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Relation between independence and correlation of uniform random variables

Correlations with a linear combination means correlation with individual variables?Geometric mean of uniform variablesHow to Test Independence of Poisson Variables?If $X$ and $Y$ are normally distributed random variables, what kind of distribution their sum follows?Distribution of X-U(0,1) conditioned on sigma algebra of Y/X, where is Y is U(0,1)?Is there a parametric joint distribution such that $X$ and $Y$ are both uniform and $mathbb{E}[Y ;|; X]$ is linear?Are two Random Variables Independent if their support has a dependency?Correlation of the sigmoid function of normal random varaiblesIntuitive reason why jointly normal and uncorrelated imply independenceConditional maximum likelihood of AR(1) UNIFORM PROCESS

1

$begingroup$

My question is fairly simple: let $X$ and $Y$ be two uncorrelated uniform random variables on $[-1,1]$. Are they independent?

I was under the impression that two random, uncorrelated variables are only necessarily independent if their joint distribution is normal, however I can't come up with a counterexample to disprove the claim I ask about. Either a counterexample or a proof would be greatly appreciated.

correlation independence uniform

share|cite|improve this question

asked 2 hours ago

PeiffapPeiffap

153

$endgroup$

add a comment | 

1

$begingroup$

My question is fairly simple: let $X$ and $Y$ be two uncorrelated uniform random variables on $[-1,1]$. Are they independent?

I was under the impression that two random, uncorrelated variables are only necessarily independent if their joint distribution is normal, however I can't come up with a counterexample to disprove the claim I ask about. Either a counterexample or a proof would be greatly appreciated.

correlation independence uniform

share|cite|improve this question

asked 2 hours ago

PeiffapPeiffap

153

$endgroup$

add a comment | 

$begingroup$

My question is fairly simple: let $X$ and $Y$ be two uncorrelated uniform random variables on $[-1,1]$. Are they independent?

I was under the impression that two random, uncorrelated variables are only necessarily independent if their joint distribution is normal, however I can't come up with a counterexample to disprove the claim I ask about. Either a counterexample or a proof would be greatly appreciated.

correlation independence uniform

share|cite|improve this question

asked 2 hours ago

PeiffapPeiffap

153

$endgroup$

share|cite|improve this question

asked 2 hours ago

PeiffapPeiffap

153

share|cite|improve this question

asked 2 hours ago

PeiffapPeiffap

153

asked 2 hours ago

PeiffapPeiffap

153

asked 2 hours ago

PeiffapPeiffap

153

1 Answer
1

active

oldest

votes

5

$begingroup$

Independent implies uncorrelated but the implication doesn't go the other way.

Uncorrelated implies independence only under certain conditions. e.g. if you have a bivariate normal, it is the case that uncorrelated implies independent (as you said).

It is easy to construct bivariate distributions with uniform margins where the variables are uncorrelated but are not independent. Here are a few examples:

1. consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.




2. take the bivariate distribution of two independent uniforms and slice it in 4 equal-size sections on each margin (yielding $4times 4=16$ pieces, each of size $frac12timesfrac12$). Now take all the probability from the 4 corner pieces and the 4 center pieces and put it evenly into the other 8 pieces.




3. Let $Y = 2|X|-1$.

In each case, the variables are uncorrelated but not independent (e.g. if $X=1$, what is $P(-0.1<Y<0.1$?)

If you specify some particular family of bivariate distributions with uniform margins it might be possible that under that formulation the only uncorrelated one is independent. Then under that condition, being uncorrelated would imply independence -- but you haven't said anything about the bivariate distribution, only about the marginal distributions.

For example, if you restrict your attention to say the Gaussian copula, then I think the only uncorrelated one has independent margins; you can readily rescale that so that each margin is on (-1,1).

