Constructing a bivariate normal from three univariate normals
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I'm trying to construct correlated bivariate normal random variables from three univariate normal random variables. I realize there is a formula for constructing a bivariate normal random variable from two univariate random normal variables, but I have reasons for wanting to adjust two previously sampled variables by a third in order to give them a correlation $rho$.
Based on the approaches for constructing bivariate normals from two univariate normals, I came up with the following approach and need help verifying its correctness.
First, imagine that we have three univariate normal variables. For simplicity here, we just assume they all have $sigma=1$.
$$ X_0 sim Normal(0, 1)$$
$$ Y_0 sim Normal(0, 1)$$
$$ Z sim Normal(0, 1)$$
Given these three univariate random variables, I construct two new random variables using the following linear combinations:
$$ X = |rho| * Z + sqrt{1-rho^2} * X_0$$
$$ Y = rho * Z + sqrt{1-rho^2} * Y_0$$
where $rho in [-1, 1]$ represents the correlation coefficient between the two univariate normals.
Can someone help me formally verify that $X$ and $Y$ are now correlated random variables with correlation $rho$?
I've convinced my self through empirical simulation. Here are plots of values sampled from $X$ and $Y$ for the cases where $rho=0$, $rho=1$, and $rho=-1$.
Plot of X vs. Y for rho of 0
Plot of X vs. Y for rho of 1
Plot of X vs. Y for rho of -1
These plots are exactly as I would expect, but it would be nice to have a formal proof based on my construction. Thanks in advance!
proof-verification random-variables normal-distribution
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add a comment |
$begingroup$
I'm trying to construct correlated bivariate normal random variables from three univariate normal random variables. I realize there is a formula for constructing a bivariate normal random variable from two univariate random normal variables, but I have reasons for wanting to adjust two previously sampled variables by a third in order to give them a correlation $rho$.
Based on the approaches for constructing bivariate normals from two univariate normals, I came up with the following approach and need help verifying its correctness.
First, imagine that we have three univariate normal variables. For simplicity here, we just assume they all have $sigma=1$.
$$ X_0 sim Normal(0, 1)$$
$$ Y_0 sim Normal(0, 1)$$
$$ Z sim Normal(0, 1)$$
Given these three univariate random variables, I construct two new random variables using the following linear combinations:
$$ X = |rho| * Z + sqrt{1-rho^2} * X_0$$
$$ Y = rho * Z + sqrt{1-rho^2} * Y_0$$
where $rho in [-1, 1]$ represents the correlation coefficient between the two univariate normals.
Can someone help me formally verify that $X$ and $Y$ are now correlated random variables with correlation $rho$?
I've convinced my self through empirical simulation. Here are plots of values sampled from $X$ and $Y$ for the cases where $rho=0$, $rho=1$, and $rho=-1$.
Plot of X vs. Y for rho of 0
Plot of X vs. Y for rho of 1
Plot of X vs. Y for rho of -1
These plots are exactly as I would expect, but it would be nice to have a formal proof based on my construction. Thanks in advance!
proof-verification random-variables normal-distribution
$endgroup$
$begingroup$
Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent.
$endgroup$
– Chris MacLellan
Jan 5 at 23:24
add a comment |
$begingroup$
I'm trying to construct correlated bivariate normal random variables from three univariate normal random variables. I realize there is a formula for constructing a bivariate normal random variable from two univariate random normal variables, but I have reasons for wanting to adjust two previously sampled variables by a third in order to give them a correlation $rho$.
Based on the approaches for constructing bivariate normals from two univariate normals, I came up with the following approach and need help verifying its correctness.
First, imagine that we have three univariate normal variables. For simplicity here, we just assume they all have $sigma=1$.
$$ X_0 sim Normal(0, 1)$$
$$ Y_0 sim Normal(0, 1)$$
$$ Z sim Normal(0, 1)$$
Given these three univariate random variables, I construct two new random variables using the following linear combinations:
$$ X = |rho| * Z + sqrt{1-rho^2} * X_0$$
$$ Y = rho * Z + sqrt{1-rho^2} * Y_0$$
where $rho in [-1, 1]$ represents the correlation coefficient between the two univariate normals.
Can someone help me formally verify that $X$ and $Y$ are now correlated random variables with correlation $rho$?
I've convinced my self through empirical simulation. Here are plots of values sampled from $X$ and $Y$ for the cases where $rho=0$, $rho=1$, and $rho=-1$.
