Lipschitz function of independent sub-Gaussian random variables












6












$begingroup$


If $Xsim mathcal{N}(0,I)$ is a Gaussian random vector, then Lipschitz functions of $X$ are sub-Gaussian with variance parameter 1 by the Tsirelson-Ibragimov-Sudakov inequality (eg. Theorem 8 here).



Suppose $X = (X_1,X_2,ldots, X_n)$ consisted of independent sub-Gaussian random variables themselves, which are not normally distributed. Does the above property still hold?






concentration-of-measure







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hi, any news on your question? I'm also interested.
    $endgroup$
    – nullgeppetto
    Jan 22 '17 at 20:02












  • $begingroup$
    No, I have not been able to resolve this yet.
    $endgroup$
    – Hedonist
    Jan 28 '17 at 16:43
















6












$begingroup$


If $Xsim mathcal{N}(0,I)$ is a Gaussian random vector, then Lipschitz functions of $X$ are sub-Gaussian with variance parameter 1 by the Tsirelson-Ibragimov-Sudakov inequality (eg. Theorem 8 here).



Suppose $X = (X_1,X_2,ldots, X_n)$ consisted of independent sub-Gaussian random variables themselves, which are not normally distributed. Does the above property still hold?






concentration-of-measure







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hi, any news on your question? I'm also interested.
    $endgroup$
    – nullgeppetto
    Jan 22 '17 at 20:02












  • $begingroup$
    No, I have not been able to resolve this yet.
    $endgroup$
    – Hedonist
    Jan 28 '17 at 16:43














6












6








6


3



$begingroup$


If $Xsim mathcal{N}(0,I)$ is a Gaussian random vector, then Lipschitz functions of $X$ are sub-Gaussian with variance parameter 1 by the Tsirelson-Ibragimov-Sudakov inequality (eg. Theorem 8 here).



Suppose $X = (X_1,X_2,ldots, X_n)$ consisted of independent sub-Gaussian random variables themselves, which are not normally distributed. Does the above property still hold?






concentration-of-measure







share|cite|improve this question









$endgroup$




If $Xsim mathcal{N}(0,I)$ is a Gaussian random vector, then Lipschitz functions of $X$ are sub-Gaussian with variance parameter 1 by the Tsirelson-Ibragimov-Sudakov inequality (eg. Theorem 8 here).



Suppose $X = (X_1,X_2,ldots, X_n)$ consisted of independent sub-Gaussian random variables themselves, which are not normally distributed. Does the above property still hold?






concentration-of-measure




concentration-of-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 22 '16 at 15:06









HedonistHedonist

650312




650312












  • $begingroup$
    Hi, any news on your question? I'm also interested.
    $endgroup$
    – nullgeppetto
    Jan 22 '17 at 20:02












  • $begingroup$
    No, I have not been able to resolve this yet.
    $endgroup$
    – Hedonist
    Jan 28 '17 at 16:43


















  • $begingroup$
    Hi, any news on your question? I'm also interested.
    $endgroup$
    – nullgeppetto
    Jan 22 '17 at 20:02












  • $begingroup$
    No, I have not been able to resolve this yet.
    $endgroup$
    – Hedonist
    Jan 28 '17 at 16:43
















$begingroup$
Hi, any news on your question? I'm also interested.
$endgroup$
– nullgeppetto
Jan 22 '17 at 20:02






$begingroup$
Hi, any news on your question? I'm also interested.
$endgroup$
– nullgeppetto
Jan 22 '17 at 20:02














$begingroup$
No, I have not been able to resolve this yet.
$endgroup$
– Hedonist
Jan 28 '17 at 16:43




$begingroup$
No, I have not been able to resolve this yet.
$endgroup$
– Hedonist
Jan 28 '17 at 16:43










2 Answers
2






active

oldest

votes


















2












$begingroup$

Try the following extension of McDiarmid’s inequality for metric spaces
with unbounded diameter:
https://arxiv.org/pdf/1309.1007.pdf






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $X$ is d-dimensional sub-gaussian vector with $Cov(X) = sigma^2 I$ and $f$ is L-Lipschitz w.r.t Euclidean norm, can we use the result of this paper? If so, what would be the sub-gaussian diameter as defined in the paper?
    $endgroup$
    – ie86
    May 25 '17 at 6:30



















0












$begingroup$

Here are two options that may suit your needs.




  1. Concentration inequality for convex functions of bounded random variables.
    If $X_1,...,X_n$ are independent taking values in in $[0,1]$ and $f$ is a quasi-convex, then [P(f(X) > m+t) le 2e^{-t^2/4},
    P(f(X) < m - t) le 2e^{-t^2/4}
    qquad
    ] where $m$ is the median of $f(X)$. See Theorem 7.12 in the book Concentration Inequalities: A Nonasymptotic Theory of Independence
    by Gábor Lugosi, Pascal Massart, and Stéphane Boucheron. It follows from the convex distance inequality due to Talagrand.


