<p>I've been meaning to post this for a long time. I came across this and thought it could add some insight. This example is from &quot;Financial Asset Pricing Theory&quot; by Munk.</p>
<p>Consider the following figure. How many assets do we need to have a complete market?
<img src="https://i.sstatic.net/xE3q3.png" alt="enter image description here" /></p>
<p>You might think that, because there are 6 different states here, we need at least 6 different assets.
In a static setting, we know that when we have <span class="math-container">$N$</span> different states, we must have <span class="math-container">$N$</span> &quot;sufficiently different assets&quot; (in the usual static setting, this means linearly independent). However, in the dynamic setting, this is not the case. Munk explains this based on two different observations:</p>
<blockquote>
<p>(i) the uncertainty is not revealed completely at once, but little by little, and (ii) we can trade dynamically in the assets. In the example there are three possible transitions of the economy from time 0 to time 1. From our one-period analysis we know that three sufficiently different assets are enough to 'span' this uncertainty. From time 1 to time 2 there are either two, three, or one possible transitions of the economy, depending on which state the economy is in at time 1. At most, we need three sufficiently different assets to span the uncertainty over this period. In total, we can generate any dividend process if we just have access to three sufficiently different assets in both periods.</p>
<p>In the case of a general multinomial tree version of a more general, finite-state discrete-time market, we can for each node in the tree define the <em>spanning number</em> as the number of branches of the subtree leaving that node. The market is then complete if, for any node in the tree, the number of linearly independent traded assets over the following period is equal to the spanning number.</p>
</blockquote>
<p>Now, in the case of a continuous-time model where the uncertainty is generated by a d-dimensional standard Brownian motion the argument is complicated, but Munk gives some insights based on the previous discussion.</p>
<blockquote>
<p>The result is quite intuitive given the following observations:</p>
<ol>
<li>For contunuous changes over an instant, only means and variances matter.</li>
<li>We can approximate the d-dimensional shock <span class="math-container">$dz_i$</span> by a random variable that takes on <span class="math-container">$d+1$</span> possible values and has the same mean and variance as <span class="math-container">$dz_t$</span>. For example, a one-dimensional shock <span class="math-container">$dz_t$</span> has mean zero and variance <span class="math-container">$dt$</span>. This is also true for a random variable <span class="math-container">$\epsilon$</span> which equals <span class="math-container">$\sqrt{dt}$</span> with a probability of <span class="math-container">$1/2$</span> and equals <span class="math-container">$-\sqrt{dt}$</span> with probability <span class="math-container">$1/2$</span>. ...</li>
<li>With continuous trading, we can adjust our exposure to the exogenous shocks at every instant.</li>
</ol>
<p>Over each instant we can thus think of the model with uncertainty generated by a d-dimensional standard Brownian motion as a discrete-time model with <span class="math-container">$d+1$</span> states. Therefore it only takes <span class="math-container">$d+1$</span> sufficiently different assets to complete the market.</p>
</blockquote>

CC BY-SA 4.0
2019-03-19T22:24:46.377Z
You can definitely get in trouble with these kind of arguments, but the discrete-time case is interesting in and of itself and is usefully suggestive for the continuous-time case. A good reference is the following: conditions for which dynamic completeness holds and conditions for convergence of discrete approximations can be found in [Anderson and Raimondo (2008)](https://onlinelibrary.wiley.com/doi/pdf/10.1111/j.1468-0262.2008.00861.x)
jmbejara
CC BY-SA 4.0
2019-03-21T19:48:52.387Z
On a related note, this paper is interesting: the law of one price is needed for dynamic completeness to imply one-period completeness. [Battauz and Ortu (2007)](https://link.springer.com/article/10.1007/s00199-005-0050-x)
jmbejara
CC BY-SA 3.0
2015-03-02T17:11:28.367Z
I am always very suspicious of this kind of loose story telling---yes, I know we do this all the time. In continuous time it's especially dubious. Sure, sounds good for the Bm case. What happens to that story when the price process is a general semimartingale? Becomes nonsense.
Michael
CC BY-SA 3.0
2015-02-28T15:17:21.630Z
2015-02-28T15:17:21.630Z
2020-06-17T09:40:54.643Z
jmbejara
<p>Mathematically, market completeness in continuous-time models does <strong>not</strong> follow from discrete-time heuristics.</p>