Plot of X vs. Y for rho of 0
Plot of X vs. Y for rho of 1
Plot of X vs. Y for rho of -1
These plots are exactly as I would expect, but it would be nice to have a formal proof based on my construction. Thanks in advance!
proof-verification random-variables normal-distribution
$endgroup$
I'm trying to construct correlated bivariate normal random variables from three univariate normal random variables. I realize there is a formula for constructing a bivariate normal random variable from two univariate random normal variables, but I have reasons for wanting to adjust two previously sampled variables by a third in order to give them a correlation $rho$.
Based on the approaches for constructing bivariate normals from two univariate normals, I came up with the following approach and need help verifying its correctness.
First, imagine that we have three univariate normal variables. For simplicity here, we just assume they all have $sigma=1$.
$$ X_0 sim Normal(0, 1)$$
$$ Y_0 sim Normal(0, 1)$$
$$ Z sim Normal(0, 1)$$
Given these three univariate random variables, I construct two new random variables using the following linear combinations:
$$ X = |rho| * Z + sqrt{1-rho^2} * X_0$$
$$ Y = rho * Z + sqrt{1-rho^2} * Y_0$$
where $rho in [-1, 1]$ represents the correlation coefficient between the two univariate normals.
Can someone help me formally verify that $X$ and $Y$ are now correlated random variables with correlation $rho$?
I've convinced my self through empirical simulation. Here are plots of values sampled from $X$ and $Y$ for the cases where $rho=0$, $rho=1$, and $rho=-1$.
Plot of X vs. Y for rho of 0
Plot of X vs. Y for rho of 1
Plot of X vs. Y for rho of -1
These plots are exactly as I would expect, but it would be nice to have a formal proof based on my construction. Thanks in advance!
proof-verification random-variables normal-distribution
proof-verification random-variables normal-distribution
asked Jan 5 at 22:36
Chris MacLellanChris MacLellan
133
133
$begingroup$
Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent.
$endgroup$
– Chris MacLellan
Jan 5 at 23:24
add a comment |
$begingroup$
Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent.
$endgroup$
– Chris MacLellan
Jan 5 at 23:24
$begingroup$
Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent.
$endgroup$
– Chris MacLellan
Jan 5 at 23:24
$begingroup$
Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent.
$endgroup$
– Chris MacLellan
Jan 5 at 23:24
add a comment |
1 Answer
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$begingroup$
I presume $X_0, Y_0$ and $Z$ are assumed to be independent? Well, if we define
$$ mathbf U := begin{bmatrix} X_0 \ Y_0 \Z end{bmatrix}, mathbf V := begin{bmatrix} X \ Yend{bmatrix},$$
then
$$ mathbf V = mathbf M mathbf U,$$
where
$$ mathbf M := begin{bmatrix} sqrt{1- rho^2} & 0 & | rho| \ 0 & sqrt{1- rho^2} & rho end{bmatrix}$$
Since $X_0, Y_0$ and $Z$ are independent with unit variance, the covariance matrix for $mathbf U$ is the $3 times 3$ identity matrix:
$$mathbb E[mathbf U mathbf U^T] = mathbf I.$$
Hence the covariance matrix for $mathbf V$ is given by
$$ mathbb E [mathbf V mathbf V^T ] = mathbf M mathbb E[mathbf U mathbf U^T ] mathbf M^T = mathbf Mmathbf M^T=begin{bmatrix} 1 & rho | rho| \ rho | rho| & 1end{bmatrix}$$
Thus the correlation coefficient between $X$ and $Y$ is
$$ rho_{X,Y} = frac{mathbb E[XY]}{sqrt{mathbb E[X^2] mathbb E[{Y^2]}}} = frac{rho| rho|}{sqrt{1 times 1}} = rho|rho|.$$
[I used the fact that $mathbb E[X] = mathbb E[Y] = 0$ here.]
So I'm afraid the correlation coefficient is not $rho$. It's ${rm sign}(rho) times |rho|^2$.
But that's easily fixed. If you redefine $mathbf M$ as $$ mathbf M := begin{bmatrix} sqrt{1- |rho |} & 0 & {rm sign}(rho)sqrt{ |rho |} \ 0 & sqrt{1- |rho | } & sqrt{|rho |} end{bmatrix},$$
then it should work out.
