An Uncertainty Principle for Markets

Today our goal is to derive an exact formulation of an uncertainty principle in markets. To that purpose we have established in earlier posts a commutation relation between demand ${d_i}$ and price ${p_i}$ of a good ${i}$ in a market. I state it again:

Prices ${p_i}$ and demands ${d_j}$ interact according to

$\displaystyle [p_i,d_j]=i \mu_i p_i \delta_{i,j} \ \ \ \ \ (1)$

for a fixed real ${\mu_i\in\mathbb{R}}$.

What I didn’t tell you so far is how measurement of market observables is supposed to work. Let me just close this gap. Measurement of an observable, e.g. the price of good ${i}$, in a market in state ${\xi}$ (e.g. in this case selling a small quantity of good ${i}$) will result in a jump of the market into a new state ${\zeta}$ being an eigenvector of the observable. The outcome of the measurement will be a real number ${\zeta_i}$ (e.g. the price), the eigenvalue of the observable corresponding to ${\zeta}$ with probability

$\displaystyle \textnormal{prob }(\zeta_i)=\frac{\left\langle \xi|\zeta \right\rangle \left\langle \zeta|\xi\right\rangle}{\|\xi\|^2}.$

For an observable ${a}$ on ${X}$ one can show that its mean value at state ${\xi\in X}$ is given as

$\displaystyle \overline{a_\xi}=\frac{\left\langle a \xi|\xi\right\rangle}{\|\xi\|^2}.$

The dispersion of an observable ${a}$ on ${X}$ is given as

$\displaystyle \overline{\left(\triangle a \right)^2_\xi}= \frac{\left\langle\left(a-\overline{a_\xi} \text{id}_X \right)^2 \xi|\xi\right\rangle}{\|\xi\|^2 }.$

Now we are in the shape to state the uncertainty principle in markets. In essence it claims that prices and demands of a good cannot be measured with arbitrary precision. Moreover, an explicit lower bound on the maximal simultaneous precision is given. Its proof is essentially a straight forward application of Cauchy-Schwarz inequality.

Proposition. For a market in state ${\xi}$ the dispersions of ${p_i}$ and ${d_i}$ satisfy

$\displaystyle \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} \geq \frac{\mu_i^2}{4} \| \sqrt{p_i} \xi\|^4.$

In the asymmetric case ${\mu_i\neq0}$, the right-hand side is strictly larger than zero.

Proof. Since dispersion and mean do not depend on the norm of a state we can, without loss of generality, assume that ${\|\xi\|=1}$ and obtain

$\displaystyle \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} = \left\langle\left(p_i - \overline{p_i} \text{id}_X \right)^2\xi|\xi\right\rangle \left\langle\left(d_i - \overline{d_i} \text{id}_X \right)^2\xi|\xi\right\rangle.$

Now Cauchy-Schwarz inequality implies

$\displaystyle \begin{array}{rcl} \overline{\left(\triangle p_i \right)^2_\xi}\, \overline{\left(\triangle d_i \right)^2_\xi} & \geq & \left\langle\left(p_i - \overline{p_i} \text{id}_X \right) \left(d_i - \overline{d_i} \text{id}_X \right) \xi|\xi\right\rangle \\ & & \qquad \times \left\langle\left(d_i - \overline{d_i} \text{id}_X \right) \left(p_i - \overline{p_i} \text{id}_X \right) \xi|\xi\right\rangle. \end{array}$

Since ${ab = \frac{1}{2}[a,b]_+ + \frac{1}{2i}i[a,b]}$ with ${[a,b]_+=ab+ba}$ we obtain

$\displaystyle \begin{array}{rcl} \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} & \geq & \left\langle\frac{1}{2} [d_i - \overline{d_i} \text{id}_X, p_i - \overline{p_i} \text{id}_X]_+\xi|\xi\right\rangle^2 \\ & & \qquad +\left\langle\frac{1}{2i} [d_i - \overline{d_i} \text{id}_X, p_i - \overline{p_i} \text{id}_X]\xi|\xi\right\rangle^2 \end{array}$

and since the first term is positive

$\displaystyle \begin{array}{rcl} \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} & \geq & \left\langle\frac{1}{2i} [d_i - \overline{d_i} \text{id}_X, p_i - \overline{p_i} \text{id}_X]\xi|\xi\right\rangle^2 \\ & \geq & \left\langle\frac{1}{2i} [d_i,p_i]\xi|\xi\right\rangle^2. \end{array}$

Now (1) and the fact that positive observables have a square root yields the final inequality

$\displaystyle \begin{array}{rcl} \overline{\left(\triangle p_i \right)^2_\xi} \, \overline{\left(\triangle d_i \right)^2_\xi} & \geq & \frac{\mu_i^2}{4} \|\sqrt{p_i} \xi\|^4. \end{array}$

Since ${\mu_i \|\sqrt{p_i} \xi\|^4}$ can only be zero if ${\mu_i}$ is zero the proposition is proved.

Admittedly, that was a bit dry, but it does the job and that is sometimes all that is necessary in mathematics. Now the pace increases and we are heading with giant leaps towards the time evolution equations for markets …