www.pudn.com > FitFunc.zip > fit_ML_laplace.m

function result = fit_ML_laplace( x,hAx ) 
% fit_ML_normal - Maximum Likelihood fit of the laplace distribution of i.i.d. samples!. 
%                  Given the samples of a laplace distribution, the PDF parameter is found 
% 
%    fits data to the probability of the form:  
%        p(x) = 1/(2*b)*exp(-abs(x-u)/b) 
%    with parameters: u,b 
% 
% format:   result = fit_ML_laplace( x,hAx ) 
% 
% input:    x   - vector, samples with laplace distribution to be parameterized 
%           hAx - handle of an axis, on which the fitted distribution is plotted 
%                 if h is given empty, a figure is created. 
% 
% output:   result  - structure with the fields 
%                      u,b    - fitted parameters 
%                      CRB_b  - Cram?r-Rao Bound for the estimator value 
%                      RMS    - RMS error of the estimation  
%                      type   - 'ML' 
% 
 
% 
% Algorithm 
% =========== 
% 
% We use the ML algorithm to estimate the PDF from the samples. 
% The laplace destribution is given by: 
% 
%    p(x;u,b) = 1/(2*b)*exp(-abs(x-u)/b) 
% 
%    where x are the samples which distribute by the function p(x;u,b) 
%            and are assumed to be i.i.d !!! 
% 
% The ML estimator is given by: 
% 
%    a         = parameters vector = [u,b] 
%    f(Xn,a)   = 1/(2*b)*exp(-abs(Xn-u)/b) 
%    L(a)      = f(X,a) = product_by_n( f(Xn,a) ) 
%              = (2*b)^(-N) * exp( - sum( abs(Xn-u) )/b ) 
%    log(L(a)) = -N*log(2*b) - sum( abs(Xn-u) )/b 
% 
%    The maximum likelihood point is found by the derivative of log(L(a)) with respect to "a": 
% 
%    diff(log(L(a)),b)  = N/(b^2) * ( sum( abs(Xn-u) )/N - b ) 
%                       = J(b) * (b_estimation - b)   
%    diff(log(L(a)),m)  = (1/b) * sum( diff( abs(Xn-u),u ) ) => can't obtain a derivative 
%                       But, u is the mean of the distribution, and therefore => u = mean(Xn) 
%                        
% 
%    Therefore, the (efficient) estimators are given by: 
% 
%               u = sum( Xn )/N 
%               b = sum( abs(Xn-u) )/N 
% 
%    The Cram?r-Rao Bounds for these estimator are: 
% 
%               VAR( m ) = ? 
%               VAR( b ) = 1/J(b) = b^2 / N 
% 
%    NOTE: the ML estimator does not detect a deviation from the model. 
%          therefore, check the RMS value ! 
% 
 
if (nargin<1) 
    error( 'fit_ML_laplace - insufficient input arguments' ); 
end 
 
% Estimation 
% ============= 
x       = x(:);                 % should be column vectors ! 
N       = length(x); 
u       = sum( x )/N; 
b       = sum(abs(x-u))/N; 
CRB_b   = b^2 / N; 
[n,x_c] = hist( x,100 ); 
n       = n / sum(n*abs(x_c(2)-x_c(1))); 
y       = 1/(2*b)*exp(-abs(x_c-u)/b); 
RMS     = sqrt( (y-n)*((y-n)')/ (x_c(2)-x_c(1))^2 / (length(x_c)-1) ); 
 
% finish summarizing results 
% ============================ 
result = struct( 'u',u,'b',b,'CRB_b',CRB_b,'RMS',RMS,'type','ML' ); 
 
% plot distribution if asked for 
% =============================== 
if (nargin>1) 
    xspan = linspace(min(x),max(x),100); 
    if ishandle( hAx ) 
        plot_laplace( xspan,result,hAx,1 ); 
    else 
        figure; 
        plot_laplace( xspan,result,gca,1 ); 
    end 
end