二維正常累積概率密度

Question

MATLAB輪廓數據我有這個二維正態分佈定義爲

mu = [0 0]; 
Sigma = [1 0.5^0.5; 0.5^0.5 1]; 

有沒有辦法時的累積概率說的是95％，以獲得輪廓數據。我不想要情節，而是（x，y）點的值導致95％輪廓。

如果有人可以幫忙，這將是非常善良。在此先感謝

Answer 1

可以使用數值解算器找到輪廓如下：

% plot the distribution 
figure; 
x = (-5:0.5:5)'; 
y = (-5:0.5:5)'; 
[X1,X2] = meshgrid(x',y'); 
X = [X1(:) X2(:)]; 
p = mvncdf(X,mu,Sigma); 
X3 = reshape(p,length(x),length(y)); 
surf(X1,X2,X3); 

x = (-5:0.1:5)'; % define the x samples at which the contour should be calculated 
y0 = zeros(size(x)); % initial guess 
y = fsolve(@(y) mvncdf([x y], mu, Sigma) - 0.95, y0); % solve your problem 
z = mvncdf([x y],mu,Sigma); % calculate the correspond cdf values 
hold on 
plot3(x(z>0.94), y(z>0.94), z(z>0.94), 'LineWidth', 5); % plot only the valid solutions, i.e. a solution does not exist for all sample points of x. 

爲了獲得所需輪廓的更好的數字表示，您可以對選定的y值重複上述方法。所以，你的線條會更好地填充整個圖表。

作爲替代，可以使用contour來計算輪廓上的要點如下：

figure 
[c, h] = contour(X1, X2, X3, [0.95 0.95]); 
c(3, :) = mvncdf(c',mu,Sigma); 

figure(1) 
plot3(c(1, :)', c(2, :)', c(3, :)', 'LineWidth', 5); 
xlim([-5 5]) 
ylim([-5 5]) 

這種方法的缺點是，你沒有控制權採樣輪廓的粗糙度。其次，這種方法使用（內插）3D cdf，這比用fsolve計算的值更不準確。

Answer 2

這將有助於？

clear all 
close all 
mu = [0 0]; 
Sigma = [1 0.5^0.5; 0.5^0.5 1]; 
x1 = -3:.2:3; x2 = -3:.2:3; 
[X1,X2] = meshgrid(x1,x2); 
F = mvnpdf([X1(:) X2(:)],mu,Sigma); 
F = reshape(F,length(x2),length(x1)); 
subplot(1,3,1) 
surf(x1,x2,F); axis square 

X = [X1(:) X2(:)]; 
p = mvncdf(X,mu,Sigma); 
P = reshape(p,31,31); 
subplot(1,3,2) 
surf(X1,X2,P); axis square 

subplot(1,3,3) 
P(P<0.95) = NaN; 
surf(X1,X2,P); axis square 

或與適當的軸

與輪廓替換衝浪