  2. View $X_i$ as a function of a standard normal. If $X_i$ can be written as $Phi(Z_i)$ where $Z_i$ is standard normal, then $f(X) = fcirc Phi(Z)$ where $Z_1,...,Z_n$ are iid standard normal. Here, the multivariate function $Phi:R^nto R^n$ applies $Phi$ on every coordinate.

    Then the Tsirelson-Ibragimov-Sudakov inequality applies to $fcirc Phi$, and the Lipschitz norm of $fcirc Phi$ is at most $|f|_{Lip} |Phi|_{Lip}$. Now, the question is whether $|Phi|_{Lip}$ is bounded by an absolute constant (and, in particular, whether $Phi$ is Lipschitz at all, otherwise $|Phi|_{Lip}=+infty$ and we do not get anything).
    Inequality $|Phi|_{Lip}<M+infty$ holds, for instance, if $X_i$ is uniformly distributed on $[0,1]$, see Theorem 5.2.10 in the book High Dimensional Probability by Roman Vershynin where this approach is described.


  3. If $X$ has density $e^{-U(x)}$ for strongly convex $U:R^nto R^n$.
    If $U$ is twice continuously differentiable and strongly convex in the sense that the Hessian $H$ of $U$ (i.e., $H_{ij} = (partial/partial x_i)(partial/partial x_i) U$ satisfies for all $xin R^n$ that $H(x) - kappa I_{ntimes n}$ is positive semi-definite, then for any 1-Lipschitz function $f$ of $X$,
    [ P( |f(X) - E[f(X)] | > t) le 2 exp(-kappa c t^2) ]
    for some absolute constant $c>0$. This is Theorem 5.2.15 in the book High Dimensional Probability by Roman Vershynin.







share|cite|improve this answer











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Try the following extension of McDiarmid’s inequality for metric spaces
    with unbounded diameter:
    https://arxiv.org/pdf/1309.1007.pdf






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If $X$ is d-dimensional sub-gaussian vector with $Cov(X) = sigma^2 I$ and $f$ is L-Lipschitz w.r.t Euclidean norm, can we use the result of this paper? If so, what would be the sub-gaussian diameter as defined in the paper?
      $endgroup$
      – ie86
      May 25 '17 at 6:30
















    2












    $begingroup$

    Try the following extension of McDiarmid’s inequality for metric spaces
    with unbounded diameter:
    https://arxiv.org/pdf/1309.1007.pdf






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If $X$ is d-dimensional sub-gaussian vector with $Cov(X) = sigma^2 I$ and $f$ is L-Lipschitz w.r.t Euclidean norm, can we use the result of this paper? If so, what would be the sub-gaussian diameter as defined in the paper?
      $endgroup$
      – ie86
      May 25 '17 at 6:30














    2












    2








    2





    $begingroup$

    Try the following extension of McDiarmid’s inequality for metric spaces
    with unbounded diameter:
    https://arxiv.org/pdf/1309.1007.pdf






    share|cite|improve this answer









    $endgroup$



    Try the following extension of McDiarmid’s inequality for metric spaces
    with unbounded diameter:
    https://arxiv.org/pdf/1309.1007.pdf







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 23 '17 at 21:41









    UriUri

    362




    362












    • $begingroup$
      If $X$ is d-dimensional sub-gaussian vector with $Cov(X) = sigma^2 I$ and $f$ is L-Lipschitz w.r.t Euclidean norm, can we use the result of this paper? If so, what would be the sub-gaussian diameter as defined in the paper?
      $endgroup$
      – ie86
      May 25 '17 at 6:30


















    • $begingroup$
      If $X$ is d-dimensional sub-gaussian vector with $Cov(X) = sigma^2 I$ and $f$ is L-Lipschitz w.r.t Euclidean norm, can we use the result of this paper? If so, what would be the sub-gaussian diameter as defined in the paper?
      $endgroup$
      – ie86
      May 25 '17 at 6:30
















    $begingroup$
    If $X$ is d-dimensional sub-gaussian vector with $Cov(X) = sigma^2 I$ and $f$ is L-Lipschitz w.r.t Euclidean norm, can we use the result of this paper? If so, what would be the sub-gaussian diameter as defined in the paper?
    $endgroup$
    – ie86
    May 25 '17 at 6:30




    $begingroup$
    If $X$ is d-dimensional sub-gaussian vector with $Cov(X) = sigma^2 I$ and $f$ is L-Lipschitz w.r.t Euclidean norm, can we use the result of this paper? If so, what would be the sub-gaussian diameter as defined in the paper?
    $endgroup$
    – ie86
    May 25 '17 at 6:30











    0












    $begingroup$

    Here are two options that may suit your needs.




