On Tue, Oct 30, 2012 at 3:12 PM, Surandokht Nikzad <sur...@gm...> wrote:
> I am afraid, I am not sure how to code/ compute it...
>
> On 12-10-30 5:08 PM, "Juan Pablo Carbajal" <aju...@gm...> wrote:
>
>>On Tue, Oct 30, 2012 at 3:03 PM, Surandokht Nikzad <sur...@gm...>
>>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
|