<p>In discrete-time, market completeness replies on only linear algebraic considerations.
This does not translate to statements about continuous-time martingales and their representation as stochastic integrals. </p>

<p>The precisely statement that gives market completeness in continuous-time models is the <strong>Martingale Representation Theorem</strong>, which says every martingale with respect to a <em>Brownian</em> filtration can be represented as an Ito integral with respect to the Brownian generating that filtration.</p>

<p>The result is <em>not</em> true for general filtrations, i.e. if <span class="math-container">$W_t$</span> is a <span class="math-container">$(\mathcal{F_t})$</span>-Brownian motion, it is not true in general that every <span class="math-container">$(\mathcal{F_t})$</span>-martingale is a <span class="math-container">$dW$</span>-integral.</p>

<p>(Standard counter-example: Let <span class="math-container">$W^{(1)}_t$</span> and <span class="math-container">$W^{(2)}_t$</span> be two independent standard Brownian motions. Then <span class="math-container">$W^{(2)}_t$</span> is a martingale with respect to the its own filtration but not adapted to the filtration generated by <span class="math-container">$W^{(1)}_t$</span>.
There cannot be an Ito integrand <span class="math-container">$\phi$</span> such that 
<span class="math-container">$$
W^{(2)}_t = \int_0^t \phi \, d W^{(1)}_s.
$$</span>
This can be shown by applying Ito's formula to <span class="math-container">$W^{(1)}_t W^{(2)}_t$</span>.
)</p>

<p>All continuous-time models where market completeness hold are formulated with minimal Brownian filtrations.
For example, in the Black-Scholes model price follows geometric Brownian motion
<span class="math-container">$$
\frac{dP}{P} = \mu dt + \sigma dW.
$$</span> 
Since <span class="math-container">$P$</span> is an exponential function of <span class="math-container">$W$</span>, the filtration generated by <span class="math-container">$P$</span> is the same as that generated by <span class="math-container">$W$</span>. All derivative payoffs are therefore adapted to the Brownian filtration. By Martingale Representation Theorem, they can be replicated/hedged/priced.
Payoffs that are not measurable with respect to the Brownian filtration cannot be replicated/hedged/priced---market is not complete with respect to such claims.</p>

<p>Similarly, in a Lucas tree economy, if the exogenous dividend process follows<br>
<span class="math-container">$$
\frac{dD}{D} = \mu dt + \sigma dW,
$$</span> 
market is complete only with respect to dividend-contingent claims.
If <span class="math-container">$W_t$</span> is a <span class="math-container">$(\mathcal{F_t})$</span>-Brownian motion with respect to some general filtration <span class="math-container">$(\mathcal{F_t})$</span>, market need not be complete with respect <span class="math-container">$(\mathcal{F_t})$</span>-adapted claims.  </p>

CC BY-SA 4.0
2020-06-02T14:05:15.527Z
2020-06-02T14:05:15.527Z
Michael
<p>I am the last person that should be answering continuous time questions like these, but if there's no one else I guess I'll give it a shot. (Any correction of my dimly remembered continuous-time finance is very welcome.)</p>

<p>My impression has always been that this is best interpreted as a consequence of the <a href="http://en.wikipedia.org/wiki/Martingale_representation_theorem">martingale representation theorem</a>. First, though, I'll loosely establish some notation. Let the probability space be generated by the $n$ independent Wiener processes $(Z_t^1,\ldots,Z_t^n)$. Let there be $n+1$ assets, where the value of the $i$th asset at $t$ is given by $S_t^i$. Assume that asset $i=0$ is a riskfree bond $dS_t^0=r_tS_t^0dt$,  while assets $i=1,\ldots,n$ are each risky and are driven by the corresponding $Z_t^i$:
$$dS_t^i=\mu_t^idt+\sigma_t^idZ_t^i$$
Assume there is a strictly positive SDF process $m_t$ normalized to $m_0=1$, such that $m_tS_t^i$ is a martingale for each $i$ (basically the definition of SDF) and where
$$dm_t=\nu_t dt+\psi_t\cdot dZ_t$$
(I use $\cdot$ as the dot product, which will be convenient.)</p>

