Certain probability statement in discrete mathematical finance

Question

Le'ts suppose the following setting:

We have a filtred probability space $(\Omega,\mathcal{F},P,\{\mathcal{F}\}_{k=0,1})$ and an adapted $\mathbb{R}^d$ valued process $S=(S^1,\dots,S^d)$. Let $\theta_1$ be an $\mathcal{F}_0$ measurable $\mathbb{R}^d$ valued process for what follows. We define $V_0(\theta_1)=\theta_1^{tr}S_0$ and $V_1(\theta_1)=\theta_1^{tr}S_1$. Now We say $S$ admits

1. no arbitrage of first kind (NA1) if it exists no $\theta_1$ s.t. $V_0(\theta_1)\le 0$ $P$-a.s., $V_1(\theta_1)\ge 0$ $P$-a.s. and $P[V_1(\theta_1)>0]>0$

2. no arbitrage of second kind (NA2) if it exists no $\theta_1$ s.t. $V_0(\theta_1)\le 0$ $P$-a.s., $V_1(\theta_1)\ge 0$ $P$-a.s. and $P[V_0(\theta_1)<0]>0$

I want to prove that (NA1) does not imply (NA2) and I'm not quite sure how to do this:

Suppose (NA1) holds, i.e. there is no such $\theta_1$. Now if (NA2) would hold, there would also be no $\theta_1$ such that $V_0(\theta_1)\le 0$ $P$-a.s., $V_1(\theta_1)\ge 0$ $P$-a.s. and $P[V_0(\theta_1)<0]>0$. I do not see how to get a contradiction to (NA1)?

As a remark they say: If $S^1_0=1$, $S^1_1\ge0$ $P$-a.s. and $P[S^1_1\not=0]=1$, then (NA1) implies (NA2). However I do also not see how this assumption on the first component of $S$ should make a different. Thanks for your help.

Answer 1

Let $d = 1$ so there is one stock, which is priced at 0 today. Then there is no way to put together a portfolio in which it is negative at time 0, and therefore no type 2 arbitrage. However, it can be 1 at time 1.