Re: piecewice pdf, problems with cdf

*To*: mathgroup at smc.vnet.net*Subject*: [mg105373] Re: [mg105365] piecewice pdf, problems with cdf*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Mon, 30 Nov 2009 06:10:05 -0500 (EST)*References*: <200911291012.FAA16385@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

I didn't normalize that correctly and it's a poor example when plotted, so here's a slightly different example: Clear[f, tmp, x, cdf] tmp[x_] = Sqrt[2] + UnitStep[x + 2] Cos@x + UnitStep[x - 3] Sin@x + UnitStep[x - 5] Exp[x - 7] - UnitStep[x - 10] Exp[x - 7]; f[x_] = tmp@x/Integrate[tmp@v, {v, -2, 10}]; cdf[x_] = Integrate[f@v, {v, -3, x}]; Plot[{f@x, cdf@x}, {x, -3, 11}, PlotRange -> All, AxesOrigin -> {0, 0}] This won't do for every problem, however. For instance: Clear[f, tmp, x, cdf] tmp[x_] = Sqrt[2] + UnitStep[x + 2] Cos@x + UnitStep[x - 3] Sin@x + UnitStep[x - 5] Log@x - UnitStep[x - 10] Log@x; f[x_] = tmp@x/Integrate[tmp@v, {v, -2, 10}] Plot[f@v, {v, -3, 10}] (Sqrt[2] - Log[x] UnitStep[-10 + x] + Log[x] UnitStep[-5 + x] + Sin[x] UnitStep[-3 + x] + Cos[x] UnitStep[2 + x])/(-5 + 12 Sqrt[2] + Cos[3] - Cos[10] + 5 Log[20] + Sin[2] + Sin[10]) So far so good, but Integrate can't do the next step! You could use NIntegrate in theory, but it's too slow in a Plot, as in your original code. NDSolve does the trick instead: Clear[cdf] cdf[x_] /; x <= -2 = 0; cdf[x_] /; x >= 10 = 1; cdf[x_] = g[x] /. First@NDSolve[{g'[x] == f@x, g[-2] == 0}, g[x], {x, -2, 10}]; Plot[{f@x, cdf@x}, {x, -2, 10}, PlotRange -> All, AxesOrigin -> {0, 0}] If the first Integrate (used to normalize "tmp") happens to fail, you can try NIntegrate, since it's only calculated once. NIntegrate[tmp@v, {v, -2, 10}] 27.1636 Bobby On Sun, 29 Nov 2009 21:32:04 -0600, DrMajorBob <btreat1 at austin.rr.com> wrote: > 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>