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Consider the multiperiod binomial/CRR model with one risky asset $S^{1}$ and a numeraire $S^{0}$. By seeing that the equivalent martingale measure is uniquely determined, we obtain that the market is complete.

A question that I have asked myself is whether adding another risky asset $S^{2}$ to the market ruins completeness. Intuitively, the extended market should not be complete because I could construct a number of equivalent martingale measures. Is there any way to argue with atoms on the underlying space? I am looking for a nice proof.

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    $\begingroup$ Completness means: (i) no transaction costs and (ii) price for every asset in every possible state of the world. Intuitively, adding another asset should not affect completeness. If we draw two binomial trees and "pretend" these are two sample spaces, we can "merge" these using Fubini's "Product Measure". $\endgroup$ – Jan Stuller Feb 15 at 11:13
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I answer from a general discrete time/discrete state model point of view. This includes the binomial tree model as a special case. In finite dimensions, you can interpret asset payoffs and returns as vectors and retreat to linear algebra.

Suppose you have $N$ states of nature and $J$ assets. Your payoff matrix is \begin{align*} A=\begin{pmatrix} X_1(\omega_1) & ... & X_J(\omega_1) \\ X_1(\omega_2) & ... & X_J(\omega_2) \\ \vdots & \ddots & \vdots \\ X_1(\omega_N) & ... & X_J(\omega_N) \end{pmatrix} \in\mathbb{R}^{N\times J}, \end{align*} where $X_i(\omega_j)$ denotes the payoff of asset $i$ in state $j$.

Completeness means to be able to hedge any payoff in the state space. Your state space is $\mathbb{R}^N$. The question is thus

Given any payoff $\mathbf{x}\in\mathbb{R}^N$, does a replicating (or hedging) portfolio, $\mathbf{q}\in\mathbb{R}^J$, exist such that $A\mathbf{q}=\mathbf{x}$? That is, can this payoff be attained by trading the available assets?

In maths language, the question is what is the rank of the matrix (= dimension of the space spanned by its columns)? The rank is just the number of linearly independent columns (asset payoffs). If the matrix $A$ has full rank, every payoff can be replicated and the market is complete. Completeness really only means to have one (linear independent) asset for each state of nature. Clearly, we only care about linear independent columns when we talk about spanning. Note that $\text{rank}(A)\leq\min\{N,J\}$.

In your case, there are two states of nature and three assets, $N=2$ and $J=3$. Thus, the matrix's rank can be at most two. If your first two assets (the risk-free bank account and the risky stock) have linearly independent payoffs (they should), then $\text{rank}(A)=2$ and the market is complete. Adding any number of assets to the model has no impact because all additional payoffs (assets) are linear combinations of the bank account and your original stock anyway. Thus, adding more assets to your market keeps it complete.

Indeed, had your market not been complete (because your first ''risky'' stock is just a scaled bank account), then adding an additional asset may actually help your model to be complete. You see, the more assets there are, the more likely it is that your model is complete.

More background on binomial trees: Using an atom system of the filtration, you split them up in one-period submodels. In each submodel, the above argument applies and thus, each submodel is complete. As a result, the entire tree (model) is complete. By the fundamental theorem of asset pricing, a complete market which does not permit for arbitrage has a unique equivalent martingale measure.

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