Constructing a bivariate normal from three univariate normals












2












$begingroup$


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!










share|cite|improve this question









$endgroup$












  • $begingroup$
    Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent.
    $endgroup$
    – Chris MacLellan
    Jan 5 at 23:24
















2












$begingroup$


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!










share|cite|improve this question









$endgroup$












  • $begingroup$
    Just to clarify based on the below solution. $X_0$, $Y_0$, and $Z$ are independent.
    $endgroup$
    – Chris MacLellan
    Jan 5 at 23:24














2












2








2





$begingroup$


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!










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $mathbb E[mathbf V mathbf V^T]$.






share|cite|improve this answer











$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063271%2fconstructing-a-bivariate-normal-from-three-univariate-normals%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    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 $mathbb E[mathbf V mathbf V^T]$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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 $mathbb E[mathbf V mathbf V^T]$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $mathbb E[mathbf V mathbf V^T]$.






        share|cite|improve this answer











        $endgroup$



        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 $mathbb E[mathbf V mathbf V^T]$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 23:11

























        answered Jan 5 at 23:04









        Kenny WongKenny Wong

        19.4k21441




        19.4k21441






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063271%2fconstructing-a-bivariate-normal-from-three-univariate-normals%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Aardman Animations

            Are they similar matrix