<p>Finally, let the $n+1$-dimensional vector $\theta_t$ be our portfolio at time $t$, such that net worth $A_t$ is given by $A_t=\theta_t\cdot S_t$. Assume that $A_0$ is fixed and that further we have
$$dA_t=\theta_t\cdot dS_t$$
Now I'll state the objective, which captures the essence of complete markets. Suppose that the world ends at time $T$, and that we want net worth $A_T$ to equal a certain stochastic $Y$, which can depend on the full history up until time $T$. Suppose that $A_0=E_0[m_TY]$, so that in a world with complete markets we could (at $t=0$) use our initial wealth $A_0$ to purchase the time $t=T$ payout $Y$. In the absence of these direct complete markets, the question is whether there is <strong>nevertheless some strategy for the portfolio</strong> $\theta_t$ that will allow us to obtain $A_T=Y$ in all states of the world. And the answer, in this setting, is yes.</p>

<p>First, one can calculate $d(m_tA_t)=\theta_t\cdot d(m_tS_t)$. Thus $m_tS_t$ being a martingale implies that $m_tA_t$ is a martingale. Thus we have $A_T=Y\Longleftrightarrow m_TA_T=m_TY$ iff
$$m_tA_t=E_t[m_TY]$$
for all $t\in[0,T]$. Note that this is true for $t=0$ by assumption; hence to get equality it is only necessary to prove that the increments are always equal on both sides.</p>

<p>Now the martingale representation theorem comes in. Since $E_t[m_TY]$ is a martingale, we can write
$$E_t[m_TY]=E_0[m_TY]+\int_0^t \phi_s\cdot dZ_s$$
for some predictable process $\phi_s$. So we need to be able to show $d(m_tA_t)=\phi_t\cdot dZ_t$. Writing
$$d(m_tA_t)=\sum_i (m_t\theta_t^i \sigma_t^i +A_t\psi_t^i)dZ_t^i$$
we see that we need $m_t\theta_t^i\sigma_t^i +A_t\psi_t^i=\phi_t^i$ for each risky asset $i=1,\ldots,n$, which we can invert to give the needed portfolio choice $\theta_t^i$:
$$\theta_t^i=\frac{\phi_t^i-A_t\psi_t^i}{m_t\sigma_t^i}$$
The riskless asset portfolio choice $\theta_t^0$ can then be backed out from $A_t=\theta_t\cdot S_t$.</p>

<p>The intuition here is simple: we need to always have $A_t$ adjust to maintain the equality $m_tA_t=E_t[m_TY]$, but both the expectation on the right and the SDF $m_t$ on the left are moving in response to the driving processes $dZ_t^i$. Hence we need to pick a portfolio $\theta_t$ such that $dA_t$ precisely offsets these movements and the equation continues to hold. And we can always do this as long as locally, our assets span all the risks $dZ_t^i$ -- which can happen more generally, even for $n$ correlated assets as long as their increments are locally linearly independent. (The case here of $n$ risky assets each drien by an independent Brownian motion is a special one.)</p>

CC BY-SA 4.0
2020-06-03T13:04:09.357Z
The key assumption is that the filtration is a given Brownian filtration, and all the martingales that arise are with respect to that Brownian filtration. Rest is a standard calculation switch between risk-neutral and physical measure. Martingale representation does not hold (therefore calculation does not go through) for general filtrations.
Michael
CC BY-SA 3.0
2014-12-17T16:08:25.240Z
Thanks.  I skimmed your answer and it looks great.  Something came up that I have to finish in next couple days, but I will have a closer look and likely accept your answer when I finish.
cc7768
CC BY-SA 3.0
2014-12-17T08:24:03.293Z
2014-12-17T08:24:03.293Z
nominally rigid
<p>In the standard discrete time economies with a finite number of states, $n$, a complete markets economy is simply an economy with $n$ independent assets (Think Ljunqvist and Sargent Chapter 8).  This is because $n$ independent assets is sufficient to span the set of states tomorrow.</p>

