Confusion in definition of Stochastic Process - Papoulis 4th Edstate space of stochastic processMarkov process vs. markov chain vs. random process vs. stochastic process vs. collection of random variablesWhat is the sample path of a stochastic processDefinition of Time SeriesAn example of stochastic processIn layman's terms: What is a stochastic process?definition of stochastic process on countable spaceRandom walk and definition of Stochastic ProcessesStochastic Processes and TrajectoriesSample function in a stochastic process

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Confusion in definition of Stochastic Process - Papoulis 4th Ed


state space of stochastic processMarkov process vs. markov chain vs. random process vs. stochastic process vs. collection of random variablesWhat is the sample path of a stochastic processDefinition of Time SeriesAn example of stochastic processIn layman's terms: What is a stochastic process?definition of stochastic process on countable spaceRandom walk and definition of Stochastic ProcessesStochastic Processes and TrajectoriesSample function in a stochastic process













0












$begingroup$


I'm reading the book Probability, Random Variable & Stochastic Processes by Papoulis & Pillai 4th Ed.



Chapter 9 starts with definition of Stochastic Process. On Pg 373 Sec 9-1 Definitions, it states




As we recall, a random variable x is a rule for assigning to every outcome $zeta$ of an experiment S a number $bfx(zeta)$. A stochastic process $bfx$$(t)$ is a rule for assigning to every $zeta$ a function $bfx$$(t, zeta)$.




Then it goes on to add the following:




Thus a stochastic process is a family of time functions depending on the parameter $zeta$ or, equivalently, a function of t and {. The domain of $zeta$ is the set of all experimental outcomes and the domain of t is a set $mathbbR$ of real numbers.




Thereafter, there is a series of contradictions that I feel with this definition. The definition means that for if conditionally say $zeta_i$ occurs then the function $bfx$$(t)$ results which on this given condition is now completely deterministic, rather than a collection of random variables defined on the same sample space. In some places in the book this the latter view is implicit. For example on pg. 375 titled $bfStatistics of Stochastic Processes$




A stochastic process is a noncountable infinity of random variables, one for each $t$.




Towards the end of pg 373,




We shall use the notation $bfx$$(t)$ to represent a stochastic
process omitting, as in the case of random variables, its dependence
on $zeta$. Thus $bfx$$(t)$ has the following interpretations:



  1. It is a family (or an ensemble) of functions x(t, $zeta$). In this interpretation, t and $zeta$ are variables.


  2. It is a single time function (or a sample of the given process). In this case, t is a variable and $zeta$ is fixed.

  3. If t is fixed and $zeta$ is variable, then x(t) is a random variable equal to the state of the given process at time t.

  4. If t and $zeta$ are fixed, then $bfx$$(t)$ is a number



For point 2, this means that a sample path $x_i(t)$ when $zeta = zeta_i$

is completely deterministic & not a collection of random variables. Say $zeta in H,T$ be the sample space. Let the stochastic process be, $x(t,H) = 1+t, tgeq 0$ & $x(t,T) = 1-t, tgeq 0$. Thus $Px(t) = 1+t = 1$, is completely deterministic.



Thus, this definition only limits itself to these type of stochastic processes.



This has left me confused.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I'm reading the book Probability, Random Variable & Stochastic Processes by Papoulis & Pillai 4th Ed.



    Chapter 9 starts with definition of Stochastic Process. On Pg 373 Sec 9-1 Definitions, it states




    As we recall, a random variable x is a rule for assigning to every outcome $zeta$ of an experiment S a number $bfx(zeta)$. A stochastic process $bfx$$(t)$ is a rule for assigning to every $zeta$ a function $bfx$$(t, zeta)$.




    Then it goes on to add the following:




    Thus a stochastic process is a family of time functions depending on the parameter $zeta$ or, equivalently, a function of t and {. The domain of $zeta$ is the set of all experimental outcomes and the domain of t is a set $mathbbR$ of real numbers.




