MATLAB Answers

integrating the product of two pdf's

13 views (last 30 days)
PChoppala
PChoppala on 5 Oct 2012
Hi
I have a small query, and will appreciate your help.
I am supposed to find the integral of the product of two probability density functions, and I request for advice on how to do this numerically (on matlab).
Please see below the simple code I wrote.
Pr is p(x1) i.e the pdf of x1 over a defined space. M is p(x2|x1) is the Markov pdf over the same space.
I need to evaluate int {p(x2|x1).p(x1)}dx1
can you please help me with this? The code is below
% Initialization of target state and covariance
% P(x1)
x1=-2;P=16;
Pr=1/(sqrt(2*pi*det(P))) * exp(-(space-x1).^2/(2*det(P)));
% Markov transition density: x=Fx+v,
% v is zero mean Gaussian with covariance 4, and F=-2
% P(x2|x1)
Q=4; F=-2;
x2=F*x1;
M=1/(sqrt(2*pi*det(Q))) * exp(-(space-(-2*x2)).^2/(2*det(Q)));
Thanks
P

  0 Comments

Sign in to comment.

Answers (2)

Tom Lane
Tom Lane on 10 Oct 2012
Here's an example that you may be able to adapt to your problem.
Suppose [x1;x2] have a multivariate normal distribution with mean [2;3] and variance matrix [1 .6;.6 1]. Then we know theoretically that the marginal distribution of x1 is normal(2,1), the marginal distribution of x2 is normal(3,1), and the distribution of x2 conditional on x1 is normal with mean 3+.6*(x1-2) and standard deviation sqrt(1-.6^2). Let's see if we can verify that.
% Marginal distribution of x2 computed directly at four points
normpdf(1:4,3,1)
ans =
0.0540 0.2420 0.3989 0.2420
% Same computed by integrating the marginal of x1 and conditional of x2
for j=1:4
quad(@(x)normpdf(x,2,1).*normpdf(j,3+.6*(x-2),sqrt(1-.6^2)),-10,10)
end
ans =
0.0540
ans =
0.2420
ans =
0.3989
ans =
0.2420

  0 Comments

Sign in to comment.


Tom Lane
Tom Lane on 8 Oct 2012
What is the "space" variable intended to represent?
Based on your integral expression, I'd expect the result to be a function of x2. I might expect your code to have calls to quad() with the upper bound of integration being the value of x2, and perhaps a loop over different possible x2 values.
You seem to use x1 to represent a variable in your text and a mean in your code. I probably would have expected to see x1,m1,P in the first density and maybe x2,x1,Q in the second.

  1 Comment

PChoppala
PChoppala on 8 Oct 2012
Apologies, the space=-10:.1:10;
I am now a bit confused on representations, you may please correct me. However, the random variable is the 'space' which is -10:0.1:10 over which x1 and x2 (means) operate.
Yeah, I was expecting the integral to be a function of x2 but am not able to figure out how to code it (or numerically understand it). I suppose it marginalizes out x1.

Sign in to comment.