Why Risk-Neutral Probabilities Must Be Positive for No-Arbitrage

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


# 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?

## Answer by spaceisdarkgreen (score 4)

https://quant.stackexchange.com/a/36136

There is one condition under which the risk neutral probability of an event can be zero: if the real world probability is zero. If not then any contract that pays off in that event must go down in price if the contract is modified as to not pay off or pay off less in that event. Otherwise, one can buy one and sell the other... it's arbitrage in the "free lottery ticket" sense that you get a nonzero probability of a payoff for free. This translates to the risk neutral probability of the event being positive.

## Answer by Pavel (score 0)

https://quant.stackexchange.com/a/36164

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:

- Borrow $S_n$ from the bank

- Purchase one share of stock

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.