1
$\begingroup$

Could anyone explain to me what does it mean by a map $f\to K_f$ and $f\to \rho(f(x_0), x_0)$ has an algebraic tail relative some measure, where $\rho$ is Prohorov metric, from this paper Paper which are expressed in equation $(5.1)$ and $(5.2)$?

$f\mapsto K_f$ has algebraic tail related to $\mu\dots (5.1)$

$f\mapsto \rho(x_0, f(x_0)$ has algebraic tail related to $\mu\dots (5.2)$

They have the definition, but I am unable to relate mathematically what $(5.1)$ and $(5.2)$ will mean from the definition $(5.2)$.

A random variable $Y$ has algebraic tail if there are positive, finite constants $\alpha, \beta$ such that \begin{align} \mathbb P(Y>y)< \frac{\alpha}{y^{\beta}} \end{align}

Thanks a lot for any help. The idea as well as the definition I am not able to grasp, especially algebraic tail related to $\mu$.

Improve this question
$\endgroup$
1
  • $\begingroup$ care to copy in equations (5.1) and (5.2) to improve presentation and readability? $\endgroup$
    – kodlu
    Commented Sep 4, 2019 at 21:46

1 Answer 1

Reset to default
1
$\begingroup$

It just means that $$\mu\left(\{f: K_f>y\}\right)<\frac{\alpha}{y^\beta}$$ and $$\mu\left(\{f: \rho(x_0,f(x_0))>y\}\right)<\frac{\alpha}{y^\beta}$$

Improve this answer
$\endgroup$
1
  • $\begingroup$ @riemann Well, "relative to $\mu$'' commonly means that we use $\mu$ as our probability measure $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Sep 6, 2019 at 1:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .