# Why risk neutral probabilities should be strictly greater than zero for no arbitrage condition?

I was recently told by a colleague that the risk neutral probabilities should ALWAYS be greater than zero to have a no arbitrage condition. Intuitively, we know probabilities cannot be < 0, but how can we prove that we need them to > 0 too?

I am assuming if this is correct, it also applies to trinomial and n-nomial trees too. Can someone please clarify?

For the binomial model with up factor $$u$$, down factor $$d$$, and interest rate $$r$$ i.e. if a security at time $$t = n$$ is worth $$S_n$$, then at time $$t = n + 1$$, $$S_{n+1} = uS_n$$ with probability $$p$$ and $$S_{n+1} = dS_n$$ with probability $$q$$; similarly, if $$X$$ is left in the bank at $$t = n$$, then at $$t = n + 1$$ the account will contain $$(1 + r)X$$. Recall the risk neutral probabilities for the binomial model: $$\tilde{p} = \frac{(1 + r) - d}{u - d}, \ \tilde{q} = \frac{u - (1 + r)}{u - d}$$ Also recall the no arbitrage condition for this model: $$d < 1 + r < u$$ This condition implies that our risk neutral probabilities $$\tilde{p} > 0$$ and $$\tilde{q} > 0$$. To answer your first question, we prove that the risk neutral probabilities must be $$> 0$$ by showing that arbitrage is possible if this is not the case. Suppose that $$1 + r < d < u$$ and we are at time $$t = n$$. This implies that $$\tilde{p} < 0$$. Our strategy will be:
1. Borrow $$S_n$$ from the bank
At time $$t = n + 1$$ we owe $$(1+r)S_n$$ to the bank. Our position in stock is worth either $$uS_n > (1 + r)S_n$$ or $$dS_n > (1 + r)S_n$$. We then sell the stock and cover what we owe to the bank, leaving us with a nonzero amount of terminal capital when we invested $$0$$. This is an arbitrage strategy. A similar strategy can be used when $$d < u < 1 + r$$, namely short stock and invest in the bank.