On Tue, Oct 30, 2012 at 3:12 PM, Surandokht Nikzad <

[hidden email]> wrote:

> I am afraid, I am not sure how to code/ compute it...

>

> On 12-10-30 5:08 PM, "Juan Pablo Carbajal" <

[hidden email]> wrote:

>

>>On Tue, Oct 30, 2012 at 3:03 PM, Surandokht Nikzad <

[hidden email]>

>>wrote:

>>> Hi there,

>>>

>>> I am beginner in Octave.

>>> I am using this for calculation kernel density for some concentrations.

>>>

>>> This is one example data:

>>>

>>> ssw ef

>>> 1.85 1.497297297

>>> 1.21 1.719008264

>>> 1.44 1.451388889

>>> 1.71 1.514619883

>>> 1.4 1.192857143

>>> 0.396 1.795454545

>>>

>>> Here is the codes:

>>>

>>> doc_ef=a(:,2);

>>> doc_ssw=a(:,1);

>>> run_doc=0:d:20;

>>> dens_est_doc_ef=kernel_dens(ones(size(run_doc)),doc_ef,run_doc);

>>> plot(run_doc,dens_est_doc_ef,'b')

>>>

>>> The integral of each distribution should be equal to one!My question is

>>>how

>>> can I avoid having plots over one in my graph. The example for graph is

>>> attached by this email.

>>>

>>> Any idea and suggestion would be appreciated.

>>>

>>> Thanks,

>>> Sue

>>>

>>>

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>>

>>As you said the integral of probability density function is one, but

>>the density function itself could be bigger then one. Think of the

>>delta distribution.

>>

http://en.wikipedia.org/wiki/Delta_distribution>>

>>If you want, you can plot the normalized frequency plot using hist

>>with the normalization argument equal to 1.

>

>

do

help hist

in your Octave prompt

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