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

**References**:**piecewice pdf, problems with cdf***From:*michael partensky <partensky@gmail.com>