<p>I had a discussion with a professor last week in which he stated that one of the conveniences of continuous time when thinking about asset pricing is that within a continuous time economy one can get complete markets simply with a risk-free bond and a risky asset (independent) for each Brownian motion in the economy.</p>

<p>He explained it as we talked, so I think I mostly understand it, but was wondering if someone would mind writing down the details?</p>

<p>I will probably spend a day or two this week on it (depends on some of the properties of the differential calculus), so if no one else answers the question then hopefully I can provide a satisfactory answer.</p>

CC BY-SA 3.0
2014-12-16T10:19:56.210Z
In the discrete time case, completeness doesn't require the number of states and number of assets to be the same, although you cannot have more states than assets. The general characterization of completeness is having a unique martingale equivalent measure, IIRC.
Michael
<macroeconomics><mathematical-economics><continuous-time><complete-markets>
Complete Markets in Continuous Time
cc7768
# Complete Markets in Continuous Time

(Asked by: username_0 on 2014-12-14)

In the standard discrete time economies with a finite number of states, $n$, a complete markets economy is simply an economy with $n$ independent assets (Think Ljunqvist and Sargent Chapter 8). This is because $n$ independent assets is sufficient to span the set of states tomorrow.

I had a discussion with a professor last week in which he stated that one of the conveniences of continuous time when thinking about asset pricing is that within a continuous time economy one can get complete markets simply with a risk-free bond and a risky asset (independent) for each Brownian motion in the economy.

He explained it as we talked, so I think I mostly understand it, but was wondering if someone would mind writing down the details?

I will probably spend a day or two this week on it (depends on some of the properties of the differential calculus), so if no one else answers the question then hopefully I can provide a satisfactory answer.

**Tags:** macroeconomics, mathematical-economics, continuous-time, complete-markets
> **Comments:**
> **username_1**: In the discrete time case, completeness doesn't require the number of states and number of assets to be the same, although you cannot have more states than assets. The general characterization of completeness is having a unique martingale equivalent measure, IIRC.  

---

## 3 answers

### Answer by username_3 on 2014-12-17. Score: 10 | Accepted answer

I am the last person that should be answering continuous time questions like these, but if there's no one else I guess I'll give it a shot. (Any correction of my dimly remembered continuous-time finance is very welcome.)

My impression has always been that this is best interpreted as a consequence of the [martingale representation theorem](http://en.wikipedia.org/wiki/Martingale_representation_theorem). First, though, I'll loosely establish some notation. Let the probability space be generated by the $n$ independent Wiener processes $(Z\_t^1,\ldots,Z\_t^n)$. Let there be $n+1$ assets, where the value of the $i$th asset at $t$ is given by $S\_t^i$. Assume that asset $i=0$ is a riskfree bond $dS\_t^0=r\_tS\_t^0dt$, while assets $i=1,\ldots,n$ are each risky and are driven by the corresponding $Z\_t^i$:
$$dS\_t^i=\mu\_t^idt+\sigma\_t^idZ\_t^i$$
Assume there is a strictly positive SDF process $m\_t$ normalized to $m\_0=1$, such that $m\_tS\_t^i$ is a martingale for each $i$ (basically the definition of SDF) and where
$$dm\_t=\nu\_t dt+\psi\_t\cdot dZ\_t$$
(I use $\cdot$ as the dot product, which will be convenient.)

Finally, let the $n+1$-dimensional vector $\theta\_t$ be our portfolio at time $t$, such that net worth $A\_t$ is given by $A\_t=\theta\_t\cdot S\_t$. Assume that $A\_0$ is fixed and that further we have
$$dA\_t=\theta\_t\cdot dS\_t$$
Now I'll state the objective, which captures the essence of complete markets. Suppose that the world ends at time $T$, and that we want net worth $A\_T$ to equal a certain stochastic $Y$, which can depend on the full history up until time $T$. Suppose that $A\_0=E\_0[m\_TY]$, so that in a world with complete markets we could (at $t=0$) use our initial wealth $A\_0$ to purchase the time $t=T$ payout $Y$. In the absence of these direct complete markets, the question is whether there is **nevertheless some strategy for the portfolio** $\theta\_t$ that will allow us to obtain $A\_T=Y$ in all states of the world. And the answer, in this setting, is yes.

