distribution according to random measures












1












$begingroup$


Is a random probability measure a probability measure, i.e. if $X_i$ is random variable on some probability space $(Omega,mathcal F,mathbb P)$ could its induced distribution be a random probability measure $mu$, say $X_isim mu$?



Then, as $mu$ is itself random, it has an own probability distribution, $musim Q$.



For example let's take the empirical measure $P_n(A)=frac{1}{n}sum_{i=1}^n delta_{X_i}(A)$. Then the empirical distribution function $F_n(x):=frac{1}{n}sum_{i=1}^n 1{X_ile x}$ is a random probability measure as it is an empirical measure indexed by the class $mathcal C={(-infty,x]: xin mathbb R}$? Can one say $X_isim P_n$?






probability probability-theory measure-theory random-variables random







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$endgroup$

















    1












    $begingroup$


    Is a random probability measure a probability measure, i.e. if $X_i$ is random variable on some probability space $(Omega,mathcal F,mathbb P)$ could its induced distribution be a random probability measure $mu$, say $X_isim mu$?



    Then, as $mu$ is itself random, it has an own probability distribution, $musim Q$.



    For example let's take the empirical measure $P_n(A)=frac{1}{n}sum_{i=1}^n delta_{X_i}(A)$. Then the empirical distribution function $F_n(x):=frac{1}{n}sum_{i=1}^n 1{X_ile x}$ is a random probability measure as it is an empirical measure indexed by the class $mathcal C={(-infty,x]: xin mathbb R}$? Can one say $X_isim P_n$?






    probability probability-theory measure-theory random-variables random







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Is a random probability measure a probability measure, i.e. if $X_i$ is random variable on some probability space $(Omega,mathcal F,mathbb P)$ could its induced distribution be a random probability measure $mu$, say $X_isim mu$?



      Then, as $mu$ is itself random, it has an own probability distribution, $musim Q$.



      For example let's take the empirical measure $P_n(A)=frac{1}{n}sum_{i=1}^n delta_{X_i}(A)$. Then the empirical distribution function $F_n(x):=frac{1}{n}sum_{i=1}^n 1{X_ile x}$ is a random probability measure as it is an empirical measure indexed by the class $mathcal C={(-infty,x]: xin mathbb R}$? Can one say $X_isim P_n$?






      probability probability-theory measure-theory random-variables random







      share|cite|improve this question









      $endgroup$




      Is a random probability measure a probability measure, i.e. if $X_i$ is random variable on some probability space $(Omega,mathcal F,mathbb P)$ could its induced distribution be a random probability measure $mu$, say $X_isim mu$?



      Then, as $mu$ is itself random, it has an own probability distribution, $musim Q$.



      For example let's take the empirical measure $P_n(A)=frac{1}{n}sum_{i=1}^n delta_{X_i}(A)$. Then the empirical distribution function $F_n(x):=frac{1}{n}sum_{i=1}^n 1{X_ile x}$ is a random probability measure as it is an empirical measure indexed by the class $mathcal C={(-infty,x]: xin mathbb R}$? Can one say $X_isim P_n$?






      probability probability-theory measure-theory random-variables random




      probability probability-theory measure-theory random-variables random






      share|cite|improve this question













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      asked Jan 3 at 17:17









      lemontreelemontree

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          1












          $begingroup$

          For a random measure, it is just the case that $Omega = mathcal{P}$ the set of all probability measures on space. For example $P_n$ in your definition.



          The empirical measure is a random variable on the space of functions. This measure is obtained via taking the push forward measure of the mapping $pi: mathcal{P}(mathbb{R}) mapsto mathcal{S}$, where $mathcal{S}$ is the collection of step functions in $mathbb{R}$.



          $Xsim mu$ is for a fixed measure $mu$. In your example, you cannot say $X_i sim P_n$ because the distribution $X_i$'s is something from which you have drawn the random variables $X_i$'s to construct the measure in the first place ($P_n$ is just a function of $X_i$'s). But, if you wanted mean that $X$ is a "sample" from the empirical measure, then you have to keep in mind that there are two steps of randomness. The first step to construct the measure, and the second is taking a sample from $P_n$ conditionally on the $X_i$'s in the first step. So one needs to write the distribution of $X$ using a conditional distribution.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Consider you have a sample $X=(X_1,dots, X_n).$ Now assume $X_imid mu sim mu$ and $musim Q$, where $mu$ is a random probability measure and $Q$ is the Distribution of the random probability measure. I think that is an example for the "two steps of randomness" you talked about. Can you explain what $X_imid mu sim mu$ means and why we can't use/ what's the difference to $X_isim mu$? Many thanks!
            $endgroup$
            – lemontree
            Jan 3 at 18:52










          • $begingroup$
            The conditional distribution will be a discrete measure. Suppose you toss a coin 6 times and you get TTHTTH. So your empirical measure $P_n$ will give mass 1/3 on head and 2/3 on tail. But $X_1$ is Bernoulli$(1/2)$. What you can do is take a new coin with respect to the observed empirical measure. The outcome $Y$ will be Bernoulli $(1/3)$ conditional on the event that there are exactly $2$ heads among $(X_1,dots,X_6)$.
            $endgroup$
            – Aragon
            Jan 3 at 19:36













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          1 Answer
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          1 Answer
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          active

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          active

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          1












          $begingroup$

          For a random measure, it is just the case that $Omega = mathcal{P}$ the set of all probability measures on space. For example $P_n$ in your definition.