Finally, I'll point out that the $X$ and $Y$ that you've constructed are guaranteed to be Gaussian, since $mathbf V$ is related to the Gaussian vector $mathbf U$ by a linear transformation. They also both have unit variance, as you can see from the diagonal elements of $mathbb E[mathbf V mathbf V^T]$.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I presume $X_0, Y_0$ and $Z$ are assumed to be independent? Well, if we define
$$ mathbf U := begin{bmatrix} X_0 \ Y_0 \Z end{bmatrix}, mathbf V := begin{bmatrix} X \ Yend{bmatrix},$$
then
$$ mathbf V = mathbf M mathbf U,$$
where
$$ mathbf M := begin{bmatrix} sqrt{1- rho^2} & 0 & | rho| \ 0 & sqrt{1- rho^2} & rho end{bmatrix}$$
Since $X_0, Y_0$ and $Z$ are independent with unit variance, the covariance matrix for $mathbf U$ is the $3 times 3$ identity matrix:
$$mathbb E[mathbf U mathbf U^T] = mathbf I.$$
Hence the covariance matrix for $mathbf V$ is given by
$$ mathbb E [mathbf V mathbf V^T ] = mathbf M mathbb E[mathbf U mathbf U^T ] mathbf M^T = mathbf Mmathbf M^T=begin{bmatrix} 1 & rho | rho| \ rho | rho| & 1end{bmatrix}$$
Thus the correlation coefficient between $X$ and $Y$ is
$$ rho_{X,Y} = frac{mathbb E[XY]}{sqrt{mathbb E[X^2] mathbb E[{Y^2]}}} = frac{rho| rho|}{sqrt{1 times 1}} = rho|rho|.$$
[I used the fact that $mathbb E[X] = mathbb E[Y] = 0$ here.]
So I'm afraid the correlation coefficient is not $rho$. It's ${rm sign}(rho) times |rho|^2$.
But that's easily fixed. If you redefine $mathbf M$ as $$ mathbf M := begin{bmatrix} sqrt{1- |rho |} & 0 & {rm sign}(rho)sqrt{ |rho |} \ 0 & sqrt{1- |rho | } & sqrt{|rho |} end{bmatrix},$$
then it should work out.
Finally, I'll point out that the $X$ and $Y$ that you've constructed are guaranteed to be Gaussian, since $mathbf V$ is related to the Gaussian vector $mathbf U$ by a linear transformation. They also both have unit variance, as you can see from the diagonal elements of $mathbb E[mathbf V mathbf V^T]$.
$endgroup$
add a comment |
$begingroup$
I presume $X_0, Y_0$ and $Z$ are assumed to be independent? Well, if we define
$$ mathbf U := begin{bmatrix} X_0 \ Y_0 \Z end{bmatrix}, mathbf V := begin{bmatrix} X \ Yend{bmatrix},$$
then
$$ mathbf V = mathbf M mathbf U,$$
where
$$ mathbf M := begin{bmatrix} sqrt{1- rho^2} & 0 & | rho| \ 0 & sqrt{1- rho^2} & rho end{bmatrix}$$
Since $X_0, Y_0$ and $Z$ are independent with unit variance, the covariance matrix for $mathbf U$ is the $3 times 3$ identity matrix:
$$mathbb E[mathbf U mathbf U^T] = mathbf I.$$
Hence the covariance matrix for $mathbf V$ is given by
$$ mathbb E [mathbf V mathbf V^T ] = mathbf M mathbb E[mathbf U mathbf U^T ] mathbf M^T = mathbf Mmathbf M^T=begin{bmatrix} 1 & rho | rho| \ rho | rho| & 1end{bmatrix}$$
Thus the correlation coefficient between $X$ and $Y$ is
$$ rho_{X,Y} = frac{mathbb E[XY]}{sqrt{mathbb E[X^2] mathbb E[{Y^2]}}} = frac{rho| rho|}{sqrt{1 times 1}} = rho|rho|.$$
[I used the fact that $mathbb E[X] = mathbb E[Y] = 0$ here.]
So I'm afraid the correlation coefficient is not $rho$. It's ${rm sign}(rho) times |rho|^2$.
But that's easily fixed. If you redefine $mathbf M$ as $$ mathbf M := begin{bmatrix} sqrt{1- |rho |} & 0 & {rm sign}(rho)sqrt{ |rho |} \ 0 & sqrt{1- |rho | } & sqrt{|rho |} end{bmatrix},$$
then it should work out.
Finally, I'll point out that the $X$ and $Y$ that you've constructed are guaranteed to be Gaussian, since $mathbf V$ is related to the Gaussian vector $mathbf U$ by a linear transformation. They also both have unit variance, as you can see from the diagonal elements of $mathbb E[mathbf V mathbf V^T]$.
