Convergence of density estimates with parzen window












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I am trying to understand why $lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i} = 0$ is necessary for convergence of Parzen density estimates. Similar question has been asked here



Intuitively we estimate density using parzen windows by squeezing a distribution into smaller windows as the number of observations increases. When $n=infty$ this distribution should converge to spike at $0$. Thus we design density kernel in a way that probability is 0 everywhere except origin after squeezing infinitely. If the density function has a finite value at someplace close to infinity it will be difficult to squeegee it to a spike at 0. This can be understood in terms of convergence of kernel density function to delta dirac function. I have tried to understand this mathetically as follows:



Convergence in mean square for parzen window assumes that limit of sequence of function in $mathbf{R}^m$ given by $epsilon^{-m} varphi(x/epsilon)$ converges to Dirac Delta function due to which we can say $bar{p}(x)=int frac{1}{V} varphi(frac{x-v}{epsilon}) p(v) dx$ converges to $p(x)$ as $epsilon to 0$.



Coming to $lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i} = 0$.



Substituting $x_i/epsilon=u_i $ for any $x >0$ in $L=lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i}$ we get



$L=lim_{||epsilon|| rightarrow 0}(prod_{i=1}^{d}x_{i})epsilon^{-m} varphi(x/epsilon)$



Since $lim_{||epsilon|| rightarrow 0}epsilon^{-m} varphi(x/epsilon)$ is delta dirac function its value should be be $0$ everywhere except $x=0$, so $L=0$



Please let me know if my understanding is correct.










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    1












    $begingroup$


    I am trying to understand why $lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i} = 0$ is necessary for convergence of Parzen density estimates. Similar question has been asked here



    Intuitively we estimate density using parzen windows by squeezing a distribution into smaller windows as the number of observations increases. When $n=infty$ this distribution should converge to spike at $0$. Thus we design density kernel in a way that probability is 0 everywhere except origin after squeezing infinitely. If the density function has a finite value at someplace close to infinity it will be difficult to squeegee it to a spike at 0. This can be understood in terms of convergence of kernel density function to delta dirac function. I have tried to understand this mathetically as follows:



    Convergence in mean square for parzen window assumes that limit of sequence of function in $mathbf{R}^m$ given by $epsilon^{-m} varphi(x/epsilon)$ converges to Dirac Delta function due to which we can say $bar{p}(x)=int frac{1}{V} varphi(frac{x-v}{epsilon}) p(v) dx$ converges to $p(x)$ as $epsilon to 0$.



    Coming to $lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i} = 0$.



    Substituting $x_i/epsilon=u_i $ for any $x >0$ in $L=lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i}$ we get



    $L=lim_{||epsilon|| rightarrow 0}(prod_{i=1}^{d}x_{i})epsilon^{-m} varphi(x/epsilon)$



    Since $lim_{||epsilon|| rightarrow 0}epsilon^{-m} varphi(x/epsilon)$ is delta dirac function its value should be be $0$ everywhere except $x=0$, so $L=0$



    Please let me know if my understanding is correct.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to understand why $lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i} = 0$ is necessary for convergence of Parzen density estimates. Similar question has been asked here



      Intuitively we estimate density using parzen windows by squeezing a distribution into smaller windows as the number of observations increases. When $n=infty$ this distribution should converge to spike at $0$. Thus we design density kernel in a way that probability is 0 everywhere except origin after squeezing infinitely. If the density function has a finite value at someplace close to infinity it will be difficult to squeegee it to a spike at 0. This can be understood in terms of convergence of kernel density function to delta dirac function. I have tried to understand this mathetically as follows:



      Convergence in mean square for parzen window assumes that limit of sequence of function in $mathbf{R}^m$ given by $epsilon^{-m} varphi(x/epsilon)$ converges to Dirac Delta function due to which we can say $bar{p}(x)=int frac{1}{V} varphi(frac{x-v}{epsilon}) p(v) dx$ converges to $p(x)$ as $epsilon to 0$.



      Coming to $lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i} = 0$.



      Substituting $x_i/epsilon=u_i $ for any $x >0$ in $L=lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i}$ we get



      $L=lim_{||epsilon|| rightarrow 0}(prod_{i=1}^{d}x_{i})epsilon^{-m} varphi(x/epsilon)$



      Since $lim_{||epsilon|| rightarrow 0}epsilon^{-m} varphi(x/epsilon)$ is delta dirac function its value should be be $0$ everywhere except $x=0$, so $L=0$



      Please let me know if my understanding is correct.










      share|cite|improve this question









      $endgroup$




      I am trying to understand why $lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i} = 0$ is necessary for convergence of Parzen density estimates. Similar question has been asked here



      Intuitively we estimate density using parzen windows by squeezing a distribution into smaller windows as the number of observations increases. When $n=infty$ this distribution should converge to spike at $0$. Thus we design density kernel in a way that probability is 0 everywhere except origin after squeezing infinitely. If the density function has a finite value at someplace close to infinity it will be difficult to squeegee it to a spike at 0. This can be understood in terms of convergence of kernel density function to delta dirac function. I have tried to understand this mathetically as follows:



      Convergence in mean square for parzen window assumes that limit of sequence of function in $mathbf{R}^m$ given by $epsilon^{-m} varphi(x/epsilon)$ converges to Dirac Delta function due to which we can say $bar{p}(x)=int frac{1}{V} varphi(frac{x-v}{epsilon}) p(v) dx$ converges to $p(x)$ as $epsilon to 0$.



      Coming to $lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i} = 0$.



      Substituting $x_i/epsilon=u_i $ for any $x >0$ in $L=lim_{||u|| rightarrow+infty}{varphi(u)}prod_{i=1}^{d}u_{i}$ we get



      $L=lim_{||epsilon|| rightarrow 0}(prod_{i=1}^{d}x_{i})epsilon^{-m} varphi(x/epsilon)$



      Since $lim_{||epsilon|| rightarrow 0}epsilon^{-m} varphi(x/epsilon)$ is delta dirac function its value should be be $0$ everywhere except $x=0$, so $L=0$



      Please let me know if my understanding is correct.







      convergence estimation dirac-delta density-function






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      asked Dec 18 '18 at 16:16









      CuriousCurious

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