Relation between independence and correlation of uniform random variables
$begingroup$
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
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
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
asked 2 hours ago
PeiffapPeiffap
153
$endgroup$
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
correlation independence uniform
asked 2 hours ago
PeiffapPeiffap
153
asked 2 hours ago
PeiffapPeiffap
153
asked 2 hours ago
PeiffapPeiffap
153
asked 2 hours ago
PeiffapPeiffap
153
asked 2 hours ago
PeiffapPeiffap
153
153
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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:
consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.
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.
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)
answered 2 hours 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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$begingroup$
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:
consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.
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.
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)
answered 2 hours 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:
consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.
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.
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)
answered 2 hours 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:
consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.
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.
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)
answered 2 hours ago
Glen_b♦Glen_b
213k22413763
$endgroup$
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:
consider an additional random variable $B$ which takes the values $pm 1$ each with probability $frac12$, independent of $X$. Then let $Y=BX$.
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.
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)
answered 2 hours ago
Glen_b♦Glen_b
213k22413763
edited 46 mins ago
answered 2 hours ago
Glen_b♦Glen_b
213k22413763
answered 2 hours ago
Glen_b♦Glen_b
213k22413763
answered 2 hours ago
Glen_b♦Glen_b
213k22413763
213k22413763
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
They make it visually clearer, yes. Thank you, again.
$endgroup$
– Peiffap
1 hour ago
add a comment |
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