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I have tried to Normalize the below Code PDF with a such Constant K But I didn't succeed , I have used this instruction: mv\[ScriptCapitalD]=ProbabilityDistribution[f[x,\[Mu]], {x, 0, \[Infinity]},Method->"Normalize"] , But no result .

MY cODE for PDF Here:(The integrand over sigma is about 0.99..), Any help ?

 f[z_?NumericQ, \[Mu]_?NumericQ] := 
      NIntegrate[
   K* Sqrt[Sqrt [ 2]*Pi]/2*Exp[-Sqrt[Sqrt[2]*Pi](z+\[Sigma])^2/\[Mu]*Sqrt[Pi*\[Sigma]]
    * Erf[\[Mu](z-\[Sigma])^2/(Sqrt[Pi*\[Sigma]])]], {\[Sigma],1, \[Infinity]}]
    pdfF[\[Mu]_?NumericQ] := PDF[ProbabilityDistribution[f[x,\[Mu] ], {x, 0, \[Infinity]}]]
    With[{n = 5},
      Show[
          Plot[
            Labeled[pdfF[#1][z], Row[{"\[Mu] = ", Rationalize[#1]}], Above], {z, 0, 5},
            PlotRange -> All,
            PlotStyle -> #2] &
        @@@
          Rest[{#, Hue[#]} & /@ Subdivide[0., 1., n]],
      ImageSize -> Large]]
    With[{n = 5}, NIntegrate[pdfF[#][z], {z, 0, 5}] & /@ (Range[n]/n)]

    Plot[Evaluate[f[z_?NumericQ, \[Mu]_?NumericQ] ], {x, 0, 10},{\[Mu],0,1}, AxesOrigin \[RightArrow] {0, 0},
    Epilog \[RightArrow] {Dashed, Line[{{0, f[z_?NumericQ, \[Mu]_?NumericQ]}, {10, f[z_?NumericQ, \[Mu]_?NumericQ]}}]}, PlotRange \[RightArrow] {{0, 10}, {0, 1}}]

This is the plot that i have got for n=5 enter image description here

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4
  • $\begingroup$ Please define K and trim the unnecessary code. Why does z only go from 0 to 5? Just executing Plot[f[x, 4], {x, 0, 20}] (with K=1) shows some oddities. $\endgroup$
    – JimB
    Commented Dec 26, 2019 at 20:33
  • $\begingroup$ @JimB ,This is what I have used for normalization but it doesn't work:mv[ScriptCapitalD] = ProbabilityDistribution[ kSqrt[10Sqrt [ 2]*Pi]/(2*Pi)*Exp[-PiSqrt[2*[Sigma]](z+[Sigma])^2 * Erf[[Mu](z-[Sigma])^2/Sqrt[Sqrt[2]*Pi*[Sigma]] ]/[Mu],{[Sigma], 1,Infinity}], {z, 0,5},{[Mu], 0,1} ,{k,1,Infinity}, Method -> "Normalize"] $\endgroup$
    – zeraoulia rafik
    Commented Dec 26, 2019 at 22:19
  • $\begingroup$ I think the issue is with the limits used for NIntegrate. Using 1 through $\infty$ doesn't allow Nintegrate to work well. I'll add an example as an extended comment. $\endgroup$
    – JimB
    Commented Dec 26, 2019 at 22:26
  • $\begingroup$ This is my PDF formula :$$\int_1^{\infty } \frac{\sqrt{10 \sqrt{2} \pi } \exp \left(-\frac{\pi \sqrt{2 \sigma } (z+\sigma )^2 \text{erf}\left(\frac{\mu (z-\sigma )^2}{\sqrt{\sqrt{2} \pi \sigma }}\right)}{\mu }\right)}{0.997266 \times2 \pi } \, d\sigma$$ $\endgroup$
    – zeraoulia rafik
    Commented Dec 26, 2019 at 22:36

1 Answer 1

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This is an extended comment.

I think the issue is that using {σ, 1, ∞} for the limits of integration doesn't allow NIntegrate to work well. (Yes, those are what the limits should be.) Here's an example.

integrand[z_, μ_, σ_] := Sqrt[Sqrt[2]*Pi]/2*
  Exp[-Sqrt[Sqrt[2]*Pi] (z + σ)^2/μ*Sqrt[Pi*σ]*Erf[μ (z - σ)^2/(Sqrt[Pi*σ])]]

Plot[integrand[4.75, 4, σ], {σ, 1, 6}, PlotRange -> All]

Integrand over sigma

Now if we integrate over two different limits of integration:

NIntegrate[integrand[4.75, 4, σ], {σ, 1, ∞}]
(* 1.5504*10^-15 )*

NIntegrate[integrand[4.75, 4, σ], {σ, 4.5, 5.1}]
(* 0.127522 *)

The latter result makes more sense. So when determining K one will need to adjust the limits of integration for different values of z and μ. However, I don't have an idea as to automate that process with reasonable limits of integration.

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  • $\begingroup$ Do you meant that can't be a PDF over the range of sigma ? , I have did the same Idea , the integral dosn't work as well for example in the Range (10,25) $\endgroup$
    – zeraoulia rafik
    Commented Dec 26, 2019 at 22:41
  • 1
    $\begingroup$ No. I'm just saying that one needs to be careful with assigning limits of integration with this function. Just using {\[sigma],1,Infinity} many times misses where the integrand is most positive and one ends up with an erroneous value of near zero. It might very well be that the integral over $z$ (not $\sigma$) is infinite. But you need to fix the issue with NIntegrate first. $\endgroup$
    – JimB
    Commented Dec 26, 2019 at 22:44
  • $\begingroup$ The value of Mu are in (0,1) $\endgroup$
    – zeraoulia rafik
    Commented Dec 27, 2019 at 19:00

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