Show that Kernel density estimate integrates to 1












1












$begingroup$


I'm trying to show that the Kernel density estimate



$$ f(x) = frac1{nb}sum_{i=1}^nK left(frac {X_i - x}bright) $$



actually integrates to 1. The $X_i$ are iid and K is a symmetric probability density function; in particular, K integrates to 1. What I get is



$$int_{-infty}^{infty} frac1{nb}sum_{i=1}^nKleft(frac {X_i - x}bright) dx$$
$$ = frac1{nb}int_{-infty}^{infty} sum_{i=1}^nKleft(frac {X_i - x}bright) dx $$



$$=frac1{nb}sum_{i=1}^nint_{-infty}^{infty} Kleft(frac {X_i - x}bright) dx$$
$$=frac1{b}int_{-infty}^{infty} Kleft(frac {X_i - x}bright) dx$$



Using as a substitution



$$ u = frac {X_i - x}b$$
$$ dx = -b du$$



we get



$$=frac1{b}int_{-infty}^{infty} K(u) (-b) du $$
$$= (-b) frac1{b}int_{-infty}^{infty} K(u) du = -1$$



Something must have gone wrong with the integration. I know that $f(x)$ is often defined differently, as



$$ f'(x) = frac1{nb}sum_{i=1}^nKleft(frac {x - X_i}bright) $$



and with this version I can show that it integrates to 1. However $f'(x)$ and $f(x)$ should be equivalent as $K$ is symmetric. So there is some flaw in my calculations (my knowledge about integration with substitution is purely heuristic).






integration statistics probability-distributions improper-integrals







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm trying to show that the Kernel density estimate



    $$ f(x) = frac1{nb}sum_{i=1}^nK left(frac {X_i - x}bright) $$



    actually integrates to 1. The $X_i$ are iid and K is a symmetric probability density function; in particular, K integrates to 1. What I get is



    $$int_{-infty}^{infty} frac1{nb}sum_{i=1}^nKleft(frac {X_i - x}bright) dx$$
    $$ = frac1{nb}int_{-infty}^{infty} sum_{i=1}^nKleft(frac {X_i - x}bright) dx $$



    $$=frac1{nb}sum_{i=1}^nint_{-infty}^{infty} Kleft(frac {X_i - x}bright) dx$$
    $$=frac1{b}int_{-infty}^{infty} Kleft(frac {X_i - x}bright) dx$$



    Using as a substitution



    $$ u = frac {X_i - x}b$$
    $$ dx = -b du$$



    we get



    $$=frac1{b}int_{-infty}^{infty} K(u) (-b) du $$
    $$= (-b) frac1{b}int_{-infty}^{infty} K(u) du = -1$$



    Something must have gone wrong with the integration. I know that $f(x)$ is often defined differently, as



    $$ f'(x) = frac1{nb}sum_{i=1}^nKleft(frac {x - X_i}bright) $$



    and with this version I can show that it integrates to 1. However $f'(x)$ and $f(x)$ should be equivalent as $K$ is symmetric. So there is some flaw in my calculations (my knowledge about integration with substitution is purely heuristic).






    integration statistics probability-distributions improper-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm trying to show that the Kernel density estimate



      $$ f(x) = frac1{nb}sum_{i=1}^nK left(frac {X_i - x}bright) $$



      actually integrates to 1. The $X_i$ are iid and K is a symmetric probability density function; in particular, K integrates to 1. What I get is



      $$int_{-infty}^{infty} frac1{nb}sum_{i=1}^nKleft(frac {X_i - x}bright) dx$$
      $$ = frac1{nb}int_{-infty}^{infty} sum_{i=1}^nKleft(frac {X_i - x}bright) dx $$



      $$=frac1{nb}sum_{i=1}^nint_{-infty}^{infty} Kleft(frac {X_i - x}bright) dx$$
      $$=frac1{b}int_{-infty}^{infty} Kleft(frac {X_i - x}bright) dx$$



      Using as a substitution



      $$ u = frac {X_i - x}b$$
      $$ dx = -b du$$



      we get



      $$=frac1{b}int_{-infty}^{infty} K(u) (-b) du $$
      $$= (-b) frac1{b}int_{-infty}^{infty} K(u) du = -1$$



      Something must have gone wrong with the integration. I know that $f(x)$ is often defined differently, as



      $$ f'(x) = frac1{nb}sum_{i=1}^nKleft(frac {x - X_i}bright) $$



      and with this version I can show that it integrates to 1. However $f'(x)$ and $f(x)$ should be equivalent as $K$ is symmetric. So there is some flaw in my calculations (my knowledge about integration with substitution is purely heuristic).






      integration statistics probability-distributions improper-integrals







      share|cite|improve this question











      $endgroup$




      I'm trying to show that the Kernel density estimate



      $$ f(x) = frac1{nb}sum_{i=1}^nK left(frac {X_i - x}bright) $$



      actually integrates to 1. The $X_i$ are iid and K is a symmetric probability density function; in particular, K integrates to 1. What I get is



      $$int_{-infty}^{infty} frac1{nb}sum_{i=1}^nKleft(frac {X_i - x}bright) dx$$
      $$ = frac1{nb}int_{-infty}^{infty} sum_{i=1}^nKleft(frac {X_i - x}bright) dx $$



      $$=frac1{nb}sum_{i=1}^nint_{-infty}^{infty} Kleft(frac {X_i - x}bright) dx$$
      $$=frac1{b}int_{-infty}^{infty} Kleft(frac {X_i - x}bright) dx$$



