Is the exponential of $-d$ a positive definite kernel for a general metric space $(X, d)$?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite
1












Let $d(cdot, cdot)$ be a distance metric on some space (not necessarily Euclidean) $mathcalX$. Is it true that
$$
K(x, y) := exp(-d(x, y))
$$
always defines a positive definite kernel? If not, are there certain conditions on $d(cdot, cdot)$ under which the result is true?







share|cite|improve this question


























    up vote
    1
    down vote

    favorite
    1












    Let $d(cdot, cdot)$ be a distance metric on some space (not necessarily Euclidean) $mathcalX$. Is it true that
    $$
    K(x, y) := exp(-d(x, y))
    $$
    always defines a positive definite kernel? If not, are there certain conditions on $d(cdot, cdot)$ under which the result is true?







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      Let $d(cdot, cdot)$ be a distance metric on some space (not necessarily Euclidean) $mathcalX$. Is it true that
      $$
      K(x, y) := exp(-d(x, y))
      $$
      always defines a positive definite kernel? If not, are there certain conditions on $d(cdot, cdot)$ under which the result is true?







      share|cite|improve this question














      Let $d(cdot, cdot)$ be a distance metric on some space (not necessarily Euclidean) $mathcalX$. Is it true that
      $$
      K(x, y) := exp(-d(x, y))
      $$
      always defines a positive definite kernel? If not, are there certain conditions on $d(cdot, cdot)$ under which the result is true?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 18 at 17:47









      user357151

      14.1k31140




      14.1k31140










      asked Aug 17 at 6:37









      p-value

      19910




      19910




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Remark: in this subject, "positive definite" typically means "positive semidefinite" (Wikipedia) and I treat it as such. One does not expect the matrix $exp(-d(x_i, x_j))$ to be strictly positive definite since some $x_i$ could be repeated.



          Counterexample



          Consider the 3-dimensional space $ell_infty^3$, i.e., the norm is $|(x, y, z)| = max(|x|, |y|, |z|)$. Let $mathcal X$ be the five-point subset of $ell_infty^3$, the coordinates of each point being a row of the matrix
          $$
          A = frac110 beginpmatrix
          0 & 0 & 1 \ 0 & 0 & -1 \ 1 & 1 & 0 \ 1 & -1 & 0 \ -1 & 1 & 0
          endpmatrix
          $$
          The distance matrix of this space is
          $$
          D = frac110 beginpmatrix 0 & 2 & 1 & 1 & 1 \ 2 & 0 & 1 & 1 & 1 \ 1 & 1 & 0 & 2 & 2 \ 1 & 1 & 2 & 0 & 2 \ 1 & 1 & 2 & 2 & 0 endpmatrix
          $$
          Applying the function $tmapsto exp(-t)$ to each element of $D$, we get a matrix $B$ which is not positive semidefinite. Specifically, the vector $v = (3, 3, -2, -2, -2)$ satisfies
          $$
          vBv^T = 30 + 42e^-1/5 - 72e^-1/10 < 0
          $$



          Sufficient condition



          The kernel $exp(-lambda d)$ is positive semidefinite for all $lambda>0$ if and only if $d$ is a a kernel of conditionally negative type, see Transforming a distance function to a kernel. In the article linked there, Schoenberg proved that a metric space admits an isometric embedding into a Hilbert space if and only if $d^2$ is a kernel of conditionally negative type.



          Conclusion: if $(X, d)$ is a metric space such that its "snowflake" $(X, d^1/2)$ admits an isometric embedding into a Hilbert space, then $exp(-d)$ is a positive semidefinite kernel. Examples of such spaces are: Hilbert space itself, $L^1$, and of course any subsets of those.



          This motivated my counterexample above: $ell_infty^3$ is one of the simplest spaces that do not embed into $L^1$.



          A contemporary source on this subject is Naor's ICM talk on $L^1$ embeddings (its main subject is the Heisenberg group, but the introduction gives an overview of the situation).






          share|cite|improve this answer
















          • 1




            Same question for $exp(-d^2)$, also with a counterexample: Gaussian kernels for arbitrary metric spaces
            – user357151
            Aug 18 at 17:45










          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',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2f2885462%2fis-the-exponential-of-d-a-positive-definite-kernel-for-a-general-metric-space%23new-answer', 'question_page');

          );

          Post as a guest






























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Remark: in this subject, "positive definite" typically means "positive semidefinite" (Wikipedia) and I treat it as such. One does not expect the matrix $exp(-d(x_i, x_j))$ to be strictly positive definite since some $x_i$ could be repeated.