---

Some R code for sampling from and plotting these bivariates (not necessarily efficiently):

n <- 100000

x <- runif(n,-1,1)

b <- rbinom(n,1,.5)*2-1

y1 <-b*x

y2 <-ifelse(0.5<abs(x)&abs(x)<1,

       runif(n,-.5,.5),

       runif(n,0.5,1)*b

      )

y3 <- 2*abs(x)-1



par(mfrow=c(1,3))

plot(x,y1,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

plot(x,y2,pch=16,cex=.5,col=rgb(.5,.5,.5,.5))

abline(h=c(-1,-.5,0,.5,1),col=4,lty=3)

abline(v=c(-1,-.5,0,.5,1),col=4,lty=3)

plot(x,y3,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

(In this formulation, $(Y_2, Y_3)$ gives a fourth example)

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

Glen_b♦Glen_b

213k22413763

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I'm struggling to see why the examples you provided still guarantee that $Y$ is uniformly distributed on $[-1, 1]$, though.
  $endgroup$
  – Peiffap
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do the plots of the bivariate densities help? In each case the shaded parts are all of constant density
  $endgroup$
  – Glen_b♦
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They make it visually clearer, yes. Thank you, again.
  $endgroup$
  – Peiffap
  1 hour ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

Independent implies uncorrelated but the implication doesn't go the other way.

Uncorrelated implies independence only under certain conditions. e.g. if you have a bivariate normal, it is the case that uncorrelated implies independent (as you said).

It is easy to construct bivariate distributions with uniform margins where the variables are uncorrelated but are not independent. Here are a few examples:

1. consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.




2. take the bivariate distribution of two independent uniforms and slice it in 4 equal-size sections on each margin (yielding $4times 4=16$ pieces, each of size $frac12timesfrac12$). Now take all the probability from the 4 corner pieces and the 4 center pieces and put it evenly into the other 8 pieces.




3. Let $Y = 2|X|-1$.

In each case, the variables are uncorrelated but not independent (e.g. if $X=1$, what is $P(-0.1<Y<0.1$?)

If you specify some particular family of bivariate distributions with uniform margins it might be possible that under that formulation the only uncorrelated one is independent. Then under that condition, being uncorrelated would imply independence -- but you haven't said anything about the bivariate distribution, only about the marginal distributions.

For example, if you restrict your attention to say the Gaussian copula, then I think the only uncorrelated one has independent margins; you can readily rescale that so that each margin is on (-1,1).

---

Some R code for sampling from and plotting these bivariates (not necessarily efficiently):

n <- 100000

x <- runif(n,-1,1)

b <- rbinom(n,1,.5)*2-1

y1 <-b*x

y2 <-ifelse(0.5<abs(x)&abs(x)<1,

       runif(n,-.5,.5),

       runif(n,0.5,1)*b

      )

y3 <- 2*abs(x)-1



par(mfrow=c(1,3))

plot(x,y1,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

plot(x,y2,pch=16,cex=.5,col=rgb(.5,.5,.5,.5))

abline(h=c(-1,-.5,0,.5,1),col=4,lty=3)

abline(v=c(-1,-.5,0,.5,1),col=4,lty=3)

plot(x,y3,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

(In this formulation, $(Y_2, Y_3)$ gives a fourth example)

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

Glen_b♦Glen_b

213k22413763

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I'm struggling to see why the examples you provided still guarantee that $Y$ is uniformly distributed on $[-1, 1]$, though.
  $endgroup$
  – Peiffap
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do the plots of the bivariate densities help? In each case the shaded parts are all of constant density
  $endgroup$
  – Glen_b♦
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They make it visually clearer, yes. Thank you, again.
  $endgroup$
  – Peiffap
  1 hour ago
  

  
  
  
  

add a comment | 

5

$begingroup$

Independent implies uncorrelated but the implication doesn't go the other way.

Uncorrelated implies independence only under certain conditions. e.g. if you have a bivariate normal, it is the case that uncorrelated implies independent (as you said).

It is easy to construct bivariate distributions with uniform margins where the variables are uncorrelated but are not independent. Here are a few examples:

1. consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.




2. take the bivariate distribution of two independent uniforms and slice it in 4 equal-size sections on each margin (yielding $4times 4=16$ pieces, each of size $frac12timesfrac12$). Now take all the probability from the 4 corner pieces and the 4 center pieces and put it evenly into the other 8 pieces.




3. Let $Y = 2|X|-1$.

In each case, the variables are uncorrelated but not independent (e.g. if $X=1$, what is $P(-0.1<Y<0.1$?)

If you specify some particular family of bivariate distributions with uniform margins it might be possible that under that formulation the only uncorrelated one is independent. Then under that condition, being uncorrelated would imply independence -- but you haven't said anything about the bivariate distribution, only about the marginal distributions.