    Thereafter, there is a series of contradictions that I feel with this definition. The definition means that for if conditionally say $zeta_i$ occurs then the function $bfx$$(t)$ results which on this given condition is now completely deterministic, rather than a collection of random variables defined on the same sample space. In some places in the book this the latter view is implicit. For example on pg. 375 titled $bfStatistics of Stochastic Processes$




    A stochastic process is a noncountable infinity of random variables, one for each $t$.




    Towards the end of pg 373,




    We shall use the notation $bfx$$(t)$ to represent a stochastic
    process omitting, as in the case of random variables, its dependence
    on $zeta$. Thus $bfx$$(t)$ has the following interpretations:



    1. It is a family (or an ensemble) of functions x(t, $zeta$). In this interpretation, t and $zeta$ are variables.


    2. It is a single time function (or a sample of the given process). In this case, t is a variable and $zeta$ is fixed.

    3. If t is fixed and $zeta$ is variable, then x(t) is a random variable equal to the state of the given process at time t.

    4. If t and $zeta$ are fixed, then $bfx$$(t)$ is a number



    For point 2, this means that a sample path $x_i(t)$ when $zeta = zeta_i$

    is completely deterministic & not a collection of random variables. Say $zeta in H,T$ be the sample space. Let the stochastic process be, $x(t,H) = 1+t, tgeq 0$ & $x(t,T) = 1-t, tgeq 0$. Thus $Px(t) = 1+t = 1$, is completely deterministic.



    Thus, this definition only limits itself to these type of stochastic processes.



    This has left me confused.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I'm reading the book Probability, Random Variable & Stochastic Processes by Papoulis & Pillai 4th Ed.



      Chapter 9 starts with definition of Stochastic Process. On Pg 373 Sec 9-1 Definitions, it states




      As we recall, a random variable x is a rule for assigning to every outcome $zeta$ of an experiment S a number $bfx(zeta)$. A stochastic process $bfx$$(t)$ is a rule for assigning to every $zeta$ a function $bfx$$(t, zeta)$.




      Then it goes on to add the following:




      Thus a stochastic process is a family of time functions depending on the parameter $zeta$ or, equivalently, a function of t and {. The domain of $zeta$ is the set of all experimental outcomes and the domain of t is a set $mathbbR$ of real numbers.




      Thereafter, there is a series of contradictions that I feel with this definition. The definition means that for if conditionally say $zeta_i$ occurs then the function $bfx$$(t)$ results which on this given condition is now completely deterministic, rather than a collection of random variables defined on the same sample space. In some places in the book this the latter view is implicit. For example on pg. 375 titled $bfStatistics of Stochastic Processes$




      A stochastic process is a noncountable infinity of random variables, one for each $t$.




      Towards the end of pg 373,




      We shall use the notation $bfx$$(t)$ to represent a stochastic
      process omitting, as in the case of random variables, its dependence
      on $zeta$. Thus $bfx$$(t)$ has the following interpretations:



      1. It is a family (or an ensemble) of functions x(t, $zeta$). In this interpretation, t and $zeta$ are variables.


      2. It is a single time function (or a sample of the given process). In this case, t is a variable and $zeta$ is fixed.

      3. If t is fixed and $zeta$ is variable, then x(t) is a random variable equal to the state of the given process at time t.

      4. If t and $zeta$ are fixed, then $bfx$$(t)$ is a number



      For point 2, this means that a sample path $x_i(t)$ when $zeta = zeta_i$

      is completely deterministic & not a collection of random variables. Say $zeta in H,T$ be the sample space. Let the stochastic process be, $x(t,H) = 1+t, tgeq 0$ & $x(t,T) = 1-t, tgeq 0$. Thus $Px(t) = 1+t = 1$, is completely deterministic.



      Thus, this definition only limits itself to these type of stochastic processes.



      This has left me confused.










      share|cite|improve this question











      $endgroup$




      I'm reading the book Probability, Random Variable & Stochastic Processes by Papoulis & Pillai 4th Ed.



