almost sure convergence and asymptotic convergence in python











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I want to plot almost sure convergence and asymptotic convergence. But, I do not have the good result (bad convergence). I don't know where is the problem. It maybe Parzen-Rosenblatt estimator. Is it correctly specified?



from matplotlib.pyplot import *
from math import *
from array import *
import numpy as np
from numpy.random import *
from scipy.misc import *
from scipy.stats import *
from scipy import *
from random import *


#N=1000
n=30
lamb=2
X=lamb*tan(pi*(np.reshape(rand(n,1),n)-0.5)) #loi de Cauchy
x=1
alpha=0.45

def k_gaussien(x):
sigma=1
if(sigma>0):
return((1/(sigma*sqrt(2*pi)))*exp(-(x**2/(2*sigma**2))))

def h(n,alpha):
for i in range(1,n):
return (i**(1-alpha))

def f_PR(x,X,alpha): # Parzen-Rosenblatt estimator
global F;
F = ones((n,1))
h_n = h(n,alpha)
for k in range(2,n):
for i in range(1,k):
F[k] = F[k] + k_gaussien((x-X[i])*i**alpha)
F[k] = F[k] * h_n
return F

# Almost sure convergence f_n(x)--> f(x) ps
figure(figsize=(20,10))
fPR=f_PR(x,X,alpha)
t=linspace(-10,10,n);
plot(cumsum(fPR)/t)
plot(t,0*linspace(1,1,n),lw=3)
plot(t,0*linspace(1,1,n),"r--",lw=3)#with Cauchy density
grid(True)
title("convergence presque sure",fontsize=20,color="blue")

# Convergence in mean N(0,e2f(x))
figure(figsize=(20,10))
x=linspace(1,n,n)
Z= sqrt(h(n,alpha))*(fPR-(1/pi)*(lamb/(lamb**2+x**2))) # sqrt(nh_n)(fn(x)-f(x))
hist(Z,bins=linspace(-10,10,50),normed=True)
title("Theoreme limite centrale",fontsize=20,color="blue")
y=linspace(-10,10,100);
v=(1/pi)*(lamb/(lamb**2+y**2)) # standard deviation (sigma)
plot(y,(1/(sqrt(2*pi)*v))*exp((-y**2)/(2*v**2)),'r',lw=3)# I substitute standard deviation(sigma^2) by cauchy density
title("convergence asymptotique", fontsize=20,color="blue")









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    up vote
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    down vote

    favorite












    I want to plot almost sure convergence and asymptotic convergence. But, I do not have the good result (bad convergence). I don't know where is the problem. It maybe Parzen-Rosenblatt estimator. Is it correctly specified?



    from matplotlib.pyplot import *
    from math import *
    from array import *
    import numpy as np
    from numpy.random import *
    from scipy.misc import *
    from scipy.stats import *
    from scipy import *
    from random import *


    #N=1000
    n=30
    lamb=2
    X=lamb*tan(pi*(np.reshape(rand(n,1),n)-0.5)) #loi de Cauchy
    x=1
    alpha=0.45

    def k_gaussien(x):
    sigma=1
    if(sigma>0):
    return((1/(sigma*sqrt(2*pi)))*exp(-(x**2/(2*sigma**2))))

    def h(n,alpha):
    for i in range(1,n):
    return (i**(1-alpha))

    def f_PR(x,X,alpha): # Parzen-Rosenblatt estimator
    global F;
    F = ones((n,1))
    h_n = h(n,alpha)
    for k in range(2,n):
    for i in range(1,k):
    F[k] = F[k] + k_gaussien((x-X[i])*i**alpha)
    F[k] = F[k] * h_n
    return F

    # Almost sure convergence f_n(x)--> f(x) ps
    figure(figsize=(20,10))
    fPR=f_PR(x,X,alpha)
    t=linspace(-10,10,n);
    plot(cumsum(fPR)/t)
    plot(t,0*linspace(1,1,n),lw=3)
    plot(t,0*linspace(1,1,n),"r--",lw=3)#with Cauchy density
    grid(True)
    title("convergence presque sure",fontsize=20,color="blue")

    # Convergence in mean N(0,e2f(x))
    figure(figsize=(20,10))
    x=linspace(1,n,n)
    Z= sqrt(h(n,alpha))*(fPR-(1/pi)*(lamb/(lamb**2+x**2))) # sqrt(nh_n)(fn(x)-f(x))
    hist(Z,bins=linspace(-10,10,50),normed=True)
    title("Theoreme limite centrale",fontsize=20,color="blue")
    y=linspace(-10,10,100);
    v=(1/pi)*(lamb/(lamb**2+y**2)) # standard deviation (sigma)
    plot(y,(1/(sqrt(2*pi)*v))*exp((-y**2)/(2*v**2)),'r',lw=3)# I substitute standard deviation(sigma^2) by cauchy density
    title("convergence asymptotique", fontsize=20,color="blue")









    share|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I want to plot almost sure convergence and asymptotic convergence. But, I do not have the good result (bad convergence). I don't know where is the problem. It maybe Parzen-Rosenblatt estimator. Is it correctly specified?



