Re: piecewice pdf, problems with cdf

*To*: mathgroup at smc.vnet.net*Subject*: [mg105374] Re: [mg105365] piecewice pdf, problems with cdf*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Mon, 30 Nov 2009 06:10:16 -0500 (EST)*References*: <200911291012.FAA16385@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

Substitute this in the second example: Clear[f, tmp, x, cdf] tmp[x_] = UnitStep[x + 2] (Sqrt[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}] I'm just trying to make it zero before -2 and always non-negative. Bobby On Sun, 29 Nov 2009 23:20:53 -0600, DrMajorBob <btreat1 at austin.rr.com> wrote: > 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>

**Re: piecewice pdf, problems with cdf**

**Re: piecewice pdf, problems with cdf**

**Re: piecewice pdf, problems with cdf**