      Chapter 9 starts with definition of Stochastic Process. On Pg 373 Sec 9-1 Definitions, it states




      As we recall, a random variable x is a rule for assigning to every outcome $zeta$ of an experiment S a number $bfx(zeta)$. A stochastic process $bfx$$(t)$ is a rule for assigning to every $zeta$ a function $bfx$$(t, zeta)$.




      Then it goes on to add the following:




      Thus a stochastic process is a family of time functions depending on the parameter $zeta$ or, equivalently, a function of t and {. The domain of $zeta$ is the set of all experimental outcomes and the domain of t is a set $mathbbR$ of real numbers.




      Thereafter, there is a series of contradictions that I feel with this definition. The definition means that for if conditionally say $zeta_i$ occurs then the function $bfx$$(t)$ results which on this given condition is now completely deterministic, rather than a collection of random variables defined on the same sample space. In some places in the book this the latter view is implicit. For example on pg. 375 titled $bfStatistics of Stochastic Processes$




      A stochastic process is a noncountable infinity of random variables, one for each $t$.




      Towards the end of pg 373,




      We shall use the notation $bfx$$(t)$ to represent a stochastic
      process omitting, as in the case of random variables, its dependence
      on $zeta$. Thus $bfx$$(t)$ has the following interpretations:



      1. It is a family (or an ensemble) of functions x(t, $zeta$). In this interpretation, t and $zeta$ are variables.


      2. It is a single time function (or a sample of the given process). In this case, t is a variable and $zeta$ is fixed.

      3. If t is fixed and $zeta$ is variable, then x(t) is a random variable equal to the state of the given process at time t.

      4. If t and $zeta$ are fixed, then $bfx$$(t)$ is a number



      For point 2, this means that a sample path $x_i(t)$ when $zeta = zeta_i$

      is completely deterministic & not a collection of random variables. Say $zeta in H,T$ be the sample space. Let the stochastic process be, $x(t,H) = 1+t, tgeq 0$ & $x(t,T) = 1-t, tgeq 0$. Thus $Px(t) = 1+t = 1$, is completely deterministic.



      Thus, this definition only limits itself to these type of stochastic processes.



      This has left me confused.







      probability statistics stochastic-processes






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 16 at 13:33







      Ricky

















      asked Mar 16 at 12:08









      RickyRicky

      16414




      16414




















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          I think the author simply explains a convenient abuse of notation.



          The stochastic process is a map $(t,zeta)mapsto x(t,zeta)$ (or equivalently, a family $(tmapsto x(t,zeta))_zeta$).



          1. Have you ever written something like "the function $t^2$ is differentiable" ? Well, $t^2$ is technically not a function, but $tmapsto t^2$ is, and that is what you meant, and everybody understood it. Similarly, you might write $x(t,zeta)$ to denote the function $(t,zeta)mapsto x(t,zeta)$. However in many reasonings, the variable $zeta$ is often fixed and it is annoying to write $zeta$ everywhere. So we agree to say that $x(t)$ denotes $x(t,zeta)$, which itself might denote the function $(t,zeta)mapsto x(t,zeta)$.


          2. If $zeta$ is fixed, you might want to look at the function $tmapsto x(t,zeta)$ (usually called a sample path). Recall that we are use to writing $x(t)$ instead of $x(t,zeta)$. So you want to have a look at the function $tmapsto x(t)$ (the variable $zeta$ is implicit). For convenience, we denote by $x(t)$ the map $tmapsto x(t)$.


          3. You also might want to have a look at the map $zetamapsto x(t,zeta)$, where $t$ is fixed. Even if $zeta$ is not fixed this time, we are really used to making it implicit and we still denote by $x(t)$ the map $zetamapsto x(t,zeta)$, which is a random variable.


          4. Of course, if $t$ and $zeta$ are fixed, then $x(t,zeta)$ is a number, which is denoted $x(t)$ where we omit $zeta$, as usual.


          In the end, depending on the context, $x(t)$ might denote a stochastic process, a sample path, a random variable or a number. Usually, the context is clear enough so that we do not get confused. If the context becomes ambiguous, then you should specify your notation (for instance write $tmapsto x(t,zeta)$ instead of $x(t)$, since the latter might be misunderstood).






