<p>In Radner's model each actor starts with an initial endowment, an asset vector <span class="math-container">$\omega$</span>. You can trade this to any asset portfolio <span class="math-container">$x$</span> that has lower or equal value, so
<span class="math-container">$$x * q \leq \omega * q.$$</span>
Denoting asset trades by <span class="math-container">$\theta = x - \omega$</span>, you can also express this as
<span class="math-container">$$
\theta * q \leq 0.
$$</span></p>

CC BY-SA 4.0
2019-04-16T21:08:31.133Z
So in the initial stage t=0, I've been assigned some quantity of each asset, and I can trade with some second party to possibly change the quantities I possess and these new quantities I end up with are the vector $x$, correct? Also, am I right to say that the only reason trade occurs is because there is uncertainty about which state will be realized in t=1?
David
CC BY-SA 4.0
2019-04-17T04:50:16.207Z
@David 1. Yes. 2. This depends on the exact modell you are using. In case the assets are not just financial but can also represent goods, trade is also possible without uncertainty, though this is not the main point of Radner. Also if you have new questions please post them as new questions.
Giskard
CC BY-SA 4.0
2019-04-16T19:19:39.063Z
2019-04-16T19:19:39.063Z
Giskard
<p>Suppose that <span class="math-container">$q$</span> is a k-tuple vector of prices for the k assets whose quantities are given by the k-tuple <span class="math-container">$\theta$</span>. I have just read that in the Radner Sequential Trade Equilibrium (not sure if this goes by another name), where the trading of assets occurs in period t=0 (as opposed to t=1), the period-zero budget can be expressed as <span class="math-container">$q*\theta$</span>, where <span class="math-container">$*$</span> denotes the inner product, and that <strong>this budget is non-positive.</strong> </p>

<p>I feel like I must be misunderstanding something about this particular setting, because I have no idea how a budget could possibly be non-positive, as opposed to non-negative. </p>

<p>Can someone give me some intuition for what's happening here?</p>

<microeconomics><general-equilibrium><asset-pricing><uncertainty><radner-equilibrium>
Budget constraint in Radner Sequential Trade Equilibria
David
# Budget constraint in Radner Sequential Trade Equilibria

(Asked by: username_0 on 2019-04-16)

Suppose that $q$ is a k-tuple vector of prices for the k assets whose quantities are given by the k-tuple $\theta$. I have just read that in the Radner Sequential Trade Equilibrium (not sure if this goes by another name), where the trading of assets occurs in period t=0 (as opposed to t=1), the period-zero budget can be expressed as $q\*\theta$, where $\*$ denotes the inner product, and that **this budget is non-positive.**

I feel like I must be misunderstanding something about this particular setting, because I have no idea how a budget could possibly be non-positive, as opposed to non-negative.

Can someone give me some intuition for what's happening here?

**Tags:** microeconomics, general-equilibrium, asset-pricing, uncertainty, radner-equilibrium

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## 1 answers

### Answer by username_1 on 2019-04-16. Score: 1 | Accepted answer

In Radner's model each actor starts with an initial endowment, an asset vector $\omega$. You can trade this to any asset portfolio $x$ that has lower or equal value, so
$$x \* q \leq \omega \* q.$$
Denoting asset trades by $\theta = x - \omega$, you can also express this as
$$
\theta \* q \leq 0.
$$

> **Comments:**
> **username_0**: So in the initial stage t=0, I've been assigned some quantity of each asset, and I can trade with some second party to possibly change the quantities I possess and these new quantities I end up with are the vector $x$, correct? Also, am I right to say that the only reason trade occurs is because there is uncertainty about which state will be realized in t=1?  
> **username_1**: @David 1. Yes. 2. This depends on the exact modell you are using. In case the assets are not just financial but can also represent goods, trade is also possible without uncertainty, though this is not the main point of Radner. Also if you have new questions please post them as new questions.  

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