          The empirical measure is a random variable on the space of functions. This measure is obtained via taking the push forward measure of the mapping $pi: mathcal{P}(mathbb{R}) mapsto mathcal{S}$, where $mathcal{S}$ is the collection of step functions in $mathbb{R}$.



          $Xsim mu$ is for a fixed measure $mu$. In your example, you cannot say $X_i sim P_n$ because the distribution $X_i$'s is something from which you have drawn the random variables $X_i$'s to construct the measure in the first place ($P_n$ is just a function of $X_i$'s). But, if you wanted mean that $X$ is a "sample" from the empirical measure, then you have to keep in mind that there are two steps of randomness. The first step to construct the measure, and the second is taking a sample from $P_n$ conditionally on the $X_i$'s in the first step. So one needs to write the distribution of $X$ using a conditional distribution.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Consider you have a sample $X=(X_1,dots, X_n).$ Now assume $X_imid mu sim mu$ and $musim Q$, where $mu$ is a random probability measure and $Q$ is the Distribution of the random probability measure. I think that is an example for the "two steps of randomness" you talked about. Can you explain what $X_imid mu sim mu$ means and why we can't use/ what's the difference to $X_isim mu$? Many thanks!
            $endgroup$
            – lemontree
            Jan 3 at 18:52










          • $begingroup$
            The conditional distribution will be a discrete measure. Suppose you toss a coin 6 times and you get TTHTTH. So your empirical measure $P_n$ will give mass 1/3 on head and 2/3 on tail. But $X_1$ is Bernoulli$(1/2)$. What you can do is take a new coin with respect to the observed empirical measure. The outcome $Y$ will be Bernoulli $(1/3)$ conditional on the event that there are exactly $2$ heads among $(X_1,dots,X_6)$.
            $endgroup$
            – Aragon
            Jan 3 at 19:36


















          1












          $begingroup$

          For a random measure, it is just the case that $Omega = mathcal{P}$ the set of all probability measures on space. For example $P_n$ in your definition.



          The empirical measure is a random variable on the space of functions. This measure is obtained via taking the push forward measure of the mapping $pi: mathcal{P}(mathbb{R}) mapsto mathcal{S}$, where $mathcal{S}$ is the collection of step functions in $mathbb{R}$.



          $Xsim mu$ is for a fixed measure $mu$. In your example, you cannot say $X_i sim P_n$ because the distribution $X_i$'s is something from which you have drawn the random variables $X_i$'s to construct the measure in the first place ($P_n$ is just a function of $X_i$'s). But, if you wanted mean that $X$ is a "sample" from the empirical measure, then you have to keep in mind that there are two steps of randomness. The first step to construct the measure, and the second is taking a sample from $P_n$ conditionally on the $X_i$'s in the first step. So one needs to write the distribution of $X$ using a conditional distribution.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Consider you have a sample $X=(X_1,dots, X_n).$ Now assume $X_imid mu sim mu$ and $musim Q$, where $mu$ is a random probability measure and $Q$ is the Distribution of the random probability measure. I think that is an example for the "two steps of randomness" you talked about. Can you explain what $X_imid mu sim mu$ means and why we can't use/ what's the difference to $X_isim mu$? Many thanks!
            $endgroup$
            – lemontree
            Jan 3 at 18:52










          • $begingroup$
            The conditional distribution will be a discrete measure. Suppose you toss a coin 6 times and you get TTHTTH. So your empirical measure $P_n$ will give mass 1/3 on head and 2/3 on tail. But $X_1$ is Bernoulli$(1/2)$. What you can do is take a new coin with respect to the observed empirical measure. The outcome $Y$ will be Bernoulli $(1/3)$ conditional on the event that there are exactly $2$ heads among $(X_1,dots,X_6)$.
            $endgroup$
            – Aragon
            Jan 3 at 19:36
















          1












          1








          1





          $begingroup$

          For a random measure, it is just the case that $Omega = mathcal{P}$ the set of all probability measures on space. For example $P_n$ in your definition.



          The empirical measure is a random variable on the space of functions. This measure is obtained via taking the push forward measure of the mapping $pi: mathcal{P}(mathbb{R}) mapsto mathcal{S}$, where $mathcal{S}$ is the collection of step functions in $mathbb{R}$.



          $Xsim mu$ is for a fixed measure $mu$. In your example, you cannot say $X_i sim P_n$ because the distribution $X_i$'s is something from which you have drawn the random variables $X_i$'s to construct the measure in the first place ($P_n$ is just a function of $X_i$'s). But, if you wanted mean that $X$ is a "sample" from the empirical measure, then you have to keep in mind that there are two steps of randomness. The first step to construct the measure, and the second is taking a sample from $P_n$ conditionally on the $X_i$'s in the first step. So one needs to write the distribution of $X$ using a conditional distribution.






          share|cite|improve this answer









          $endgroup$



          For a random measure, it is just the case that $Omega = mathcal{P}$ the set of all probability measures on space. For example $P_n$ in your definition.