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I have edited my question for clarity. My doubt is that the definition is very restrictive.
            $endgroup$
            – Ricky
            Mar 16 at 13:34










          • $begingroup$
            "the definition is very restrictive". Which definition? Point 2? It's not really a definition. It's more like one of the 4 possible ways of understanding the term $x(t)$. Depending on the context, $x(t)$ means different things. Each situation described by points 1 to 4 is different. Of course in the case where $zeta$ is fixed, everything is deterministic (sort of), but that concerns only points 2 and 4.
            $endgroup$
            – Will
            Mar 16 at 14:22










          Your Answer





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          1 Answer
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          1 Answer
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          active

          oldest

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          active

          oldest

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          0












          $begingroup$

          I think the author simply explains a convenient abuse of notation.



          The stochastic process is a map $(t,zeta)mapsto x(t,zeta)$ (or equivalently, a family $(tmapsto x(t,zeta))_zeta$).



          1. Have you ever written something like "the function $t^2$ is differentiable" ? Well, $t^2$ is technically not a function, but $tmapsto t^2$ is, and that is what you meant, and everybody understood it. Similarly, you might write $x(t,zeta)$ to denote the function $(t,zeta)mapsto x(t,zeta)$. However in many reasonings, the variable $zeta$ is often fixed and it is annoying to write $zeta$ everywhere. So we agree to say that $x(t)$ denotes $x(t,zeta)$, which itself might denote the function $(t,zeta)mapsto x(t,zeta)$.


          2. If $zeta$ is fixed, you might want to look at the function $tmapsto x(t,zeta)$ (usually called a sample path). Recall that we are use to writing $x(t)$ instead of $x(t,zeta)$. So you want to have a look at the function $tmapsto x(t)$ (the variable $zeta$ is implicit). For convenience, we denote by $x(t)$ the map $tmapsto x(t)$.


          3. You also might want to have a look at the map $zetamapsto x(t,zeta)$, where $t$ is fixed. Even if $zeta$ is not fixed this time, we are really used to making it implicit and we still denote by $x(t)$ the map $zetamapsto x(t,zeta)$, which is a random variable.


          4. Of course, if $t$ and $zeta$ are fixed, then $x(t,zeta)$ is a number, which is denoted $x(t)$ where we omit $zeta$, as usual.


          In the end, depending on the context, $x(t)$ might denote a stochastic process, a sample path, a random variable or a number. Usually, the context is clear enough so that we do not get confused. If the context becomes ambiguous, then you should specify your notation (for instance write $tmapsto x(t,zeta)$ instead of $x(t)$, since the latter might be misunderstood).






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I have edited my question for clarity. My doubt is that the definition is very restrictive.
            $endgroup$
            – Ricky
            Mar 16 at 13:34










          • $begingroup$
            "the definition is very restrictive". Which definition? Point 2? It's not really a definition. It's more like one of the 4 possible ways of understanding the term $x(t)$. Depending on the context, $x(t)$ means different things. Each situation described by points 1 to 4 is different. Of course in the case where $zeta$ is fixed, everything is deterministic (sort of), but that concerns only points 2 and 4.
            $endgroup$
            – Will
            Mar 16 at 14:22















          0












          $begingroup$

          I think the author simply explains a convenient abuse of notation.



          The stochastic process is a map $(t,zeta)mapsto x(t,zeta)$ (or equivalently, a family $(tmapsto x(t,zeta))_zeta$).



          1. Have you ever written something like "the function $t^2$ is differentiable" ? Well, $t^2$ is technically not a function, but $tmapsto t^2$ is, and that is what you meant, and everybody understood it. Similarly, you might write $x(t,zeta)$ to denote the function $(t,zeta)mapsto x(t,zeta)$. However in many reasonings, the variable $zeta$ is often fixed and it is annoying to write $zeta$ everywhere. So we agree to say that $x(t)$ denotes $x(t,zeta)$, which itself might denote the function $(t,zeta)mapsto x(t,zeta)$.


