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.
convergence estimation dirac-delta density-function
asked Dec 18 '18 at 16:16
CuriousCurious
889516
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add a comment |
$begingroup$
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.
convergence estimation dirac-delta density-function
asked Dec 18 '18 at 16:16
CuriousCurious
889516
$endgroup$
add a comment |
$begingroup$
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.
convergence estimation dirac-delta density-function
asked Dec 18 '18 at 16:16
CuriousCurious
889516
$endgroup$
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
convergence estimation dirac-delta density-function
asked Dec 18 '18 at 16:16
CuriousCurious
889516
asked Dec 18 '18 at 16:16
CuriousCurious
889516
asked Dec 18 '18 at 16:16
CuriousCurious
889516
asked Dec 18 '18 at 16:16
CuriousCurious
889516
asked Dec 18 '18 at 16:16
CuriousCurious
889516
889516
add a comment |
add a comment |
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