# I want to estimate bivariate normal distribution and having trouble estimating it. How can i make it possible?

3 views (last 30 days)

Show older comments

i want to estimate nomal probability density function of 2 variables (i.e. bivariate normal distribution ). i used this formula from Wikipedia:

my variables 'x' and 'y' are:

I write this code:

Covariance_matrix = cov(x,y); % estimated ro,sigma1 and sigma2 using this

ro = -0.0717;

mu1 = 0.1068;

mu2 = 0.8556;

sigma1 = 1.3250e-04;

sigma2 = 9.1330e-04;

sd1 = 0.0115;

sd2 = 0.0302;

Pdf = 1/(2*pi*sd1*sd2*sqrt(1-(ro^2))) * exp(-1/2*(1-ro^2)) * ((((x-mu1)/sigma1).^2) + (((y-mu2)/sigma2).^2)) - (2*ro*((x-mu1)*(y-mu2)/sd1*sd2));

My result is:

first of all i want to know am i going write and my second question is that i learnt probability ranges between 0 and 1 but in this case probability density values are very high so how is it possible? Please help.

##### 0 Comments

### Answers (1)

Image Analyst
on 10 Oct 2014

You tried to do too much in one line and messed up the parentheses. Split it apart into separate terms so that doesn't happen:

clc; % Clear the command window.

close all; % Close all figures (except those of imtool.)

imtool close all; % Close all imtool figures.

clear; % Erase all existing variables.

workspace; % Make sure the workspace panel is showing.

fontSize = 20;

format compact;

xl = linspace(-2, 2, 30);

yl = linspace(-2, 2, 30);

[x,y] = meshgrid(xl,yl);

Covariance_matrix = cov(x,y) % estimated ro,sigma1 and sigma2 using this

rho = -0.0717;

mu1 = 0.1068;

mu2 = 0.8556;

sigmaX = .51;

sigmaY = .91;

amplitude = 1 / (2 * pi* sigmaX * sigmaY * sqrt(1-rho^2))

factor = -1 / (2 * (1-rho^2))

term1 = ((x - mu1) / sigmaX) .^ 2;

term2 = ((y - mu2) / sigmaY) .^ 2;

term3 = 2 * rho * (x-mu1) * (y-mu2) / (sigmaX * sigmaY);

Pdf = amplitude * exp(factor * (term1 + term2 - term3));

surf(Pdf)

##### 4 Comments

PRAKASH JOSHI
on 21 Jul 2022

Image Analyst
on 21 Jul 2022

@PRAKASH JOSHI I'd probably take the result above and multiply it by a Rotation Matrix:

### See Also

### Categories

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!