I'm studying Pavliotis' Stochastic Processes and I am having trouble with one of the exercises.Specifically the first exercise of chapter two is prove that the Markov property in the sense off the immediate future being independent of the past given the present, i.e. P ( X n + 1 | X 1 , … , X n ) = P ( X n + 1 | X n ) implies that arbitrary futures are independent of the past given the present, i.e. P ( X n + m | X 1 , … , X n ) = P ( X n + m | X n ).Conceptually I imagine a proof using induction along the lines of: Assume the equality holds for some m, then P ( X n + m + 1 | X 1 , … , X n ) = ∫ P ( X n + m + 1 , X n + m = x | X 1 , … , X n ) d x = ∫ P ( X n + m + 1 | X n + m = x ) P ( X n + m = x | X 1 , … , X n ) d x = ∫ P ( X n + m + 1 | X n + m = x ) P ( X n + m = x | X n ) d x = ∫ P ( X n + m + 1 , X n + m = x | X n ) d x = P ( X n + m + 1 | X n ) However, this is based on intuition and my experience non-measure-theoretic probability theory. I cannot justify these steps when I think of P as a measure and P ( X = x ) meaning P { X − 1 ( x ) }. In fact I have had trouble finding a definition of conditional probability that is at all helpful. Most authors seem to provide the abstract definition in terms of conditional expectation with respect to σ-algebra, but I haven't found any resources that show how to work with this definition.So my question is: how (if it all) are these steps, particularly the first two equalities justified from a measure theoretic perspective?

Question

I&#039;m studying Pavliotis&#039; Stochastic Processes and I am having trouble with one of the exercises.Specifically the first exercise of chapter two is prove that the Markov property in the sense off the immediate future being independent of the past given the present, i.e.   P  (      X          n      +      1            |        X    1    ,  …  ,      X    n    )  =  P  (      X          n      +      1            |        X    n    ) implies that arbitrary futures are independent of the past given the present, i.e.   P  (      X          n      +      m            |        X    1    ,  …  ,      X    n    )  =  P  (      X          n      +      m            |        X    n    ).Conceptually I imagine a proof using induction along the lines of: Assume the equality holds for some m, then                    P        (                  X                      n            +            m            +            1                                    |                          X          1                ,        …        ,                  X          n                )        =        ∫        P        (                  X                      n            +            m            +            1                          ,                  X                      n            +            m                          =        x                  |                          X          1                ,        …        ,                  X          n                )        d        x                            =        ∫        P        (                  X                      n            +            m            +            1                                    |                          X                      n            +            m                          =        x        )        P        (                  X                      n            +            m                          =        x                  |                          X          1                ,        …        ,                  X          n                )        d        x                            =        ∫        P        (                  X                      n            +            m            +            1                                    |                          X                      n            +            m                          =        x        )        P        (                  X                      n            +            m                          =        x                  |                          X          n                )        d        x                            =        ∫        P        (                  X                      n            +            m            +            1                          ,                  X                      n            +            m                          =        x                  |                          X          n                )        d        x        =        P        (                  X                      n            +            m            +            1                                    |                          X          n                )            However, this is based on intuition and my experience non-measure-theoretic probability theory. I cannot justify these steps when I think of   P as a measure and   P  (  X  =  x  ) meaning   P  {      X          −      1        (  x  )  }. In fact I have had trouble finding a definition of conditional probability that is at all helpful. Most authors seem to provide the abstract definition in terms of conditional expectation with respect to   σ-algebra, but I haven&#039;t found any resources that show how to work with this definition.So my question is: how (if it all) are these steps, particularly the first two equalities justified from a measure theoretic perspective?

SweallySnicles3 · Accepted Answer

A Markov process       X    t   is a stochastic process which can be described in terms of a transition operator       T          s      ,      t       where       X    t    =      T          s      ,      t            X    s  . For the particular case of Markov chains, the operator is a matrix.For your original question, we are given       X          n      +      1        =      T          n      ,      n      +      1            X    n   Therefore       X          n      +      m        =      T          n      .      m            X    n   where       T          n      ,      m        =      T          m      −      1.      m        ×  .  .  .  ×      T          n      ,      n      +      1       is the product of given operators.

Holetaug · Accepted Answer

General formula   P  (  A      |    B  )  =  ∫  P  (  A      |    B  =  x  )      f    B    (  x  )  d  x, where       f    B    (  x  ) is pdf. In the rhs of your first equation you are missing a pdf for       X          m      +      n       although you put something in on the next line, which may be the pdf.

I'm studying Pavliotis' Stochastic Processes and I am having trouble with one of the exercises. Sp

Answered question

Answer & Explanation

New Questions in Pre-Algebra