Concentration of measure is a small area of statistics and probability theory that proved inequalities regarding the statistical properties of sets of random variables that exclude one of those random variables in the set. For example, $X^{(i)}$ is the set of random variables $X_1, X_2, \dots, X_n$ that excludes one of those variables, $X_i$, that was previously contained in the full set. In other words, set $X^{(i)}$ excludes variable $X_i$.

Concentration of measure phenomena arise in various settings, often studied by separate communities: over Gaussian space, Riemannian manifolds, discrete product spaces, and algebraic structures, but the Gaussian-probability setting is likely to be most relatable to us.

Do concentration of measure phenomena, or concentration inequalities, exist anywhere in financial economics or financial mathematics? If so, how are they useful?

  • 1
    $\begingroup$ Concentration inequalies are heavily used in the theory underlying machine learning. Thus, you may likely find them in applications of machine learning to problems in finance, which certainly do exist. $\endgroup$ – Tobsn Nov 10 '20 at 13:52
  • 1
    $\begingroup$ Concentration of measure is exploitable as the other poster said, in Machine learning. In particular, low dimensional problems are easy due to few dimensions, high dimensional problems are easy due to measure concentration (for example, as the dimension goes to infinity, the surface area of the sphere dominates the volume of the ball (assuming it is valid to compare n, n-1 dimensional measures). The medium dimensional problems are problematic. $\endgroup$ – rubikscube09 Nov 10 '20 at 23:10
  • 1
    $\begingroup$ However, I'm unaware as to non-theoretical applications (theory here being say, some sort of proof of convergence, or asymptotic bounds). I know they figure heavily when one wants to study the geometry of banach spaces, which is unfortunately a beautiful, but inactive field of mathematics. $\endgroup$ – rubikscube09 Nov 10 '20 at 23:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.