          Counterexample



          Consider the 3-dimensional space $ell_infty^3$, i.e., the norm is $|(x, y, z)| = max(|x|, |y|, |z|)$. Let $mathcal X$ be the five-point subset of $ell_infty^3$, the coordinates of each point being a row of the matrix
          $$
          A = frac110 beginpmatrix
          0 & 0 & 1 \ 0 & 0 & -1 \ 1 & 1 & 0 \ 1 & -1 & 0 \ -1 & 1 & 0
          endpmatrix
          $$
          The distance matrix of this space is
          $$
          D = frac110 beginpmatrix 0 & 2 & 1 & 1 & 1 \ 2 & 0 & 1 & 1 & 1 \ 1 & 1 & 0 & 2 & 2 \ 1 & 1 & 2 & 0 & 2 \ 1 & 1 & 2 & 2 & 0 endpmatrix
          $$
          Applying the function $tmapsto exp(-t)$ to each element of $D$, we get a matrix $B$ which is not positive semidefinite. Specifically, the vector $v = (3, 3, -2, -2, -2)$ satisfies
          $$
          vBv^T = 30 + 42e^-1/5 - 72e^-1/10 < 0
          $$



          Sufficient condition



          The kernel $exp(-lambda d)$ is positive semidefinite for all $lambda>0$ if and only if $d$ is a a kernel of conditionally negative type, see Transforming a distance function to a kernel. In the article linked there, Schoenberg proved that a metric space admits an isometric embedding into a Hilbert space if and only if $d^2$ is a kernel of conditionally negative type.



          Conclusion: if $(X, d)$ is a metric space such that its "snowflake" $(X, d^1/2)$ admits an isometric embedding into a Hilbert space, then $exp(-d)$ is a positive semidefinite kernel. Examples of such spaces are: Hilbert space itself, $L^1$, and of course any subsets of those.



          This motivated my counterexample above: $ell_infty^3$ is one of the simplest spaces that do not embed into $L^1$.



          A contemporary source on this subject is Naor's ICM talk on $L^1$ embeddings (its main subject is the Heisenberg group, but the introduction gives an overview of the situation).






          share|cite|improve this answer
















          • 1




            Same question for $exp(-d^2)$, also with a counterexample: Gaussian kernels for arbitrary metric spaces
            – user357151
            Aug 18 at 17:45














          up vote
          1
          down vote



          accepted










          Remark: in this subject, "positive definite" typically means "positive semidefinite" (Wikipedia) and I treat it as such. One does not expect the matrix $exp(-d(x_i, x_j))$ to be strictly positive definite since some $x_i$ could be repeated.



          Counterexample



          Consider the 3-dimensional space $ell_infty^3$, i.e., the norm is $|(x, y, z)| = max(|x|, |y|, |z|)$. Let $mathcal X$ be the five-point subset of $ell_infty^3$, the coordinates of each point being a row of the matrix
          $$
          A = frac110 beginpmatrix
          0 & 0 & 1 \ 0 & 0 & -1 \ 1 & 1 & 0 \ 1 & -1 & 0 \ -1 & 1 & 0
          endpmatrix
          $$
          The distance matrix of this space is
          $$
          D = frac110 beginpmatrix 0 & 2 & 1 & 1 & 1 \ 2 & 0 & 1 & 1 & 1 \ 1 & 1 & 0 & 2 & 2 \ 1 & 1 & 2 & 0 & 2 \ 1 & 1 & 2 & 2 & 0 endpmatrix
          $$
          Applying the function $tmapsto exp(-t)$ to each element of $D$, we get a matrix $B$ which is not positive semidefinite. Specifically, the vector $v = (3, 3, -2, -2, -2)$ satisfies
          $$
          vBv^T = 30 + 42e^-1/5 - 72e^-1/10 < 0
          $$



          Sufficient condition



          The kernel $exp(-lambda d)$ is positive semidefinite for all $lambda>0$ if and only if $d$ is a a kernel of conditionally negative type, see Transforming a distance function to a kernel. In the article linked there, Schoenberg proved that a metric space admits an isometric embedding into a Hilbert space if and only if $d^2$ is a kernel of conditionally negative type.



          Conclusion: if $(X, d)$ is a metric space such that its "snowflake" $(X, d^1/2)$ admits an isometric embedding into a Hilbert space, then $exp(-d)$ is a positive semidefinite kernel. Examples of such spaces are: Hilbert space itself, $L^1$, and of course any subsets of those.



          This motivated my counterexample above: $ell_infty^3$ is one of the simplest spaces that do not embed into $L^1$.



          A contemporary source on this subject is Naor's ICM talk on $L^1$ embeddings (its main subject is the Heisenberg group, but the introduction gives an overview of the situation).






          share|cite|improve this answer
















          • 1




            Same question for $exp(-d^2)$, also with a counterexample: Gaussian kernels for arbitrary metric spaces
            – user357151
            Aug 18 at 17:45












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Remark: in this subject, "positive definite" typically means "positive semidefinite" (Wikipedia) and I treat it as such. One does not expect the matrix $exp(-d(x_i, x_j))$ to be strictly positive definite since some $x_i$ could be repeated.