First, one can calculate $d(m\_tA\_t)=\theta\_t\cdot d(m\_tS\_t)$. Thus $m\_tS\_t$ being a martingale implies that $m\_tA\_t$ is a martingale. Thus we have $A\_T=Y\Longleftrightarrow m\_TA\_T=m\_TY$ iff
$$m\_tA\_t=E\_t[m\_TY]$$
for all $t\in[0,T]$. Note that this is true for $t=0$ by assumption; hence to get equality it is only necessary to prove that the increments are always equal on both sides.

Now the martingale representation theorem comes in. Since $E\_t[m\_TY]$ is a martingale, we can write
$$E\_t[m\_TY]=E\_0[m\_TY]+\int\_0^t \phi\_s\cdot dZ\_s$$
for some predictable process $\phi\_s$. So we need to be able to show $d(m\_tA\_t)=\phi\_t\cdot dZ\_t$. Writing
$$d(m\_tA\_t)=\sum\_i (m\_t\theta\_t^i \sigma\_t^i +A\_t\psi\_t^i)dZ\_t^i$$
we see that we need $m\_t\theta\_t^i\sigma\_t^i +A\_t\psi\_t^i=\phi\_t^i$ for each risky asset $i=1,\ldots,n$, which we can invert to give the needed portfolio choice $\theta\_t^i$:
$$\theta\_t^i=\frac{\phi\_t^i-A\_t\psi\_t^i}{m\_t\sigma\_t^i}$$
The riskless asset portfolio choice $\theta\_t^0$ can then be backed out from $A\_t=\theta\_t\cdot S\_t$.

The intuition here is simple: we need to always have $A\_t$ adjust to maintain the equality $m\_tA\_t=E\_t[m\_TY]$, but both the expectation on the right and the SDF $m\_t$ on the left are moving in response to the driving processes $dZ\_t^i$. Hence we need to pick a portfolio $\theta\_t$ such that $dA\_t$ precisely offsets these movements and the equation continues to hold. And we can always do this as long as locally, our assets span all the risks $dZ\_t^i$ -- which can happen more generally, even for $n$ correlated assets as long as their increments are locally linearly independent. (The case here of $n$ risky assets each drien by an independent Brownian motion is a special one.)

> **Comments:**
> **username_0**: Thanks.  I skimmed your answer and it looks great.  Something came up that I have to finish in next couple days, but I will have a closer look and likely accept your answer when I finish.  
> **username_1**: The key assumption is that the filtration is a given Brownian filtration, and all the martingales that arise are with respect to that Brownian filtration. Rest is a standard calculation switch between risk-neutral and physical measure. Martingale representation does not hold (therefore calculation does not go through) for general filtrations.  

---

### Answer by username_2 on 2015-02-28. Score: 6

I've been meaning to post this for a long time. I came across this and thought it could add some insight. This example is from "Financial Asset Pricing Theory" by Munk.

Consider the following figure. How many assets do we need to have a complete market?
![enter image description here](https://i.sstatic.net/xE3q3.png)

You might think that, because there are 6 different states here, we need at least 6 different assets.
In a static setting, we know that when we have $N$ different states, we must have $N$ "sufficiently different assets" (in the usual static setting, this means linearly independent). However, in the dynamic setting, this is not the case. Munk explains this based on two different observations:

> (i) the uncertainty is not revealed completely at once, but little by little, and (ii) we can trade dynamically in the assets. In the example there are three possible transitions of the economy from time 0 to time 1. From our one-period analysis we know that three sufficiently different assets are enough to 'span' this uncertainty. From time 1 to time 2 there are either two, three, or one possible transitions of the economy, depending on which state the economy is in at time 1. At most, we need three sufficiently different assets to span the uncertainty over this period. In total, we can generate any dividend process if we just have access to three sufficiently different assets in both periods.
>
> In the case of a general multinomial tree version of a more general, finite-state discrete-time market, we can for each node in the tree define the *spanning number* as the number of branches of the subtree leaving that node. The market is then complete if, for any node in the tree, the number of linearly independent traded assets over the following period is equal to the spanning number.

