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

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$begingroup$

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?

probability probability-theory random-variables

share|cite|improve this question

edited Mar 15 at 18:44

Mike Earnest

25.5k22151

asked Mar 14 at 17:31

AnasAnas

254

$endgroup$

add a comment | 

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?

probability probability-theory random-variables

share|cite|improve this question

edited Mar 15 at 18:44

Mike Earnest

25.5k22151

asked Mar 14 at 17:31

AnasAnas

254

$endgroup$

add a comment | 

$begingroup$

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?

probability probability-theory random-variables

share|cite|improve this question

edited Mar 15 at 18:44

Mike Earnest

25.5k22151

asked Mar 14 at 17:31

AnasAnas

254

$endgroup$

share|cite|improve this question

edited Mar 15 at 18:44

Mike Earnest

25.5k22151

asked Mar 14 at 17:31

AnasAnas

254

share|cite|improve this question

edited Mar 15 at 18:44

Mike Earnest

25.5k22151

asked Mar 14 at 17:31

AnasAnas

254

edited Mar 15 at 18:44

Mike Earnest

25.5k22151

edited Mar 15 at 18:44

Mike Earnest

25.5k22151

asked Mar 14 at 17:31

AnasAnas

254

asked Mar 14 at 17:31

AnasAnas

254

1 Answer
1

active

oldest

votes

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

answered Mar 14 at 17:35

jmerryjmerry

15.3k1632

$endgroup$

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  $begingroup$
  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
  

  
  
  
  

 | 
show 2 more comments

Your Answer

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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

answered Mar 14 at 17:35

jmerryjmerry

15.3k1632

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

 | 
show 2 more comments

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

answered Mar 14 at 17:35

jmerryjmerry

15.3k1632

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

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

answered Mar 14 at 17:35

jmerryjmerry

15.3k1632

$endgroup$

share|cite|improve this answer

answered Mar 14 at 17:35

jmerryjmerry

15.3k1632

answered Mar 14 at 17:35

jmerryjmerry

15.3k1632

answered Mar 14 at 17:35

jmerryjmerry

15.3k1632