$endgroup$
add a comment |
$begingroup$
I presume $X_0, Y_0$ and $Z$ are assumed to be independent? Well, if we define
$$ mathbf U := begin{bmatrix} X_0 \ Y_0 \Z end{bmatrix}, mathbf V := begin{bmatrix} X \ Yend{bmatrix},$$
then
$$ mathbf V = mathbf M mathbf U,$$
where
$$ mathbf M := begin{bmatrix} sqrt{1- rho^2} & 0 & | rho| \ 0 & sqrt{1- rho^2} & rho end{bmatrix}$$
Since $X_0, Y_0$ and $Z$ are independent with unit variance, the covariance matrix for $mathbf U$ is the $3 times 3$ identity matrix:
$$mathbb E[mathbf U mathbf U^T] = mathbf I.$$
Hence the covariance matrix for $mathbf V$ is given by
$$ mathbb E [mathbf V mathbf V^T ] = mathbf M mathbb E[mathbf U mathbf U^T ] mathbf M^T = mathbf Mmathbf M^T=begin{bmatrix} 1 & rho | rho| \ rho | rho| & 1end{bmatrix}$$
Thus the correlation coefficient between $X$ and $Y$ is
$$ rho_{X,Y} = frac{mathbb E[XY]}{sqrt{mathbb E[X^2] mathbb E[{Y^2]}}} = frac{rho| rho|}{sqrt{1 times 1}} = rho|rho|.$$
[I used the fact that $mathbb E[X] = mathbb E[Y] = 0$ here.]
So I'm afraid the correlation coefficient is not $rho$. It's ${rm sign}(rho) times |rho|^2$.
But that's easily fixed. If you redefine $mathbf M$ as $$ mathbf M := begin{bmatrix} sqrt{1- |rho |} & 0 & {rm sign}(rho)sqrt{ |rho |} \ 0 & sqrt{1- |rho | } & sqrt{|rho |} end{bmatrix},$$
then it should work out.
Finally, I'll point out that the $X$ and $Y$ that you've constructed are guaranteed to be Gaussian, since $mathbf V$ is related to the Gaussian vector $mathbf U$ by a linear transformation. They also both have unit variance, as you can see from the diagonal elements of $mathbb E[mathbf V mathbf V^T]$.
$endgroup$
I presume $X_0, Y_0$ and $Z$ are assumed to be independent? Well, if we define
$$ mathbf U := begin{bmatrix} X_0 \ Y_0 \Z end{bmatrix}, mathbf V := begin{bmatrix} X \ Yend{bmatrix},$$
then
$$ mathbf V = mathbf M mathbf U,$$
where
$$ mathbf M := begin{bmatrix} sqrt{1- rho^2} & 0 & | rho| \ 0 & sqrt{1- rho^2} & rho end{bmatrix}$$
Since $X_0, Y_0$ and $Z$ are independent with unit variance, the covariance matrix for $mathbf U$ is the $3 times 3$ identity matrix:
$$mathbb E[mathbf U mathbf U^T] = mathbf I.$$
Hence the covariance matrix for $mathbf V$ is given by
$$ mathbb E [mathbf V mathbf V^T ] = mathbf M mathbb E[mathbf U mathbf U^T ] mathbf M^T = mathbf Mmathbf M^T=begin{bmatrix} 1 & rho | rho| \ rho | rho| & 1end{bmatrix}$$
Thus the correlation coefficient between $X$ and $Y$ is
$$ rho_{X,Y} = frac{mathbb E[XY]}{sqrt{mathbb E[X^2] mathbb E[{Y^2]}}} = frac{rho| rho|}{sqrt{1 times 1}} = rho|rho|.$$
[I used the fact that $mathbb E[X] = mathbb E[Y] = 0$ here.]
So I'm afraid the correlation coefficient is not $rho$. It's ${rm sign}(rho) times |rho|^2$.
But that's easily fixed. If you redefine $mathbf M$ as $$ mathbf M := begin{bmatrix} sqrt{1- |rho |} & 0 & {rm sign}(rho)sqrt{ |rho |} \ 0 & sqrt{1- |rho | } & sqrt{|rho |} end{bmatrix},$$
then it should work out.
Finally, I'll point out that the $X$ and $Y$ that you've constructed are guaranteed to be Gaussian, since $mathbf V$ is related to the Gaussian vector $mathbf U$ by a linear transformation. They also both have unit variance, as you can see from the diagonal elements of $mathbb E[mathbf V mathbf V^T]$.
edited Jan 5 at 23:11
answered Jan 5 at 23:04
Kenny WongKenny Wong
19.4k21441
19.4k21441
add a comment |
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$begingroup$
Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent.
$endgroup$
– Chris MacLellan
Jan 5 at 23:24