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If $N$ is independent of each $X_i$, is $N$ independent of $sum_i X_i$?


What is this operation on random variables called?Conditional probability given independent random variablesQuestion about definition of independent discrete random variables.If $N$ is an integral random variable, can $S_N$ be expressed in terms of the $S_n$ and $P(N=n)$?Find distribution of rv X_N where N is independent rv and each X_i~exp(lambda_i)Conditional Expectation of random sum of independent random variables (when $N$ and $X_i$ are dependent)Find the value of $mathbbE(X_1+X_2+ldots+X_N)$ of i.i.d random variables $X_i$s.Size of $Omega$ to have independent +1/-1 random variablesDistribution of the Number of Distinct Items in a Sequence of Independent Random VariablesMoment-generating function of $m$ independent variables













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Let $X_1,X_2, ...,X_n$ a sequence of discrete random variables i.i.d, and $N>0$ a positive discrete random variable, such that $N$ and $X_i$ are independent for $i=1,...,n$.



Can we say that $X_1+...+X_n$ and $N$ are independent?










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    Let $X_1,X_2, ...,X_n$ a sequence of discrete random variables i.i.d, and $N>0$ a positive discrete random variable, such that $N$ and $X_i$ are independent for $i=1,...,n$.



    Can we say that $X_1+...+X_n$ and $N$ are independent?










    share|cite|improve this question











    $endgroup$














      1












      1








      1


      1



      $begingroup$


      Let $X_1,X_2, ...,X_n$ a sequence of discrete random variables i.i.d, and $N>0$ a positive discrete random variable, such that $N$ and $X_i$ are independent for $i=1,...,n$.



      Can we say that $X_1+...+X_n$ and $N$ are independent?










      share|cite|improve this question











      $endgroup$




      Let $X_1,X_2, ...,X_n$ a sequence of discrete random variables i.i.d, and $N>0$ a positive discrete random variable, such that $N$ and $X_i$ are independent for $i=1,...,n$.



      Can we say that $X_1+...+X_n$ and $N$ are independent?







      probability probability-theory random-variables






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 15 at 18:44









      Mike Earnest

      25.5k22151




      25.5k22151










      asked Mar 14 at 17:31









      AnasAnas

      254




      254




















          1 Answer
          1






          active

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          2












          $begingroup$

          No. An example:



          Let the $X_i$ each be the discrete uniform distribution on $0,frac1n,frac2n,dots,fracn-1n$ and let $N$ be the fractional part of $X_1+X_2+cdots+X_n$. It's independent of any one $X_i$, identically distributed to them, and completely determined by the sum.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I think $X_i$ are meant to be discrete. Not much change in your example though.
            $endgroup$
            – Radost
            Mar 14 at 17:38










          • $begingroup$
            I noticed that after writing it. Already edited.
            $endgroup$
            – jmerry
            Mar 14 at 17:39











          • $begingroup$
            How do you justify that $N$ so defined is independent of the $X_i$ ?
            $endgroup$
            – Yves Daoust
            Mar 14 at 17:41











          • $begingroup$
            Can we say that $[N=n]$ is independent of $[X_1+...X_n=k]$ ?
            $endgroup$
            – Anas
            Mar 14 at 17:45











          • $begingroup$
            @YvesDaoust: Even if you lock down exactly what $X_1,X_2,dots,X_n-1$ are, adding another discrete uniform to that evens out the probabilities of the fractional part to $frac1n$ at each possible value. That's the independence.
            $endgroup$
            – jmerry
            Mar 14 at 17:56










          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          2












          $begingroup$

          No. An example:



          Let the $X_i$ each be the discrete uniform distribution on $0,frac1n,frac2n,dots,fracn-1n$ and let $N$ be the fractional part of $X_1+X_2+cdots+X_n$. It's independent of any one $X_i$, identically distributed to them, and completely determined by the sum.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I think $X_i$ are meant to be discrete. Not much change in your example though.
            $endgroup$
            – Radost
            Mar 14 at 17:38










          • $begingroup$
            I noticed that after writing it. Already edited.
            $endgroup$
            – jmerry
            Mar 14 at 17:39











          • $begingroup$
            How do you justify that $N$ so defined is independent of the $X_i$ ?
            $endgroup$
            – Yves Daoust
            Mar 14 at 17:41











          • $begingroup$
            Can we say that $[N=n]$ is independent of $[X_1+...X_n=k]$ ?
            $endgroup$
            – Anas
            Mar 14 at 17:45











          • $begingroup$
            @YvesDaoust: Even if you lock down exactly what $X_1,X_2,dots,X_n-1$ are, adding another discrete uniform to that evens out the probabilities of the fractional part to $frac1n$ at each possible value. That's the independence.
            $endgroup$
            – jmerry
            Mar 14 at 17:56















          2












          $begingroup$

          No. An example:



          Let the $X_i$ each be the discrete uniform distribution on $0,frac1n,frac2n,dots,fracn-1n$ and let $N$ be the fractional part of $X_1+X_2+cdots+X_n$. It's independent of any one $X_i$, identically distributed to them, and completely determined by the sum.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I think $X_i$ are meant to be discrete. Not much change in your example though.
            $endgroup$
            – Radost
            Mar 14 at 17:38










          • $begingroup$
            I noticed that after writing it. Already edited.
            $endgroup$
            – jmerry
            Mar 14 at 17:39











