's define a value p s.t.
$$p = \frac{e^{rT}-d}{u-d}$$
So, the price $f$ is
$$f = e^{rT}(pf_u + (1-p)f_d)$$
where $f_u$ is the option payoff value in case of "up" (growing value according to a factor $u>1$) and $f_d$ is the
option payoff value in case of "down" (decreasing value according to a factor $0<d<1$).
The question:
Prove that probability $0\leq p \leq 1$.
My attempt, proof with reductio ad absurdum:
Let's start from $p < 0$
$$\frac{e^{rT}-d}{u-d} < 0$$
But $u-d>0$ so
$$ e^{rT}-d < 0$$
id est
$$ e^{rT}<d $$
$$rT<\ln d$$
and we know that $\ln d<0$, because $0<d<1$
But $T>0$ surely because it's a time
and $r$ would become $<0$, but this is not possible because it's an interest rate ($0\leq r\leq 1$).
Up to now, proved that $p < 0$ is not possible.
But it's the round to show that $p>1$ is not possible.
With similar passages I obtain
$$rT>\ln d$$
with $\ln d<0$
But this inequality is possible, so I'm stuck.
Can someone help me? Thanks.
It follows from the definition of $p$ that $0\leq p\leq 1$ is equivalent to $d\leq e^{rT}\leq u$.
Next, note that if either $e^{rT}<d$ or $e^{rT}>u$ there will be an arbitrage opportunity. In the first case the "risky" asset is guaranteed to grow in value more quickly than the "risk-free" asset, and in the second case it is guaranteed to grow more slowly. In both cases we could set up a portfolio having value zero at time zero which is guaranteed to have a positive value at time $T$.
So under the assumption that there are no arbitrage opportunities, we must have $d\leq e^{rT}\leq u$.