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From: carl on 12 Feb 2010 09:18
In my matlab code I use the mvnpdf function to compute the probability for a
set of samples based on the following covariance matrix (Sigma) and mean
(mu):

mu =

0.9556 0.2994 0.2569

Sigma =

0.0082 0.0052 0.0067
0.0052 0.0171 0.0199
0.0067 0.0199 0.0241



prob = mvnpdf(X , mu, Sigma);

where X is a 1000*3 matrix containing various samples.

If I do:

max(prob)

I get:


197.3913

And if I do:

sum(prob)

I get:

1.2987e+006


I would expect that max(prob) would be 1 (since probabilites lies in the
range [0;1]) and that max(prob) would also be 1 since the integral should
sum to 1.

I assume that me problem is that I mix the term:

probability density function
http://en.wikipedia.org/wiki/Probability_density_function

with the term:

probability distribution
http://en.wikipedia.org/wiki/Discrete_probability_distribution

But maybe someone could give me a hint to better understand the difference?

From: Brian Borchers on 12 Feb 2010 10:32
A probability density function (pdf) has to be integrated over some
range of x values to obtain a probability. If the pdf is nonzero over
a narrow range it's quite easy for the maximum of the pdf to be larger
than one, even though the integral of the pdf from x=-infinity to x=
+infinity is 1.

For example, consider a random variable X that is uniformly
distributed on the interval [0,1/2]. The pdf is

f(x)=2 0<=x<=1/2.
f(x)=0 x<0 or x>1/2.

The probability that x is between 0 and 0.1 is

P(0<=x<=0.1)=int(f,x=0..0.1)=0.2.
From: Peter Perkins on 12 Feb 2010 10:33
On 2/12/2010 9:18 AM, carl wrote:
> In my matlab code I use the mvnpdf function to compute the probability
> for a set of samples based on the following covariance matrix (Sigma)
> and mean (mu):

> I would expect that max(prob) would be 1 (since probabilites lies in the
> range [0;1]) and that max(prob) would also be 1 since the integral
> should sum to 1.

The integral should _integrate_ to 1. For a discrete distribution, the individual probabilies must be less than or equal to one, because they must sum to one. But for a continuous distribution, it's an integral. Consider the uniform distribution on the interval [0,a]. It's uniform, so the density is a constant, call it c, and the integrl of the density over the interval [0,a], (a-0)*c == a*c, has to equal one. Thus the density has to be 1/a. When a is .5, say, the density is larger than one.

More towards your specific question: try plotting normpdf(linspace(-1,1,1001),0,.1) and normcdf(linspace(-1,1,1001),0,.1) and compare the two.

Probabilities have to be less than 1, Densities can be anything, even infinite (at individual points). Hope this helps.
From: carl on 12 Feb 2010 11:04

"Brian Borchers" <borchers.brian(a)gmail.com> wrote in message
news:5a9bc554-5cf9-47fd-a610-b98217017a7a(a)g28g2000prb.googlegroups.com...
>A probability density function (pdf) has to be integrated over some
> range of x values to obtain a probability. If the pdf is nonzero over
> a narrow range it's quite easy for the maximum of the pdf to be larger
> than one, even though the integral of the pdf from x=-infinity to x=
> +infinity is 1.
>
> For example, consider a random variable X that is uniformly
> distributed on the interval [0,1/2]. The pdf is
>
> f(x)=2 0<=x<=1/2.
> f(x)=0 x<0 or x>1/2.
>
> The probability that x is between 0 and 0.1 is
>
> P(0<=x<=0.1)=int(f,x=0..0.1)=0.2.

Ok I thought that mvnpdf corresponded to this expression:

http://upload.wikimedia.org/math/a/d/4/ad4c63257208b495d1084a74a15e0113.png

which in the litterature is both referred to as the multivariate gaussian
distribution, multivariate probability density and probability mass
function.

And that the plot would look like this:

http://upload.wikimedia.org/wikipedia/commons/7/74/Normal_Distribution_PDF.svg

depending on the parameters sigma and the mean. But cleary that this is not
mvnpdf. Maybe its best to implement things from scratch to understand how
they work.

From: John D'Errico on 12 Feb 2010 11:21
"carl" <carl@.com> wrote in message <4b757bff$0$283$14726298(a)news.sunsite.dk>...
>
> "Brian Borchers" <borchers.brian(a)gmail.com> wrote in message
> news:5a9bc554-5cf9-47fd-a610-b98217017a7a(a)g28g2000prb.googlegroups.com...
> >A probability density function (pdf) has to be integrated over some
> > range of x values to obtain a probability. If the pdf is nonzero over
> > a narrow range it's quite easy for the maximum of the pdf to be larger
> > than one, even though the integral of the pdf from x=-infinity to x=
> > +infinity is 1.
> >
> > For example, consider a random variable X that is uniformly
> > distributed on the interval [0,1/2]. The pdf is
> >
> > f(x)=2 0<=x<=1/2.
> > f(x)=0 x<0 or x>1/2.
> >
> > The probability that x is between 0 and 0.1 is
> >
> > P(0<=x<=0.1)=int(f,x=0..0.1)=0.2.
>
> Ok I thought that mvnpdf corresponded to this expression:
>
> http://upload.wikimedia.org/math/a/d/4/ad4c63257208b495d1084a74a15e0113.png
>
> which in the litterature is both referred to as the multivariate gaussian
> distribution, multivariate probability density and probability mass
> function.
>
> And that the plot would look like this:
>
> http://upload.wikimedia.org/wikipedia/commons/7/74/Normal_Distribution_PDF.svg
>
> depending on the parameters sigma and the mean. But cleary that this is not
> mvnpdf. Maybe its best to implement things from scratch to understand how
> they work.

It does correspond to that expression. But you need
to appreciate that that expression can easily be larger
than 1.

Only the integral must be 1.

John
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