          2. If $zeta$ is fixed, you might want to look at the function $tmapsto x(t,zeta)$ (usually called a sample path). Recall that we are use to writing $x(t)$ instead of $x(t,zeta)$. So you want to have a look at the function $tmapsto x(t)$ (the variable $zeta$ is implicit). For convenience, we denote by $x(t)$ the map $tmapsto x(t)$.


          3. You also might want to have a look at the map $zetamapsto x(t,zeta)$, where $t$ is fixed. Even if $zeta$ is not fixed this time, we are really used to making it implicit and we still denote by $x(t)$ the map $zetamapsto x(t,zeta)$, which is a random variable.


          4. Of course, if $t$ and $zeta$ are fixed, then $x(t,zeta)$ is a number, which is denoted $x(t)$ where we omit $zeta$, as usual.


          In the end, depending on the context, $x(t)$ might denote a stochastic process, a sample path, a random variable or a number. Usually, the context is clear enough so that we do not get confused. If the context becomes ambiguous, then you should specify your notation (for instance write $tmapsto x(t,zeta)$ instead of $x(t)$, since the latter might be misunderstood).






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I have edited my question for clarity. My doubt is that the definition is very restrictive.
            $endgroup$
            – Ricky
            Mar 16 at 13:34










          • $begingroup$
            "the definition is very restrictive". Which definition? Point 2? It's not really a definition. It's more like one of the 4 possible ways of understanding the term $x(t)$. Depending on the context, $x(t)$ means different things. Each situation described by points 1 to 4 is different. Of course in the case where $zeta$ is fixed, everything is deterministic (sort of), but that concerns only points 2 and 4.
            $endgroup$
            – Will
            Mar 16 at 14:22













          0












          0








          0





          $begingroup$

          I think the author simply explains a convenient abuse of notation.



          The stochastic process is a map $(t,zeta)mapsto x(t,zeta)$ (or equivalently, a family $(tmapsto x(t,zeta))_zeta$).



          1. Have you ever written something like "the function $t^2$ is differentiable" ? Well, $t^2$ is technically not a function, but $tmapsto t^2$ is, and that is what you meant, and everybody understood it. Similarly, you might write $x(t,zeta)$ to denote the function $(t,zeta)mapsto x(t,zeta)$. However in many reasonings, the variable $zeta$ is often fixed and it is annoying to write $zeta$ everywhere. So we agree to say that $x(t)$ denotes $x(t,zeta)$, which itself might denote the function $(t,zeta)mapsto x(t,zeta)$.


          2. If $zeta$ is fixed, you might want to look at the function $tmapsto x(t,zeta)$ (usually called a sample path). Recall that we are use to writing $x(t)$ instead of $x(t,zeta)$. So you want to have a look at the function $tmapsto x(t)$ (the variable $zeta$ is implicit). For convenience, we denote by $x(t)$ the map $tmapsto x(t)$.


          3. You also might want to have a look at the map $zetamapsto x(t,zeta)$, where $t$ is fixed. Even if $zeta$ is not fixed this time, we are really used to making it implicit and we still denote by $x(t)$ the map $zetamapsto x(t,zeta)$, which is a random variable.


          4. Of course, if $t$ and $zeta$ are fixed, then $x(t,zeta)$ is a number, which is denoted $x(t)$ where we omit $zeta$, as usual.


          In the end, depending on the context, $x(t)$ might denote a stochastic process, a sample path, a random variable or a number. Usually, the context is clear enough so that we do not get confused. If the context becomes ambiguous, then you should specify your notation (for instance write $tmapsto x(t,zeta)$ instead of $x(t)$, since the latter might be misunderstood).






          share|cite|improve this answer









          $endgroup$



          I think the author simply explains a convenient abuse of notation.



          The stochastic process is a map $(t,zeta)mapsto x(t,zeta)$ (or equivalently, a family $(tmapsto x(t,zeta))_zeta$).



