<p>Here is a simple fact: In your notation, the model under consideration is complete if and only if the matrix</p>
<p><span class="math-container">\begin{bmatrix}
1+R &amp; 1+R \\
d   &amp; u 
\end{bmatrix}</span></p>
<p>is one-to-one, i.e. <span class="math-container">$d \neq u$</span>. (Equivalently, its transpose is onto, which is what is shown in your quoted text.)</p>
<p>No-arbitrage holds if and only if <span class="math-container">$d \leq 1+R \leq u$</span> and <span class="math-container">$d &lt; u$</span>. (By assumption, <span class="math-container">$d \neq u$</span>. Otherwise you have two risk-free assets, and model either trivially collapes or trivially admits arbitrage.) Therefore NA implies market completeness, because <span class="math-container">$d &lt; u$</span> implies <span class="math-container">$d \neq u$</span>. Note that the no-arbitrage condition---that the risk-free return lies in the interval of risky return---is not really used.</p>
<p>In general, market completeness under no-arbitrage (uniqueness of risk-neutral measure) is a property in addition to no-arbitrage (existence of a risk-neutral measure). In the binomial case, however, there is really one &quot;degree of freedom&quot; and completeness holds trivially as long as the model does not collapse.</p>
<p>Leaving the binomial setting, this need not be true. For example, consider a model with <span class="math-container">$n$</span> states and <span class="math-container">$n$</span> assets, with <span class="math-container">$n &gt; 2$</span>. The model could have no-arbitrage but not be complete. Similarly, consider a model with <span class="math-container">$n$</span> states and <span class="math-container">$m$</span> assets, with <span class="math-container">$m &gt; n &gt; 2$</span>. The model could be complete (in fact, with redundant assets) but admits arbitrage.</p>

CC BY-SA 4.0
2022-11-14T20:38:39.367Z
Soooooooo complete can hold even for $d < u \le 1+R$ or $1+R \le d < u$ ?
BCLC
CC BY-SA 4.0
2021-09-23T08:52:56.870Z
2021-09-30T20:50:04.380Z
2021-09-30T20:50:04.380Z
Michael
<p>Edit: Can complete hold even if <span class="math-container">$d &lt; u \le 1+R$</span> or <span class="math-container">$1+R \le d &lt; u$</span> ?</p>
<hr />
<p>In Tomas Björk's Arbitrage Theory in Continuous Time, there exists this proposition</p>
<blockquote>
<p><img src="https://i.sstatic.net/8htwR.png" alt="enter image description here" /></p>
</blockquote>
<p>It seems that to show that the model is complete, we must show that the claims are reachable, i.e. we must find replicating portfolios for each claim.</p>
<p>Which part of finding the replicating portfolio makes use of the assumption, where I understand the assumption is equivalent to <span class="math-container">$d &lt; 1+R &lt; u$</span> (or <span class="math-container">$d \le 1+R \le u$</span>, but <span class="math-container">$d&lt;u$</span>)? All I see so far is the <span class="math-container">$d&lt;u$</span>, but that's nothing really special: of course the stock price <span class="math-container">$u$</span> if it would go up from its current price should be higher than the stock price <span class="math-container">$d$</span> if it would go down from the its current price.</p>

<finance><financial-economics><asset-pricing><arbitrage><complete-markets>
Arbitrage free implies complete market in general binomial model?
BCLC
# Arbitrage free implies complete market in general binomial model?

(Asked by: username_0 on 2021-09-14)

Edit: Can complete hold even if $d < u \le 1+R$ or $1+R \le d < u$ ?

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In Tomas Björk's Arbitrage Theory in Continuous Time, there exists this proposition

> ![enter image description here](https://i.sstatic.net/8htwR.png)

It seems that to show that the model is complete, we must show that the claims are reachable, i.e. we must find replicating portfolios for each claim.

Which part of finding the replicating portfolio makes use of the assumption, where I understand the assumption is equivalent to $d < 1+R < u$ (or $d \le 1+R \le u$, but $d<u$)? All I see so far is the $d<u$, but that's nothing really special: of course the stock price $u$ if it would go up from its current price should be higher than the stock price $d$ if it would go down from the its current price.

**Tags:** finance, financial-economics, asset-pricing, arbitrage, complete-markets


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## 1 answers

### Answer by username_1 on 2021-09-23. Score: 2

Here is a simple fact: In your notation, the model under consideration is complete if and only if the matrix

\begin{bmatrix}
1+R & 1+R \\
d & u
\end{bmatrix}

is one-to-one, i.e. $d \neq u$. (Equivalently, its transpose is onto, which is what is shown in your quoted text.)

No-arbitrage holds if and only if $d \leq 1+R \leq u$ and $d < u$. (By assumption, $d \neq u$. Otherwise you have two risk-free assets, and model either trivially collapes or trivially admits arbitrage.) Therefore NA implies market completeness, because $d < u$ implies $d \neq u$. Note that the no-arbitrage condition---that the risk-free return lies in the interval of risky return---is not really used.

In general, market completeness under no-arbitrage (uniqueness of risk-neutral measure) is a property in addition to no-arbitrage (existence of a risk-neutral measure). In the binomial case, however, there is really one "degree of freedom" and completeness holds trivially as long as the model does not collapse.

Leaving the binomial setting, this need not be true. For example, consider a model with $n$ states and $n$ assets, with $n > 2$. The model could have no-arbitrage but not be complete. Similarly, consider a model with $n$ states and $m$ assets, with $m > n > 2$. The model could be complete (in fact, with redundant assets) but admits arbitrage.

> **Comments:**
> **username_0**: Soooooooo complete can hold even for $d < u \le 1+R$ or $1+R \le d < u$ ?  

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