Now, in the case of a continuous-time model where the uncertainty is generated by a d-dimensional standard Brownian motion the argument is complicated, but Munk gives some insights based on the previous discussion.

> The result is quite intuitive given the following observations:
>
> 1. For contunuous changes over an instant, only means and variances matter.
> 2. We can approximate the d-dimensional shock $dz\_i$ by a random variable that takes on $d+1$ possible values and has the same mean and variance as $dz\_t$. For example, a one-dimensional shock $dz\_t$ has mean zero and variance $dt$. This is also true for a random variable $\epsilon$ which equals $\sqrt{dt}$ with a probability of $1/2$ and equals $-\sqrt{dt}$ with probability $1/2$. ...
> 3. With continuous trading, we can adjust our exposure to the exogenous shocks at every instant.
>
> Over each instant we can thus think of the model with uncertainty generated by a d-dimensional standard Brownian motion as a discrete-time model with $d+1$ states. Therefore it only takes $d+1$ sufficiently different assets to complete the market.

> **Comments:**
> **username_1**: I am always very suspicious of this kind of loose story telling---yes, I know we do this all the time. In continuous time it's especially dubious. Sure, sounds good for the Bm case. What happens to that story when the price process is a general semimartingale? Becomes nonsense.  
> **username_2**: You can definitely get in trouble with these kind of arguments, but the discrete-time case is interesting in and of itself and is usefully suggestive for the continuous-time case. A good reference is the following: conditions for which dynamic completeness holds and conditions for convergence of discrete approximations can be found in [Anderson and Raimondo (2008)](https://onlinelibrary.wiley.com/doi/pdf/10.1111/j.1468-0262.2008.00861.x)  
> **username_2**: On a related note, this paper is interesting: the law of one price is needed for dynamic completeness to imply one-period completeness. [Battauz and Ortu (2007)](https://link.springer.com/article/10.1007/s00199-005-0050-x)  

---

### Answer by username_1 on 2020-06-02. Score: 2

Mathematically, market completeness in continuous-time models does **not** follow from discrete-time heuristics.

In discrete-time, market completeness replies on only linear algebraic considerations.
This does not translate to statements about continuous-time martingales and their representation as stochastic integrals.

The precisely statement that gives market completeness in continuous-time models is the **Martingale Representation Theorem**, which says every martingale with respect to a *Brownian* filtration can be represented as an Ito integral with respect to the Brownian generating that filtration.

The result is *not* true for general filtrations, i.e. if $W\_t$ is a $(\mathcal{F\_t})$-Brownian motion, it is not true in general that every $(\mathcal{F\_t})$-martingale is a $dW$-integral.

(Standard counter-example: Let $W^{(1)}\_t$ and $W^{(2)}\_t$ be two independent standard Brownian motions. Then $W^{(2)}\_t$ is a martingale with respect to the its own filtration but not adapted to the filtration generated by $W^{(1)}\_t$.
There cannot be an Ito integrand $\phi$ such that
$$
W^{(2)}\_t = \int\_0^t \phi \, d W^{(1)}\_s.
$$
This can be shown by applying Ito's formula to $W^{(1)}\_t W^{(2)}\_t$.
)

All continuous-time models where market completeness hold are formulated with minimal Brownian filtrations.
For example, in the Black-Scholes model price follows geometric Brownian motion
$$
\frac{dP}{P} = \mu dt + \sigma dW.
$$
Since $P$ is an exponential function of $W$, the filtration generated by $P$ is the same as that generated by $W$. All derivative payoffs are therefore adapted to the Brownian filtration. By Martingale Representation Theorem, they can be replicated/hedged/priced.
Payoffs that are not measurable with respect to the Brownian filtration cannot be replicated/hedged/priced---market is not complete with respect to such claims.

Similarly, in a Lucas tree economy, if the exogenous dividend process follows  
$$
\frac{dD}{D} = \mu dt + \sigma dW,
$$
market is complete only with respect to dividend-contingent claims.
If $W\_t$ is a $(\mathcal{F\_t})$-Brownian motion with respect to some general filtration $(\mathcal{F\_t})$, market need not be complete with respect $(\mathcal{F\_t})$-adapted claims.


---