          1. Have you ever written something like "the function $t^2$ is differentiable" ? Well, $t^2$ is technically not a function, but $tmapsto t^2$ is, and that is what you meant, and everybody understood it. Similarly, you might write $x(t,zeta)$ to denote the function $(t,zeta)mapsto x(t,zeta)$. However in many reasonings, the variable $zeta$ is often fixed and it is annoying to write $zeta$ everywhere. So we agree to say that $x(t)$ denotes $x(t,zeta)$, which itself might denote the function $(t,zeta)mapsto x(t,zeta)$.


          2. If $zeta$ is fixed, you might want to look at the function $tmapsto x(t,zeta)$ (usually called a sample path). Recall that we are use to writing $x(t)$ instead of $x(t,zeta)$. So you want to have a look at the function $tmapsto x(t)$ (the variable $zeta$ is implicit). For convenience, we denote by $x(t)$ the map $tmapsto x(t)$.


          3. You also might want to have a look at the map $zetamapsto x(t,zeta)$, where $t$ is fixed. Even if $zeta$ is not fixed this time, we are really used to making it implicit and we still denote by $x(t)$ the map $zetamapsto x(t,zeta)$, which is a random variable.


          4. Of course, if $t$ and $zeta$ are fixed, then $x(t,zeta)$ is a number, which is denoted $x(t)$ where we omit $zeta$, as usual.


          In the end, depending on the context, $x(t)$ might denote a stochastic process, a sample path, a random variable or a number. Usually, the context is clear enough so that we do not get confused. If the context becomes ambiguous, then you should specify your notation (for instance write $tmapsto x(t,zeta)$ instead of $x(t)$, since the latter might be misunderstood).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 16 at 13:15









          WillWill

          5015




          5015











          • $begingroup$
            I have edited my question for clarity. My doubt is that the definition is very restrictive.
            $endgroup$
            – Ricky
            Mar 16 at 13:34










          • $begingroup$
            "the definition is very restrictive". Which definition? Point 2? It's not really a definition. It's more like one of the 4 possible ways of understanding the term $x(t)$. Depending on the context, $x(t)$ means different things. Each situation described by points 1 to 4 is different. Of course in the case where $zeta$ is fixed, everything is deterministic (sort of), but that concerns only points 2 and 4.
            $endgroup$
            – Will
            Mar 16 at 14:22
















          • $begingroup$
            I have edited my question for clarity. My doubt is that the definition is very restrictive.
            $endgroup$
            – Ricky
            Mar 16 at 13:34










          • $begingroup$
            "the definition is very restrictive". Which definition? Point 2? It's not really a definition. It's more like one of the 4 possible ways of understanding the term $x(t)$. Depending on the context, $x(t)$ means different things. Each situation described by points 1 to 4 is different. Of course in the case where $zeta$ is fixed, everything is deterministic (sort of), but that concerns only points 2 and 4.
            $endgroup$
            – Will
            Mar 16 at 14:22















          $begingroup$
          I have edited my question for clarity. My doubt is that the definition is very restrictive.
          $endgroup$
          – Ricky
          Mar 16 at 13:34




          $begingroup$
          I have edited my question for clarity. My doubt is that the definition is very restrictive.
          $endgroup$
          – Ricky
          Mar 16 at 13:34












          $begingroup$
          "the definition is very restrictive". Which definition? Point 2? It's not really a definition. It's more like one of the 4 possible ways of understanding the term $x(t)$. Depending on the context, $x(t)$ means different things. Each situation described by points 1 to 4 is different. Of course in the case where $zeta$ is fixed, everything is deterministic (sort of), but that concerns only points 2 and 4.
          $endgroup$
          – Will
          Mar 16 at 14:22




          $begingroup$
          "the definition is very restrictive". Which definition? Point 2? It's not really a definition. It's more like one of the 4 possible ways of understanding the term $x(t)$. Depending on the context, $x(t)$ means different things. Each situation described by points 1 to 4 is different. Of course in the case where $zeta$ is fixed, everything is deterministic (sort of), but that concerns only points 2 and 4.
          $endgroup$
          – Will
          Mar 16 at 14:22

















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