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Attempt So Far:
1) First Part:
I have shown that the market is arbitrage-free since the only possible portfolio for which $V_1^h\geq0 \ $ given that $V_0^h=0 \ $ is $h=(0,0,0)$ and this clearly means that $V_1^h=0$ which contradicts (iii). Concluding that there is no arbitrage strategy.

2) Second part:

We can write $\pmb{u_1}=(1.5,1.05)$, $\pmb{u_2}=(1,1.125)$ and $\pmb{u_3}=(1,1.11)$ such that the Hadamard product (elementwise multiplication) of $\pmb{u_{i}}_{,\ 1\leq i< \leq 3} \ \ $ and $\ (S_0^1,S_0^2)$ produces the possible $(S_1^1,S_1^2)$ found above with the given probabilities.

Fix a contingent claim $X=F(S_1^1,S_1^2)$ and I want to show that there exist $h^*=(x^*,y^*,z^*)$ that replicates X. Then $V_1^{h^*}=\Phi(\pmb{u_{i}}_{,\ 1\leq i< \leq 3} \ )=X.$ Thus, we've:

$$V_1^{h^*}= \left\{ \begin{array}{ll} \ 1.1x^* +15y^*+ 21S_0^2z^*=\Phi(\pmb{u_1}),& \text{with probability } \ 0.5 \\ 1.1x^* +10y^*+ 22.5S_0^2z^*=\Phi(\pmb{u_2}), &\text{with probability } \ 0.25 \\ 1.1x^* +10y^*+22.2S_0^2z^*=\Phi(\pmb{u_3}), & \text{with probability } \ 0.25 \end{array} \right. $$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left \{ \begin{array}{ll} \ 1.1x^* +1.5S_0^1y^*+ 1.05S_0^2z^*=\Phi(\pmb{u_1}),& \text{with probability } \ 0.5 \\ 1.1x^* +S_0^1y^*+ 1.25S_0^2z^*=\Phi(\pmb{u_2}), &\text{with probability } \ 0.25 \\ 1.1x^* +S_0^1y^*+ 1.11S_0^2z^*=\Phi(\pmb{u_3}), & \text{with probability } \ 0.25. \end{array} \right.$$

From here, I'm unsure of how to proceed. I'm not supposed to solve the system to find $x^*, \ y^* \ $ and $z^*$ because the question clearly states that I shouldn't need to do that. That said, I'm not sure how to establish that the system above always has a solution without solving the system.

  • $\begingroup$ For the second part you might want to use the 2nd Fundamental Theorem of Asset Pricing. That is, an arbitrage-free market is complete iff the risk-neutral measure is unique. (Reminder: market completeness means that every contingent claim can be replicated) So, if you can show that $Q$ is unique, you have shown that your claim can be replicated. $\endgroup$
    – R. Rayl
    May 13, 2021 at 8:27
  • $\begingroup$ For a quadratic matrix A you could use the fact that the system Ax=b has a unique solution if A is invertible, which can here be confirmed by e.g. calculating the determinant. $\endgroup$ Jun 7, 2022 at 15:42

1 Answer 1


It sounds like you have shown the first part, so I will concentrate on the second part.

We will use the Second Fundamental Theorem of Asset Pricing: An arbitrage-free market is complete if and only if the risk-neutral measure is unique.

Note that market completeness means that any contingent claim can be replicated by a portfolio. So, if you can show that $Q$ is unique, you have shown that your claim can be replicated.

To show $Q$ is unique, consider the price movements of each asset. We know that under $Q$ the discounted expected (under $Q$) price of the each asset must equal the current price of the asset. So

\begin{align} \mathbb{E}_Q[S_1^1] \frac{1}{1.1} &= 10 \\ \mathbb{E}_Q[S_1^2] \frac{1}{1.1} &= 20 \end{align}

So using the possible values that we are given for $S^1_1$ and $S^2_1$, we can write:

\begin{align} 15 q_1 + 10q_2 + 10q_3 &= 11 \\ 21q_1 + 22.5q_2 + 22.2q_3 &= 22 \\ q_1 + q_2 + q_3 &= 1 \end{align} where $q_1, q_2, q_3$ correspond to the $P$-probabilities $0.5, 0.25, 0.25$. Note that we have three equations with three unknowns. Hence, we have a unique solution, namely \begin{align} q_1 = 0.2, \; q_2 = 0.1\dot{3}, \; q_3 = 0.\dot{6} \end{align}

Thus, $Q$ in unique. This implies our market is complete (since you already showed it is arbitrage-free). Therefore, any contingent claim can be replicated (i.e the replicating portfolio exists).


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