    1. Concentration inequality for convex functions of bounded random variables.
      If $X_1,...,X_n$ are independent taking values in in $[0,1]$ and $f$ is a quasi-convex, then [P(f(X) > m+t) le 2e^{-t^2/4},
      P(f(X) < m - t) le 2e^{-t^2/4}
      qquad
      ] where $m$ is the median of $f(X)$. See Theorem 7.12 in the book Concentration Inequalities: A Nonasymptotic Theory of Independence
      by Gábor Lugosi, Pascal Massart, and Stéphane Boucheron. It follows from the convex distance inequality due to Talagrand.


    2. View $X_i$ as a function of a standard normal. If $X_i$ can be written as $Phi(Z_i)$ where $Z_i$ is standard normal, then $f(X) = fcirc Phi(Z)$ where $Z_1,...,Z_n$ are iid standard normal. Here, the multivariate function $Phi:R^nto R^n$ applies $Phi$ on every coordinate.

      Then the Tsirelson-Ibragimov-Sudakov inequality applies to $fcirc Phi$, and the Lipschitz norm of $fcirc Phi$ is at most $|f|_{Lip} |Phi|_{Lip}$. Now, the question is whether $|Phi|_{Lip}$ is bounded by an absolute constant (and, in particular, whether $Phi$ is Lipschitz at all, otherwise $|Phi|_{Lip}=+infty$ and we do not get anything).
      Inequality $|Phi|_{Lip}<M+infty$ holds, for instance, if $X_i$ is uniformly distributed on $[0,1]$, see Theorem 5.2.10 in the book High Dimensional Probability by Roman Vershynin where this approach is described.


    3. If $X$ has density $e^{-U(x)}$ for strongly convex $U:R^nto R^n$.
      If $U$ is twice continuously differentiable and strongly convex in the sense that the Hessian $H$ of $U$ (i.e., $H_{ij} = (partial/partial x_i)(partial/partial x_i) U$ satisfies for all $xin R^n$ that $H(x) - kappa I_{ntimes n}$ is positive semi-definite, then for any 1-Lipschitz function $f$ of $X$,
      [ P( |f(X) - E[f(X)] | > t) le 2 exp(-kappa c t^2) ]
      for some absolute constant $c>0$. This is Theorem 5.2.15 in the book High Dimensional Probability by Roman Vershynin.







    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Here are two options that may suit your needs.




      1. Concentration inequality for convex functions of bounded random variables.
        If $X_1,...,X_n$ are independent taking values in in $[0,1]$ and $f$ is a quasi-convex, then [P(f(X) > m+t) le 2e^{-t^2/4},
        P(f(X) < m - t) le 2e^{-t^2/4}
        qquad
        ] where $m$ is the median of $f(X)$. See Theorem 7.12 in the book Concentration Inequalities: A Nonasymptotic Theory of Independence
        by Gábor Lugosi, Pascal Massart, and Stéphane Boucheron. It follows from the convex distance inequality due to Talagrand.


      2. View $X_i$ as a function of a standard normal. If $X_i$ can be written as $Phi(Z_i)$ where $Z_i$ is standard normal, then $f(X) = fcirc Phi(Z)$ where $Z_1,...,Z_n$ are iid standard normal. Here, the multivariate function $Phi:R^nto R^n$ applies $Phi$ on every coordinate.

        Then the Tsirelson-Ibragimov-Sudakov inequality applies to $fcirc Phi$, and the Lipschitz norm of $fcirc Phi$ is at most $|f|_{Lip} |Phi|_{Lip}$. Now, the question is whether $|Phi|_{Lip}$ is bounded by an absolute constant (and, in particular, whether $Phi$ is Lipschitz at all, otherwise $|Phi|_{Lip}=+infty$ and we do not get anything).
        Inequality $|Phi|_{Lip}<M+infty$ holds, for instance, if $X_i$ is uniformly distributed on $[0,1]$, see Theorem 5.2.10 in the book High Dimensional Probability by Roman Vershynin where this approach is described.


      3. If $X$ has density $e^{-U(x)}$ for strongly convex $U:R^nto R^n$.
        If $U$ is twice continuously differentiable and strongly convex in the sense that the Hessian $H$ of $U$ (i.e., $H_{ij} = (partial/partial x_i)(partial/partial x_i) U$ satisfies for all $xin R^n$ that $H(x) - kappa I_{ntimes n}$ is positive semi-definite, then for any 1-Lipschitz function $f$ of $X$,
        [ P( |f(X) - E[f(X)] | > t) le 2 exp(-kappa c t^2) ]
        for some absolute constant $c>0$. This is Theorem 5.2.15 in the book High Dimensional Probability by Roman Vershynin.







      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Here are two options that may suit your needs.




