Skip to main content
deleted 13 characters in body
Source Link
Did
  • 5.7k
  • 1
  • 30
  • 36

consider the following mappings, G and T,

$y(s) = \[Gx\](s)=\exp\left[\sum_{s'}p(s'|s)\log x(s') \right]$$y(s) = Gx(s)=\exp\left[\sum_{s'}p(s'|s)\log x(s') \right]$

$z(s) = \[Ty\](s)=\sum_{s'}q(s'|s)y(s')e^{-r(s')}$$z(s) = Ty(s)=\sum_{s'}q(s'|s)y(s')e^{-r(s')}$

where $0< x(s)\leq 1$ ,$r(s)<0$ , $s,s'\in \{1,2,...,N\}$, and $p(s'|s)$ and $q(s'|s)$ are normalized conditional distributions.

(the first mapping is a generalized geometric mean, and the second is an arithmetic mean with some discount)

my question is - does iterating these mappings, i.e., $x_{t+1} = T\[G(x_t)\]$$x_{t+1} = TG(x_t)$, converges to a unique solution  ?

consider the following mappings, G and T,

$y(s) = \[Gx\](s)=\exp\left[\sum_{s'}p(s'|s)\log x(s') \right]$

$z(s) = \[Ty\](s)=\sum_{s'}q(s'|s)y(s')e^{-r(s')}$

where $0< x(s)\leq 1$ ,$r(s)<0$ , $s,s'\in \{1,2,...,N\}$, and $p(s'|s)$ and $q(s'|s)$ are normalized conditional distributions.

(the first mapping is a generalized geometric mean, and the second is an arithmetic mean with some discount)

my question is - does iterating these mappings, i.e., $x_{t+1} = T\[G(x_t)\]$, converges to a unique solution  ?

consider the following mappings, G and T,

$y(s) = Gx(s)=\exp\left[\sum_{s'}p(s'|s)\log x(s') \right]$

$z(s) = Ty(s)=\sum_{s'}q(s'|s)y(s')e^{-r(s')}$

where $0< x(s)\leq 1$ ,$r(s)<0$ , $s,s'\in \{1,2,...,N\}$, and $p(s'|s)$ and $q(s'|s)$ are normalized conditional distributions.

(the first mapping is a generalized geometric mean, and the second is an arithmetic mean with some discount)

my question is - does iterating these mappings, i.e., $x_{t+1} = TG(x_t)$, converges to a unique solution?

improved title, added probability tag
Link
Gerald Edgar
  • 41.1k
  • 5
  • 125
  • 219

a unique solution ? iteration involving conditional distributions

removed inappropriate tag, stuck on real-analysis tag
Link
Yemon Choi
  • 25.8k
  • 9
  • 69
  • 156
Source Link
Loading