For example, if you restrict your attention to say the Gaussian copula, then I think the only uncorrelated one has independent margins; you can readily rescale that so that each margin is on (-1,1).

---

Some R code for sampling from and plotting these bivariates (not necessarily efficiently):

n <- 100000

x <- runif(n,-1,1)

b <- rbinom(n,1,.5)*2-1

y1 <-b*x

y2 <-ifelse(0.5<abs(x)&abs(x)<1,

       runif(n,-.5,.5),

       runif(n,0.5,1)*b

      )

y3 <- 2*abs(x)-1



par(mfrow=c(1,3))

plot(x,y1,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

plot(x,y2,pch=16,cex=.5,col=rgb(.5,.5,.5,.5))

abline(h=c(-1,-.5,0,.5,1),col=4,lty=3)

abline(v=c(-1,-.5,0,.5,1),col=4,lty=3)

plot(x,y3,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

(In this formulation, $(Y_2, Y_3)$ gives a fourth example)

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

Glen_b♦Glen_b

213k22413763

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I'm struggling to see why the examples you provided still guarantee that $Y$ is uniformly distributed on $[-1, 1]$, though.
  $endgroup$
  – Peiffap
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do the plots of the bivariate densities help? In each case the shaded parts are all of constant density
  $endgroup$
  – Glen_b♦
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They make it visually clearer, yes. Thank you, again.
  $endgroup$
  – Peiffap
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Independent implies uncorrelated but the implication doesn't go the other way.

Uncorrelated implies independence only under certain conditions. e.g. if you have a bivariate normal, it is the case that uncorrelated implies independent (as you said).

It is easy to construct bivariate distributions with uniform margins where the variables are uncorrelated but are not independent. Here are a few examples:

1. consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.




2. take the bivariate distribution of two independent uniforms and slice it in 4 equal-size sections on each margin (yielding $4times 4=16$ pieces, each of size $frac12timesfrac12$). Now take all the probability from the 4 corner pieces and the 4 center pieces and put it evenly into the other 8 pieces.




3. Let $Y = 2|X|-1$.

In each case, the variables are uncorrelated but not independent (e.g. if $X=1$, what is $P(-0.1<Y<0.1$?)

If you specify some particular family of bivariate distributions with uniform margins it might be possible that under that formulation the only uncorrelated one is independent. Then under that condition, being uncorrelated would imply independence -- but you haven't said anything about the bivariate distribution, only about the marginal distributions.

For example, if you restrict your attention to say the Gaussian copula, then I think the only uncorrelated one has independent margins; you can readily rescale that so that each margin is on (-1,1).

---

Some R code for sampling from and plotting these bivariates (not necessarily efficiently):

n <- 100000

x <- runif(n,-1,1)

b <- rbinom(n,1,.5)*2-1

y1 <-b*x

y2 <-ifelse(0.5<abs(x)&abs(x)<1,

       runif(n,-.5,.5),

       runif(n,0.5,1)*b

      )

y3 <- 2*abs(x)-1



par(mfrow=c(1,3))

plot(x,y1,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

plot(x,y2,pch=16,cex=.5,col=rgb(.5,.5,.5,.5))

abline(h=c(-1,-.5,0,.5,1),col=4,lty=3)

abline(v=c(-1,-.5,0,.5,1),col=4,lty=3)

plot(x,y3,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

(In this formulation, $(Y_2, Y_3)$ gives a fourth example)

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

Glen_b♦Glen_b

213k22413763

$endgroup$

n <- 100000

x <- runif(n,-1,1)

b <- rbinom(n,1,.5)*2-1

y1 <-b*x

y2 <-ifelse(0.5<abs(x)&abs(x)<1,

       runif(n,-.5,.5),

       runif(n,0.5,1)*b

      )

y3 <- 2*abs(x)-1



par(mfrow=c(1,3))

plot(x,y1,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

plot(x,y2,pch=16,cex=.5,col=rgb(.5,.5,.5,.5))

abline(h=c(-1,-.5,0,.5,1),col=4,lty=3)

abline(v=c(-1,-.5,0,.5,1),col=4,lty=3)

plot(x,y3,pch=16,cex=.3,col=rgb(.5,.5,.5,.5))

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

Glen_b♦Glen_b

213k22413763

answered 1 hour ago

Glen_b♦Glen_b

213k22413763

answered 1 hour ago

Glen_b♦Glen_b

213k22413763