        1. Concentration inequality for convex functions of bounded random variables.
          If $X_1,...,X_n$ are independent taking values in in $[0,1]$ and $f$ is a quasi-convex, then [P(f(X) > m+t) le 2e^{-t^2/4},
          P(f(X) < m - t) le 2e^{-t^2/4}
          qquad
          ] where $m$ is the median of $f(X)$. See Theorem 7.12 in the book Concentration Inequalities: A Nonasymptotic Theory of Independence
          by Gábor Lugosi, Pascal Massart, and Stéphane Boucheron. It follows from the convex distance inequality due to Talagrand.


        2. View $X_i$ as a function of a standard normal. If $X_i$ can be written as $Phi(Z_i)$ where $Z_i$ is standard normal, then $f(X) = fcirc Phi(Z)$ where $Z_1,...,Z_n$ are iid standard normal. Here, the multivariate function $Phi:R^nto R^n$ applies $Phi$ on every coordinate.

          Then the Tsirelson-Ibragimov-Sudakov inequality applies to $fcirc Phi$, and the Lipschitz norm of $fcirc Phi$ is at most $|f|_{Lip} |Phi|_{Lip}$. Now, the question is whether $|Phi|_{Lip}$ is bounded by an absolute constant (and, in particular, whether $Phi$ is Lipschitz at all, otherwise $|Phi|_{Lip}=+infty$ and we do not get anything).
          Inequality $|Phi|_{Lip}<M+infty$ holds, for instance, if $X_i$ is uniformly distributed on $[0,1]$, see Theorem 5.2.10 in the book High Dimensional Probability by Roman Vershynin where this approach is described.


        3. If $X$ has density $e^{-U(x)}$ for strongly convex $U:R^nto R^n$.
          If $U$ is twice continuously differentiable and strongly convex in the sense that the Hessian $H$ of $U$ (i.e., $H_{ij} = (partial/partial x_i)(partial/partial x_i) U$ satisfies for all $xin R^n$ that $H(x) - kappa I_{ntimes n}$ is positive semi-definite, then for any 1-Lipschitz function $f$ of $X$,
          [ P( |f(X) - E[f(X)] | > t) le 2 exp(-kappa c t^2) ]
          for some absolute constant $c>0$. This is Theorem 5.2.15 in the book High Dimensional Probability by Roman Vershynin.







        share|cite|improve this answer











        $endgroup$



        Here are two options that may suit your needs.




        1. Concentration inequality for convex functions of bounded random variables.
          If $X_1,...,X_n$ are independent taking values in in $[0,1]$ and $f$ is a quasi-convex, then [P(f(X) > m+t) le 2e^{-t^2/4},
          P(f(X) < m - t) le 2e^{-t^2/4}
          qquad
          ] where $m$ is the median of $f(X)$. See Theorem 7.12 in the book Concentration Inequalities: A Nonasymptotic Theory of Independence
          by Gábor Lugosi, Pascal Massart, and Stéphane Boucheron. It follows from the convex distance inequality due to Talagrand.


        2. View $X_i$ as a function of a standard normal. If $X_i$ can be written as $Phi(Z_i)$ where $Z_i$ is standard normal, then $f(X) = fcirc Phi(Z)$ where $Z_1,...,Z_n$ are iid standard normal. Here, the multivariate function $Phi:R^nto R^n$ applies $Phi$ on every coordinate.

          Then the Tsirelson-Ibragimov-Sudakov inequality applies to $fcirc Phi$, and the Lipschitz norm of $fcirc Phi$ is at most $|f|_{Lip} |Phi|_{Lip}$. Now, the question is whether $|Phi|_{Lip}$ is bounded by an absolute constant (and, in particular, whether $Phi$ is Lipschitz at all, otherwise $|Phi|_{Lip}=+infty$ and we do not get anything).
          Inequality $|Phi|_{Lip}<M+infty$ holds, for instance, if $X_i$ is uniformly distributed on $[0,1]$, see Theorem 5.2.10 in the book High Dimensional Probability by Roman Vershynin where this approach is described.


        3. If $X$ has density $e^{-U(x)}$ for strongly convex $U:R^nto R^n$.
          If $U$ is twice continuously differentiable and strongly convex in the sense that the Hessian $H$ of $U$ (i.e., $H_{ij} = (partial/partial x_i)(partial/partial x_i) U$ satisfies for all $xin R^n$ that $H(x) - kappa I_{ntimes n}$ is positive semi-definite, then for any 1-Lipschitz function $f$ of $X$,
          [ P( |f(X) - E[f(X)] | > t) le 2 exp(-kappa c t^2) ]
          for some absolute constant $c>0$. This is Theorem 5.2.15 in the book High Dimensional Probability by Roman Vershynin.








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 '18 at 0:39

























        answered Nov 30 '18 at 23:37









        jlewkjlewk

        765




        765






























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