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Conditioning bivariate gaussian distribution

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Hi,
I have a bivariate normal distribution as follows f(x):
m = 0;
c = [0.5 0.8; 0.8 2.0];
x1 = -4:0.2:4;
x2 = -4:0.2:4;
[X1, X2] = meshgrid(x1,x2);
X = X1(:)';
Y = X2(:)';
fun = @(X, Y) 1/(2*pi*(det(c))^(0.5))* exp(-(0.5)*sum(([X; Y]-m).*(inv(c)*([X; Y]-m))));
i = @(X)integral(@(Y)fun(X,Y),-inf,inf,'ArrayValued',true);
fplot(i)
And this gives output:
hw2q7.jpg
I want to find the f(x / y = 1.5)
I have tried to find f(x) and f(y) and then filter out y = 1.5 to get the x values from the distribution. but this method is not working and giving errors as follows:
mu_x = 0;
c = [0.5 0.8; 0.8 2.0];
x1 = -4:0.2:4;
x2 = -4:0.2:4;
[X1,X2] = meshgrid(x1,x2);
X = X1(:)';
Y = X2(:)';
fun = @(X, Y) 1/(2*pi*(det(c))^(0.5))* exp(-(0.5)*sum(([X; Y]-mu_x).*(inv(c)*([X; Y]-mu_x))));
px = @(X)integral(@(Y)fun(X,Y),-inf,inf,'ArrayValued',true);
py = @(Y)integral(@(X)fun(X,Y),-inf,inf,'ArrayValued',true);
%vq1 = interp1(-3:0.2:3,px,-3:0.2:3)
px = px([-3:0.2:3])
p = [px([-3:0.2:3]); py([-3:0.2:3])]
fplot(py)
The error :
Error using vertcat
Dimensions of arrays being concatenated are not consistent.
Error in pg2>@(X,Y)1/(2*pi*(det(c))^(0.5))*exp(-(1/2)*sum(([X;Y]-mu_x).*(inv(c)*([X;Y]-mu_x)))) (line 109)
fun = @(X, Y) 1/(2*pi*(det(c))^(0.5))* exp(-(1/2)*sum(([X; Y]-mu_x).*(inv(c)*([X; Y]-mu_x))));
Error in pg2>@(Y)fun(X,Y) (line 110)
px = @(X)integral(@(Y)fun(X,Y),-inf,inf,'ArrayValued',true);
Error in integralCalc/iterateArrayValued (line 156)
fxj = FUN(t(1)).*w(1);
Error in integralCalc/vadapt (line 130)
[q,errbnd] = iterateArrayValued(u,tinterval,pathlen);
Error in integralCalc (line 103)
[q,errbnd] = vadapt(@minusInfToInfInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in pg2>@(X)integral(@(Y)fun(X,Y),-inf,inf,'ArrayValued',true) (line 110)
px = @(X)integral(@(Y)fun(X,Y),-inf,inf,'ArrayValued',true);
Error in pg2>dist (line 113)
px = px([-4:0.2:4])
How do i get the values f(x) and f(y) from px and py in range -4:0.2:4 so that i can find f(x / y = 1.5)?

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Accepted Answer

Bruno Luong
Bruno Luong on 15 Sep 2019
Edited: Bruno Luong on 15 Sep 2019
There a few issues with your code.
First INTEGRAL is integration of a scalar function, you cannot integrate a vector function. So in 1D you need to loop in the parameters (x) or (y).
Second parameters ans variable are no longer have the same length, you can not do vercat them.
Here is a modified code
mu_x = 0;
c = [0.5 0.8; 0.8 2.0];
% this block is irrelevant!!!
% x1 = -4:0.2:4;
% x2 = -4:0.2:4;
% [X1,X2] = meshgrid(x1,x2);
% X = X1(:)';
% Y = X2(:)';
fun = @(X, Y) gfun(X,Y,mu_x,c); % nested defined bellow
px = @(X) arrayfun(@(x) integral(@(y)fun(x,y),-inf,inf,'ArrayValued',true), X);
py = @(Y) arrayfun(@(y) integral(@(x)fun(x,y),-inf,inf,'ArrayValued',true), Y);
Pxy = [px([-3:0.2:3]); py([-3:0.2:3])]
close all
plot(Pxy')
%%
function f = gfun(X,Y,mu,c)
[X,Y] = ndgrid(X(:)',Y(:)'); % expanding the scalar to match the vector, regardless which is which
XY = [X; Y];
f = 1/(2*pi*(det(c))^(0.5))* exp(-(0.5)*sum((XY-mu).*(c\(XY-mu))));
end
gaussian.png

  6 Comments

Show 3 older comments
Aishwarya Radhakrishnan
Aishwarya Radhakrishnan on 16 Sep 2019
Hi Bruno,
I have used the above methods and tried to find the conditional independence. Is this the correct procedure?
function [] = BivariateGaussianDistributions_ConditionalProbability()
fig = figure('Name','Marginalize','NumberTitle','off');
mu = 0;
c = [0.5 0.8; 0.8 2.0];
fun = @(X, Y) distfun(X,Y,mu,c); % nested defined bellow
py = @(Y) arrayfun(@(y) integral(@(x)fun(x,y),-inf,inf,'ArrayValued',true), Y);
Py2 = py([1.5]); % P(y=1.5)
[X1,X2] = meshgrid((1.5).*[-4:0.02:4],[-4:0.02:4]); % x = 1.5*y
X = [X1(:) X2(:)]';
PXY = 1/(2*pi*(det(c))^(0.5))* exp(-(1/2)*sum((X-mu).*(pinv(c)*(X-mu))));
PXY = reshape(PXY,size(X1)); % P(X,Y = 2)
PX_Y2 = PXY / Py2; % P(X | Y = 2) = P(X,Y = 2) / P(y=1.5)
plot(X1,PX_Y2)
title('Approximate Univariate Normal Distribution');
xlabel('x');
ylabel('Approximate p(x)');
end
function f = distfun(X,Y,mu,c)
[X,Y] = ndgrid(X(:)',Y(:)'); % expanding the scalar to match the vector, regardless which is which
XY = [X; Y];
f = 1/(2*pi*(det(c))^(0.5))* exp(-(0.5)*sum((XY-mu).*(c\(XY-mu))));
end
For this i'm getting the following curve:
st.jpg
Bruno Luong
Bruno Luong on 17 Sep 2019
No idea what you compute in your script and what want to plot.
The condition probability is simply the PDF projected on the line
m = 0;
c = [0.5 0.8; 0.8 2.0];
fun = @(X, Y) 1/(2*pi*(det(c))^(0.5))* exp(-(0.5)*sum(([X; Y]-m).*(c\([X; Y]-m))));
fun_condi_x = @(x1) fun(x1,1.5*x1);
x1_c = -4:0.05:4;
close all
plot(x1_c,fun_condi_x(x1_c)/integral(fun_condi_x,-Inf,Inf));
xlabel('x1');
ylabel('PDF of f(x1,x2) conditional x2/x1=1.5')
x1 = -4:0.05:4;
x2 = -4:0.05:4;
[X1, X2] = meshgrid(x1,x2);
Z = fun(X1(:)',X2(:)');
Z = reshape(Z,size(X1));
% First figure plot is the restriction of Z on the white line
% (normalized so that the integral == 1, since it's a PDF if we consider x1 as
% an independent parametrization variable.)
figure
imagesc(x1,x2,Z);
set(gca,'Ydir','normal')
hold on
x2_c = 1.5*x1_c;
plot(x1_c,x2_c,'w','linewidth',2);
g1.png
g2.png

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