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Re: piecewice pdf, problems with cdf

  • To: mathgroup at smc.vnet.net
  • Subject: [mg105372] Re: [mg105365] piecewice pdf, problems with cdf
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Mon, 30 Nov 2009 06:09:52 -0500 (EST)
  • References: <200911291012.FAA16385@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

I really meant to post this solution:

Clear[x, f1, cdf]
f1[x_] = Piecewise[{{x^2/9, 0 <= x <= 3}}];
cdf[x_?Negative] = 0;
cdf[x_] /; x >= 3 = 1;
cdf[x_] = Assuming[0 < x < 3, Integrate[f1@v, {v, -Infinity, x}]]
Plot[cdf@x, {x, -5, 5}, PlotRange -> All]

If there are many pieces, UnitStep is your friend. For instance, here's a  
function that changes at -2, +3, +5, and +10, where I start with something  
arbitrary and normalize by dividing by the integral:

Clear[f, tmp, x, cdf]
tmp[x_] =
   UnitStep[x + 2] Sin@x + UnitStep[x - 3] Cos@x +
    UnitStep[x - 5] Exp@x - UnitStep[x - 10] Exp@x;
f[x_] = tmp@x/Integrate[tmp@v, {v, 0, 10}]
cdf[x_] = Integrate[f@v, {v, 0, x}]

Integrate is a SYMBOLIC solver, so it's inherently limited. It doesn't  
understand functions built with If.

Bobby

Plot[cdf@x, {x, -3, 11}, PlotRange -> All]

On Sun, 29 Nov 2009 20:07:42 -0600, michael partensky  
<partensky at gmail.com> wrote:

> Thanks, Bobby.
>
> Two questions though.
> (1) Why some other definitions  would not work with Integrate.
> I know the derivative has an issue with Delta function, but what's the
> problem with the  Integrate given that all values of the integrand (pdf)  
> are
> defined.
> At least, it should work if each piece is handled individually.
> (2) What if we  have many pieces? Should we enter all them via "Assuming"
> (which could be inconvenient). After all, the assumptions reproduce  
> x-ranges
> that are already in  the function definition. Can we directly  Integrate  
> pdf
> to produce  a piecewise CDF function that will properly return for any  
> value
> of x (automatically locating it in a proper piece) ?
>
> Thanks
> Michael
>
>
>
> On Sun, Nov 29, 2009 at 7:49 PM, DrMajorBob <btreat1 at austin.rr.com>  
> wrote:
>
>> If it's a piecewise function... use Piecewise!!
>>
>> Clear[x, f1, cdf]
>> f1[x_] = Piecewise[{{x^2/9, 0 <= x <= 3}}];
>> cdf[x_] =
>>  Piecewise[{{Assuming[0 < x <= 3,
>>      Integrate[f1@v, {v, -Infinity, x}]], 0 < x <= 3}}];
>>
>> Plot[cdf@x, {x, -5, 5}]
>>
>> Bobby
>>
>>
>> On Sun, 29 Nov 2009 04:12:35 -0600, michael partensky  
>> <partensky at gmail.com>
>> wrote:
>>
>>  Hi! Teaching the continuous distributions, I needed to introduce the
>>> piecewise functions.
>>> Here is the example that did not work well:
>>>
>>> In[56]:= f1[x_] /; 0 < x <= 3 := 1/9  x ^2;
>>> f1[x_] := 0;
>>>
>>> Plot[f1[x],{x,-1,4}] works fine. However, the results for cdf are
>>> ambiguous
>>> In[57]:= cdf[x_] := Integrate[f1[v], {v, -\[Infinity], x}]
>>>
>>> In[59]:= cdf[1]
>>> Out[59]= 0
>>>
>>> I thought that may be the second definition (for some reason) overwrote
>>> the
>>> first, but apparently this was not the case.
>>>
>>> Then I tried using Which,
>>>
>>> f1[x_] := Which[0 < x <= 3, x^2/9, x <= 0 || x > 3, 0];
>>>
>>> Plot[f2[x], {x, -1, 4}] worked fine.
>>>
>>> However, Plotting CDF is very slow.
>>>
>>> What is the reason for the first error and how to accelerate (compile?)
>>>  the
>>> second?
>>>
>>> Thanks
>>> Michael
>>>
>>> PS: I was aware about the issues with the derivatives of Piecewise
>>> functions, but expected  integration to be safe. What did i do wrong?
>>>
>>> On Tue, Nov 24, 2009 at 5:50 AM, Bill Rowe <readnews at sbcglobal.net>
>>> wrote:
>>>
>>>  On 11/23/09 at 6:53 AM, lshifr at gmail.com (Leonid Shifrin) wrote:
>>>>
>>>> >Hi, Michael.
>>>>
>>>> >Your solution is indeed very memory -  hungry due to the mapping of
>>>> >Permutations, as you mentioned. The total number of permutations can
>>>> >be easily deduced from the list of multiplicities of elements in a
>>>> >given partition: n!/(n1!n2!...nk!), where n1, ..., nk are
>>>> >multiplicities of elements, and n is the length of the partition:
>>>> >n=n1+...+nk. The multiplicities can be obtained by Tally. The
>>>> >following modification can be realistically used in a much wider
>>>> >region of the problem's parameter space than your original one, and
>>>> >may possibly serve your needs.
>>>>
>>>> >In[1]:=
>>>> >Clear[outsNew];
>>>> >outsNew[sum_, thr_] :=
>>>> >Total[Factorial[Length[#]]/
>>>> >Times @@ Factorial[Tally[#][[All, 2]]] & /@
>>>> >Cases[IntegerPartitions[sum, thr, {1, 2, 3, 4, 5, 6}],
>>>> >Table[_, {thr}]]];
>>>>
>>>> The above can be made more efficient by using the form of
>>>> IntegerPartitions that only generates partitions of a given
>>>> length. That is
>>>>
>>>> outs[sum_, thr_] :=
>>>>   Total[Factorial[Length[#]]/
>>>>      Times @@ Factorial[Tally[#][[All, 2]]] & /@
>>>>     IntegerPartitions[sum, {thr}, {1, 2, 3, 4, 5, 6}]]
>>>>
>>>> does the same thing a bit more efficiently
>>>>
>>>>
>>>>
>>>>
>>>
>>
>> --
>> DrMajorBob at yahoo.com
>>


-- 
DrMajorBob at yahoo.com


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