slicesample
              Draws nsamples samples from a target stationary distribution pdf using slice sampling of Radford M. Neal.
Input:
Next, several property-value pairs can or must be specified, they are:
(Required properties) One of:
or
The following input property/pair values may be needed depending on the desired outut:
Outputs:
Example : Sampling from a normal distribution
| start = 1; nsamples = 1e3; pdf = @(x) exp (-.5 * x .^ 2) / (pi ^ .5 * 2 ^ .5); [smpl, accept] = slicesample (start, nsamples, "pdf", pdf, "thin", 4); histfit (smpl); | 
See also: rand, mhsample, randsample
Source Code: slicesample
| 
 ## Define function to sample
 d = 2;
 mu = [-1; 2];
 rand ("seed", 5)  # for reproducibility
 Sigma = rand (d);
 Sigma = (Sigma + Sigma');
 Sigma += eye (d)*abs (eigs (Sigma, 1, "sa")) * 1.1;
 pdf = @(x)(2*pi)^(-d/2)*det(Sigma)^-.5*exp(-.5*sum((x.'-mu).*(Sigma\(x.'-mu)),1));
 ## Inputs
 start = ones (1,2);
 nsamples = 500;
 K = 500;
 m = 10;
 rande ("seed", 4);  rand ("seed", 5)  # for reproducibility
 [smpl, accept] = slicesample (start, nsamples, "pdf", pdf, "burnin", K, "thin", m, "width", [20, 30]);
 figure;
 hold on;
 plot (smpl(:,1), smpl(:,2), 'x');
 [x, y] = meshgrid (linspace (-6,4), linspace(-3,7));
 z = reshape (pdf ([x(:), y(:)]), size(x));
 mesh (x, y, z, "facecolor", "None");
 ## Using sample points to find the volume of half a sphere with radius of .5
 f = @(x) ((.25-(x(:,1)+1).^2-(x(:,2)-2).^2).^.5.*(((x(:,1)+1).^2+(x(:,2)-2).^2)<.25)).';
 int = mean (f (smpl) ./ pdf (smpl));
 errest = std (f (smpl) ./ pdf (smpl)) / nsamples^.5;
 trueerr = abs (2/3*pi*.25^(3/2)-int);
 fprintf ("Monte Carlo integral estimate int f(x) dx = %f\n", int);
 fprintf ("Monte Carlo integral error estimate %f\n", errest);
 fprintf ("The actual error %f\n", trueerr);
 mesh (x,y,reshape (f([x(:), y(:)]), size(x)), "facecolor", "None");
Monte Carlo integral estimate int f(x) dx = 0.228408
Monte Carlo integral error estimate 0.029831
The actual error 0.033392
                     | 
 
                  | 
 ## Integrate truncated normal distribution to find normilization constant
 pdf = @(x) exp (-.5*x.^2)/(pi^.5*2^.5);
 nsamples = 1e3;
 rande ("seed", 4);  rand ("seed", 5)  # for reproducibility
 [smpl, accept] = slicesample (1, nsamples, "pdf", pdf, "thin", 4);
 f = @(x) exp (-.5 * x .^ 2) .* (x >= -2 & x <= 2);
 x = linspace (-3, 3, 1000);
 area (x, f(x));
 xlabel ("x");
 ylabel ("f(x)");
 int = mean (f (smpl) ./ pdf (smpl));
 errest = std (f (smpl) ./ pdf (smpl)) / nsamples ^ 0.5;
 trueerr = abs (erf (2 ^ 0.5) * 2 ^ 0.5 * pi ^ 0.5 - int);
 fprintf("Monte Carlo integral estimate int f(x) dx = %f\n", int);
 fprintf("Monte Carlo integral error estimate %f\n", errest);
 fprintf("The actual error %f\n", trueerr);
Monte Carlo integral estimate int f(x) dx = 2.376284
Monte Carlo integral error estimate 0.017608
The actual error 0.016292
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