          • $begingroup$
            How do you justify that $N$ so defined is independent of the $X_i$ ?
            $endgroup$
            – Yves Daoust
            Mar 14 at 17:41











          • $begingroup$
            Can we say that $[N=n]$ is independent of $[X_1+...X_n=k]$ ?
            $endgroup$
            – Anas
            Mar 14 at 17:45











          • $begingroup$
            @YvesDaoust: Even if you lock down exactly what $X_1,X_2,dots,X_n-1$ are, adding another discrete uniform to that evens out the probabilities of the fractional part to $frac1n$ at each possible value. That's the independence.
            $endgroup$
            – jmerry
            Mar 14 at 17:56













          2












          2








          2





          $begingroup$

          No. An example:



          Let the $X_i$ each be the discrete uniform distribution on $0,frac1n,frac2n,dots,fracn-1n$ and let $N$ be the fractional part of $X_1+X_2+cdots+X_n$. It's independent of any one $X_i$, identically distributed to them, and completely determined by the sum.






          share|cite|improve this answer









          $endgroup$



          No. An example:



          Let the $X_i$ each be the discrete uniform distribution on $0,frac1n,frac2n,dots,fracn-1n$ and let $N$ be the fractional part of $X_1+X_2+cdots+X_n$. It's independent of any one $X_i$, identically distributed to them, and completely determined by the sum.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 14 at 17:35









          jmerryjmerry

          15.3k1632




          15.3k1632











          • $begingroup$
            I think $X_i$ are meant to be discrete. Not much change in your example though.
            $endgroup$
            – Radost
            Mar 14 at 17:38










          • $begingroup$
            I noticed that after writing it. Already edited.
            $endgroup$
            – jmerry
            Mar 14 at 17:39











          • $begingroup$
            How do you justify that $N$ so defined is independent of the $X_i$ ?
            $endgroup$
            – Yves Daoust
            Mar 14 at 17:41











          • $begingroup$
            Can we say that $[N=n]$ is independent of $[X_1+...X_n=k]$ ?
            $endgroup$
            – Anas
            Mar 14 at 17:45











          • $begingroup$
            @YvesDaoust: Even if you lock down exactly what $X_1,X_2,dots,X_n-1$ are, adding another discrete uniform to that evens out the probabilities of the fractional part to $frac1n$ at each possible value. That's the independence.
            $endgroup$
            – jmerry
            Mar 14 at 17:56
















          • $begingroup$
            I think $X_i$ are meant to be discrete. Not much change in your example though.
            $endgroup$
            – Radost
            Mar 14 at 17:38










          • $begingroup$
            I noticed that after writing it. Already edited.
            $endgroup$
            – jmerry
            Mar 14 at 17:39











          • $begingroup$
            How do you justify that $N$ so defined is independent of the $X_i$ ?
            $endgroup$
            – Yves Daoust
            Mar 14 at 17:41











          • $begingroup$
            Can we say that $[N=n]$ is independent of $[X_1+...X_n=k]$ ?
            $endgroup$
            – Anas
            Mar 14 at 17:45











          • $begingroup$
            @YvesDaoust: Even if you lock down exactly what $X_1,X_2,dots,X_n-1$ are, adding another discrete uniform to that evens out the probabilities of the fractional part to $frac1n$ at each possible value. That's the independence.
            $endgroup$
            – jmerry
            Mar 14 at 17:56















          $begingroup$
          I think $X_i$ are meant to be discrete. Not much change in your example though.
          $endgroup$
          – Radost
          Mar 14 at 17:38




          $begingroup$
          I think $X_i$ are meant to be discrete. Not much change in your example though.
          $endgroup$
          – Radost
          Mar 14 at 17:38












          $begingroup$
          I noticed that after writing it. Already edited.
          $endgroup$
          – jmerry
          Mar 14 at 17:39





          $begingroup$
          I noticed that after writing it. Already edited.
          $endgroup$
          – jmerry
          Mar 14 at 17:39













          $begingroup$
          How do you justify that $N$ so defined is independent of the $X_i$ ?
          $endgroup$
          – Yves Daoust
          Mar 14 at 17:41





          $begingroup$
          How do you justify that $N$ so defined is independent of the $X_i$ ?
          $endgroup$
          – Yves Daoust
          Mar 14 at 17:41













          $begingroup$
          Can we say that $[N=n]$ is independent of $[X_1+...X_n=k]$ ?
          $endgroup$
          – Anas
          Mar 14 at 17:45





          $begingroup$
          Can we say that $[N=n]$ is independent of $[X_1+...X_n=k]$ ?
          $endgroup$
          – Anas
          Mar 14 at 17:45













          $begingroup$
          @YvesDaoust: Even if you lock down exactly what $X_1,X_2,dots,X_n-1$ are, adding another discrete uniform to that evens out the probabilities of the fractional part to $frac1n$ at each possible value. That's the independence.
          $endgroup$
          – jmerry
          Mar 14 at 17:56




          $begingroup$
          @YvesDaoust: Even if you lock down exactly what $X_1,X_2,dots,X_n-1$ are, adding another discrete uniform to that evens out the probabilities of the fractional part to $frac1n$ at each possible value. That's the independence.
          $endgroup$
          – jmerry
          Mar 14 at 17:56

















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