          Counterexample



          Consider the 3-dimensional space $ell_infty^3$, i.e., the norm is $|(x, y, z)| = max(|x|, |y|, |z|)$. Let $mathcal X$ be the five-point subset of $ell_infty^3$, the coordinates of each point being a row of the matrix
          $$
          A = frac110 beginpmatrix
          0 & 0 & 1 \ 0 & 0 & -1 \ 1 & 1 & 0 \ 1 & -1 & 0 \ -1 & 1 & 0
          endpmatrix
          $$
          The distance matrix of this space is
          $$
          D = frac110 beginpmatrix 0 & 2 & 1 & 1 & 1 \ 2 & 0 & 1 & 1 & 1 \ 1 & 1 & 0 & 2 & 2 \ 1 & 1 & 2 & 0 & 2 \ 1 & 1 & 2 & 2 & 0 endpmatrix
          $$
          Applying the function $tmapsto exp(-t)$ to each element of $D$, we get a matrix $B$ which is not positive semidefinite. Specifically, the vector $v = (3, 3, -2, -2, -2)$ satisfies
          $$
          vBv^T = 30 + 42e^-1/5 - 72e^-1/10 < 0
          $$



          Sufficient condition



          The kernel $exp(-lambda d)$ is positive semidefinite for all $lambda>0$ if and only if $d$ is a a kernel of conditionally negative type, see Transforming a distance function to a kernel. In the article linked there, Schoenberg proved that a metric space admits an isometric embedding into a Hilbert space if and only if $d^2$ is a kernel of conditionally negative type.



          Conclusion: if $(X, d)$ is a metric space such that its "snowflake" $(X, d^1/2)$ admits an isometric embedding into a Hilbert space, then $exp(-d)$ is a positive semidefinite kernel. Examples of such spaces are: Hilbert space itself, $L^1$, and of course any subsets of those.



          This motivated my counterexample above: $ell_infty^3$ is one of the simplest spaces that do not embed into $L^1$.



          A contemporary source on this subject is Naor's ICM talk on $L^1$ embeddings (its main subject is the Heisenberg group, but the introduction gives an overview of the situation).






          share|cite|improve this answer












          Remark: in this subject, "positive definite" typically means "positive semidefinite" (Wikipedia) and I treat it as such. One does not expect the matrix $exp(-d(x_i, x_j))$ to be strictly positive definite since some $x_i$ could be repeated.



          Counterexample



          Consider the 3-dimensional space $ell_infty^3$, i.e., the norm is $|(x, y, z)| = max(|x|, |y|, |z|)$. Let $mathcal X$ be the five-point subset of $ell_infty^3$, the coordinates of each point being a row of the matrix
          $$
          A = frac110 beginpmatrix
          0 & 0 & 1 \ 0 & 0 & -1 \ 1 & 1 & 0 \ 1 & -1 & 0 \ -1 & 1 & 0
          endpmatrix
          $$
          The distance matrix of this space is
          $$
          D = frac110 beginpmatrix 0 & 2 & 1 & 1 & 1 \ 2 & 0 & 1 & 1 & 1 \ 1 & 1 & 0 & 2 & 2 \ 1 & 1 & 2 & 0 & 2 \ 1 & 1 & 2 & 2 & 0 endpmatrix
          $$
          Applying the function $tmapsto exp(-t)$ to each element of $D$, we get a matrix $B$ which is not positive semidefinite. Specifically, the vector $v = (3, 3, -2, -2, -2)$ satisfies
          $$
          vBv^T = 30 + 42e^-1/5 - 72e^-1/10 < 0
          $$



          Sufficient condition



          The kernel $exp(-lambda d)$ is positive semidefinite for all $lambda>0$ if and only if $d$ is a a kernel of conditionally negative type, see Transforming a distance function to a kernel. In the article linked there, Schoenberg proved that a metric space admits an isometric embedding into a Hilbert space if and only if $d^2$ is a kernel of conditionally negative type.



          Conclusion: if $(X, d)$ is a metric space such that its "snowflake" $(X, d^1/2)$ admits an isometric embedding into a Hilbert space, then $exp(-d)$ is a positive semidefinite kernel. Examples of such spaces are: Hilbert space itself, $L^1$, and of course any subsets of those.



          This motivated my counterexample above: $ell_infty^3$ is one of the simplest spaces that do not embed into $L^1$.



          A contemporary source on this subject is Naor's ICM talk on $L^1$ embeddings (its main subject is the Heisenberg group, but the introduction gives an overview of the situation).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 18 at 17:38









          user357151

          14.1k31140




          14.1k31140







          • 1




            Same question for $exp(-d^2)$, also with a counterexample: Gaussian kernels for arbitrary metric spaces
            – user357151
            Aug 18 at 17:45












          • 1




            Same question for $exp(-d^2)$, also with a counterexample: Gaussian kernels for arbitrary metric spaces
            – user357151
            Aug 18 at 17:45







          1




          1




          Same question for $exp(-d^2)$, also with a counterexample: Gaussian kernels for arbitrary metric spaces
          – user357151
          Aug 18 at 17:45




          Same question for $exp(-d^2)$, also with a counterexample: Gaussian kernels for arbitrary metric spaces
          – user357151
          Aug 18 at 17:45












           

          draft saved


          draft discarded


























           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2885462%2fis-the-exponential-of-d-a-positive-definite-kernel-for-a-general-metric-space%23new-answer', 'question_page');

          );

          Post as a guest













































































          這個網誌中的熱門文章

          How to combine Bézier curves to a surface?

          Why am i infinitely getting the same tweet with the Twitter Search API?

          Carbon dioxide