      from matplotlib.pyplot import *
      from math import *
      from array import *
      import numpy as np
      from numpy.random import *
      from scipy.misc import *
      from scipy.stats import *
      from scipy import *
      from random import *


      #N=1000
      n=30
      lamb=2
      X=lamb*tan(pi*(np.reshape(rand(n,1),n)-0.5)) #loi de Cauchy
      x=1
      alpha=0.45

      def k_gaussien(x):
      sigma=1
      if(sigma>0):
      return((1/(sigma*sqrt(2*pi)))*exp(-(x**2/(2*sigma**2))))

      def h(n,alpha):
      for i in range(1,n):
      return (i**(1-alpha))

      def f_PR(x,X,alpha): # Parzen-Rosenblatt estimator
      global F;
      F = ones((n,1))
      h_n = h(n,alpha)
      for k in range(2,n):
      for i in range(1,k):
      F[k] = F[k] + k_gaussien((x-X[i])*i**alpha)
      F[k] = F[k] * h_n
      return F

      # Almost sure convergence f_n(x)--> f(x) ps
      figure(figsize=(20,10))
      fPR=f_PR(x,X,alpha)
      t=linspace(-10,10,n);
      plot(cumsum(fPR)/t)
      plot(t,0*linspace(1,1,n),lw=3)
      plot(t,0*linspace(1,1,n),"r--",lw=3)#with Cauchy density
      grid(True)
      title("convergence presque sure",fontsize=20,color="blue")

      # Convergence in mean N(0,e2f(x))
      figure(figsize=(20,10))
      x=linspace(1,n,n)
      Z= sqrt(h(n,alpha))*(fPR-(1/pi)*(lamb/(lamb**2+x**2))) # sqrt(nh_n)(fn(x)-f(x))
      hist(Z,bins=linspace(-10,10,50),normed=True)
      title("Theoreme limite centrale",fontsize=20,color="blue")
      y=linspace(-10,10,100);
      v=(1/pi)*(lamb/(lamb**2+y**2)) # standard deviation (sigma)
      plot(y,(1/(sqrt(2*pi)*v))*exp((-y**2)/(2*v**2)),'r',lw=3)# I substitute standard deviation(sigma^2) by cauchy density
      title("convergence asymptotique", fontsize=20,color="blue")









      share|improve this question















      I want to plot almost sure convergence and asymptotic convergence. But, I do not have the good result (bad convergence). I don't know where is the problem. It maybe Parzen-Rosenblatt estimator. Is it correctly specified?



      from matplotlib.pyplot import *
      from math import *
      from array import *
      import numpy as np
      from numpy.random import *
      from scipy.misc import *
      from scipy.stats import *
      from scipy import *
      from random import *


      #N=1000
      n=30
      lamb=2
      X=lamb*tan(pi*(np.reshape(rand(n,1),n)-0.5)) #loi de Cauchy
      x=1
      alpha=0.45

      def k_gaussien(x):
      sigma=1
      if(sigma>0):
      return((1/(sigma*sqrt(2*pi)))*exp(-(x**2/(2*sigma**2))))

      def h(n,alpha):
      for i in range(1,n):
      return (i**(1-alpha))

      def f_PR(x,X,alpha): # Parzen-Rosenblatt estimator
      global F;
      F = ones((n,1))
      h_n = h(n,alpha)
      for k in range(2,n):
      for i in range(1,k):
      F[k] = F[k] + k_gaussien((x-X[i])*i**alpha)
      F[k] = F[k] * h_n
      return F

      # Almost sure convergence f_n(x)--> f(x) ps
      figure(figsize=(20,10))
      fPR=f_PR(x,X,alpha)
      t=linspace(-10,10,n);
      plot(cumsum(fPR)/t)
      plot(t,0*linspace(1,1,n),lw=3)
      plot(t,0*linspace(1,1,n),"r--",lw=3)#with Cauchy density
      grid(True)
      title("convergence presque sure",fontsize=20,color="blue")

      # Convergence in mean N(0,e2f(x))
      figure(figsize=(20,10))
      x=linspace(1,n,n)
      Z= sqrt(h(n,alpha))*(fPR-(1/pi)*(lamb/(lamb**2+x**2))) # sqrt(nh_n)(fn(x)-f(x))
      hist(Z,bins=linspace(-10,10,50),normed=True)
      title("Theoreme limite centrale",fontsize=20,color="blue")
      y=linspace(-10,10,100);
      v=(1/pi)*(lamb/(lamb**2+y**2)) # standard deviation (sigma)
      plot(y,(1/(sqrt(2*pi)*v))*exp((-y**2)/(2*v**2)),'r',lw=3)# I substitute standard deviation(sigma^2) by cauchy density
      title("convergence asymptotique", fontsize=20,color="blue")






      python simulation






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      edited Nov 11 at 17:48









      Thierry Lathuille

      7,12182630




      7,12182630










      asked Nov 11 at 17:46









      Sa Majesté

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