          The empirical measure is a random variable on the space of functions. This measure is obtained via taking the push forward measure of the mapping $pi: mathcal{P}(mathbb{R}) mapsto mathcal{S}$, where $mathcal{S}$ is the collection of step functions in $mathbb{R}$.



          $Xsim mu$ is for a fixed measure $mu$. In your example, you cannot say $X_i sim P_n$ because the distribution $X_i$'s is something from which you have drawn the random variables $X_i$'s to construct the measure in the first place ($P_n$ is just a function of $X_i$'s). But, if you wanted mean that $X$ is a "sample" from the empirical measure, then you have to keep in mind that there are two steps of randomness. The first step to construct the measure, and the second is taking a sample from $P_n$ conditionally on the $X_i$'s in the first step. So one needs to write the distribution of $X$ using a conditional distribution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 18:05









          AragonAragon

          363




          363












          • $begingroup$
            Consider you have a sample $X=(X_1,dots, X_n).$ Now assume $X_imid mu sim mu$ and $musim Q$, where $mu$ is a random probability measure and $Q$ is the Distribution of the random probability measure. I think that is an example for the "two steps of randomness" you talked about. Can you explain what $X_imid mu sim mu$ means and why we can't use/ what's the difference to $X_isim mu$? Many thanks!
            $endgroup$
            – lemontree
            Jan 3 at 18:52










          • $begingroup$
            The conditional distribution will be a discrete measure. Suppose you toss a coin 6 times and you get TTHTTH. So your empirical measure $P_n$ will give mass 1/3 on head and 2/3 on tail. But $X_1$ is Bernoulli$(1/2)$. What you can do is take a new coin with respect to the observed empirical measure. The outcome $Y$ will be Bernoulli $(1/3)$ conditional on the event that there are exactly $2$ heads among $(X_1,dots,X_6)$.
            $endgroup$
            – Aragon
            Jan 3 at 19:36




















          • $begingroup$
            Consider you have a sample $X=(X_1,dots, X_n).$ Now assume $X_imid mu sim mu$ and $musim Q$, where $mu$ is a random probability measure and $Q$ is the Distribution of the random probability measure. I think that is an example for the "two steps of randomness" you talked about. Can you explain what $X_imid mu sim mu$ means and why we can't use/ what's the difference to $X_isim mu$? Many thanks!
            $endgroup$
            – lemontree
            Jan 3 at 18:52










          • $begingroup$
            The conditional distribution will be a discrete measure. Suppose you toss a coin 6 times and you get TTHTTH. So your empirical measure $P_n$ will give mass 1/3 on head and 2/3 on tail. But $X_1$ is Bernoulli$(1/2)$. What you can do is take a new coin with respect to the observed empirical measure. The outcome $Y$ will be Bernoulli $(1/3)$ conditional on the event that there are exactly $2$ heads among $(X_1,dots,X_6)$.
            $endgroup$
            – Aragon
            Jan 3 at 19:36


















          $begingroup$
          Consider you have a sample $X=(X_1,dots, X_n).$ Now assume $X_imid mu sim mu$ and $musim Q$, where $mu$ is a random probability measure and $Q$ is the Distribution of the random probability measure. I think that is an example for the "two steps of randomness" you talked about. Can you explain what $X_imid mu sim mu$ means and why we can't use/ what's the difference to $X_isim mu$? Many thanks!
          $endgroup$
          – lemontree
          Jan 3 at 18:52




          $begingroup$
          Consider you have a sample $X=(X_1,dots, X_n).$ Now assume $X_imid mu sim mu$ and $musim Q$, where $mu$ is a random probability measure and $Q$ is the Distribution of the random probability measure. I think that is an example for the "two steps of randomness" you talked about. Can you explain what $X_imid mu sim mu$ means and why we can't use/ what's the difference to $X_isim mu$? Many thanks!
          $endgroup$
          – lemontree
          Jan 3 at 18:52












          $begingroup$
          The conditional distribution will be a discrete measure. Suppose you toss a coin 6 times and you get TTHTTH. So your empirical measure $P_n$ will give mass 1/3 on head and 2/3 on tail. But $X_1$ is Bernoulli$(1/2)$. What you can do is take a new coin with respect to the observed empirical measure. The outcome $Y$ will be Bernoulli $(1/3)$ conditional on the event that there are exactly $2$ heads among $(X_1,dots,X_6)$.
          $endgroup$
          – Aragon
          Jan 3 at 19:36






          $begingroup$
          The conditional distribution will be a discrete measure. Suppose you toss a coin 6 times and you get TTHTTH. So your empirical measure $P_n$ will give mass 1/3 on head and 2/3 on tail. But $X_1$ is Bernoulli$(1/2)$. What you can do is take a new coin with respect to the observed empirical measure. The outcome $Y$ will be Bernoulli $(1/3)$ conditional on the event that there are exactly $2$ heads among $(X_1,dots,X_6)$.
          $endgroup$
          – Aragon
          Jan 3 at 19:36




















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