给出PDF格式的随机值

Question

我必须根据确定的概率密度函数（截断的拉普拉斯）生成随机值：

    Sigma_Phi = 1;
B = 1/(1-exp(-sqrt(2)*pi/Sigma_Phi));
Phi = -60:0.001:60;

for iter=1:length(Phi)
    if Phi(iter) < pi && Phi(iter)>=0
        P(iter) = ((B*exp(-abs(sqrt(2)*Phi(iter)/Sigma_Phi)))/(sqrt(2)*Sigma_Phi));
    elseif Phi(iter) >= -pi && Phi(iter)<=0
        P(iter) = ((B*exp(-abs(sqrt(2)*Phi(iter)/Sigma_Phi)))/(sqrt(2)*Sigma_Phi));
    else
        P(iter)=0;
    end
end

然后我必须在PDF之后生成随机值，但我不知道该怎么做。

Answer 1

以下是使用inverse transform sampling的简单通用解决方案：

% for reproducibility
rng(333) 

Sigma_Phi = 1;
B = 1/(1-exp(-sqrt(2)*pi/Sigma_Phi));
Phi = -10:0.001:10;

P = nan(length(Phi),1);
for iter=1:length(Phi)
    if Phi(iter) < pi && Phi(iter)>=0
        P(iter) = ((B*exp(-abs(sqrt(2)*Phi(iter)/Sigma_Phi)))/(sqrt(2)*Sigma_Phi));
    elseif Phi(iter) >= -pi && Phi(iter)<=0
        P(iter) = ((B*exp(-abs(sqrt(2)*Phi(iter)/Sigma_Phi)))/(sqrt(2)*Sigma_Phi));
    else
        P(iter)=0;
    end
end

% create cdf
cdf         = cumtrapz(Phi, P);

% keep only the unique values: needed for interpolation
idx_mid = (Phi < pi) & (Phi >= -pi);
cdf = cdf(idx_mid);
Phi = Phi(idx_mid);
P   = P(idx_mid);

% number of required random draws
n = 1e4;
% generate uniformly distributed random numbers from [0,1]
r = rand(n,1);

% generate random numbers from the desired pdf; inverse transform sampling
laplrnd = interp1(cdf, Phi, r);

% Verfication plot
[f,x] = hist(laplrnd,100);

bar(x,f/trapz(x,f))
hold on
plot(Phi, P, 'red', 'Linewidth', 1.2)
legend('histogram from random values', 'analytical pdf')