3
$\begingroup$

Let

  • $(\Omega_i,\mathcal A_i)$ be a probability space
  • $\mu$ be a probability measure on $(\Omega_1,\mathcal A_1)$
  • $\kappa$ be a Markov kernel with source $(\Omega_1,\mathcal A_1)$ and target $(\Omega_2,\mathcal A_2)$

Note that $$(\mu\kappa)(A_2):=\int\mu({ d}\omega_1)\kappa(\omega_1,A_2)\;\;\;\text{for }A_2\in\mathcal A_2$$ is a probability measure on $(\Omega,\mathcal A_2)$.

I've read that, by the Cauchy-Schwarz inequality, $$\int\mu({\rm d}\omega_1)\int\kappa(\omega_1,{\rm d}\omega_2)f(\omega_2)\le\int f\:{d}(\mu\kappa)\tag1$$ for all $\mathcal A_2$-measurable $f:\Omega_2\to[0,\infty)$. However, it's obvious that we've got equality in $(1)$ for any elementary $\mathcal A_2$-measurable $f$, so shouldn't we trivial obtain equality for all $\mathcal A_2$-measurable $f$? What am I missing? I don't see how the Cauchy-Schwarz inequality is needed here.

$\endgroup$
  • 1
    $\begingroup$ You are absolutely right. What you have read makes no sense at all. There is no need for C-S and the inequality is actually an equality for any non-negative measurable function $f$. $\endgroup$ – Kabo Murphy Oct 26 '18 at 23:30

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.