      Using as a substitution



      $$ u = frac {X_i - x}b$$
      $$ dx = -b du$$



      we get



      $$=frac1{b}int_{-infty}^{infty} K(u) (-b) du $$
      $$= (-b) frac1{b}int_{-infty}^{infty} K(u) du = -1$$



      Something must have gone wrong with the integration. I know that $f(x)$ is often defined differently, as



      $$ f'(x) = frac1{nb}sum_{i=1}^nKleft(frac {x - X_i}bright) $$



      and with this version I can show that it integrates to 1. However $f'(x)$ and $f(x)$ should be equivalent as $K$ is symmetric. So there is some flaw in my calculations (my knowledge about integration with substitution is purely heuristic).






      integration statistics probability-distributions improper-integrals




      integration statistics probability-distributions improper-integrals






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      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 9 at 16:28









      Chinnapparaj R

      5,4331928




      5,4331928










      asked Jan 9 at 16:26









      CariieCariie

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          $begingroup$

          $require{cancel}$
          begin{align}
          int_{-infty}^{+infty} f(x)~{rm d}x &= frac{1}{nb}sum_{i = 1}^n int_{-infty}^{+infty} Kleft(frac{X_i - x}{b}right){rm d}x & text{with } ~u = frac{X_i - x}{b},~{rm d}u = -{rm d}x/b \
          &= -frac{1}{n}sum_{i=1}^n int_{+infty}^{-infty} K(u)~{rm du} \
          &= frac{1}{n}sum_{i=1}^n cancelto{1}{int_{-infty}^{+infty}K(u){rm d}u} \
          &= frac{1}{cancel{n}} cancel{n} = 1
          end{align}



          The part you need to pay attention to is the limits, when $xto +infty$ you have $u to -infty$ and vice versa. So the limits of the integral change of order






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            $begingroup$

            $require{cancel}$
            begin{align}
            int_{-infty}^{+infty} f(x)~{rm d}x &= frac{1}{nb}sum_{i = 1}^n int_{-infty}^{+infty} Kleft(frac{X_i - x}{b}right){rm d}x & text{with } ~u = frac{X_i - x}{b},~{rm d}u = -{rm d}x/b \
            &= -frac{1}{n}sum_{i=1}^n int_{+infty}^{-infty} K(u)~{rm du} \
            &= frac{1}{n}sum_{i=1}^n cancelto{1}{int_{-infty}^{+infty}K(u){rm d}u} \
            &= frac{1}{cancel{n}} cancel{n} = 1
            end{align}



            The part you need to pay attention to is the limits, when $xto +infty$ you have $u to -infty$ and vice versa. So the limits of the integral change of order






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $require{cancel}$
              begin{align}
              int_{-infty}^{+infty} f(x)~{rm d}x &= frac{1}{nb}sum_{i = 1}^n int_{-infty}^{+infty} Kleft(frac{X_i - x}{b}right){rm d}x & text{with } ~u = frac{X_i - x}{b},~{rm d}u = -{rm d}x/b \
              &= -frac{1}{n}sum_{i=1}^n int_{+infty}^{-infty} K(u)~{rm du} \
              &= frac{1}{n}sum_{i=1}^n cancelto{1}{int_{-infty}^{+infty}K(u){rm d}u} \
              &= frac{1}{cancel{n}} cancel{n} = 1
              end{align}



              The part you need to pay attention to is the limits, when $xto +infty$ you have $u to -infty$ and vice versa. So the limits of the integral change of order






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $require{cancel}$
                begin{align}
                int_{-infty}^{+infty} f(x)~{rm d}x &= frac{1}{nb}sum_{i = 1}^n int_{-infty}^{+infty} Kleft(frac{X_i - x}{b}right){rm d}x & text{with } ~u = frac{X_i - x}{b},~{rm d}u = -{rm d}x/b \
                &= -frac{1}{n}sum_{i=1}^n int_{+infty}^{-infty} K(u)~{rm du} \
                &= frac{1}{n}sum_{i=1}^n cancelto{1}{int_{-infty}^{+infty}K(u){rm d}u} \
                &= frac{1}{cancel{n}} cancel{n} = 1
                end{align}



                The part you need to pay attention to is the limits, when $xto +infty$ you have $u to -infty$ and vice versa. So the limits of the integral change of order






                share|cite|improve this answer









                $endgroup$



                $require{cancel}$
                begin{align}
                int_{-infty}^{+infty} f(x)~{rm d}x &= frac{1}{nb}sum_{i = 1}^n int_{-infty}^{+infty} Kleft(frac{X_i - x}{b}right){rm d}x & text{with } ~u = frac{X_i - x}{b},~{rm d}u = -{rm d}x/b \
                &= -frac{1}{n}sum_{i=1}^n int_{+infty}^{-infty} K(u)~{rm du} \
                &= frac{1}{n}sum_{i=1}^n cancelto{1}{int_{-infty}^{+infty}K(u){rm d}u} \
                &= frac{1}{cancel{n}} cancel{n} = 1
                end{align}



                The part you need to pay attention to is the limits, when $xto +infty$ you have $u to -infty$ and vice versa. So the limits of the integral change of order







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 9 at 16:37









                caveraccaverac

                14.5k31130




                14.5k31130






























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