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Topic: How to combine an If statement with Statistics
Replies: 1   Last Post: Jul 18, 2013 2:44 AM

 Bob Hanlon Posts: 906 Registered: 10/29/11
Re: How to combine an If statement with Statistics
Posted: Jul 18, 2013 2:44 AM

off = 140;(*Expected Number of Off Counts*)
=E1 = .3;(*Ratio of On/Off Counts*)
Non = RandomVariate[PoissonDistribution[=E1*off], 100000];
Noff = RandomVariate[PoissonDistribution[off], 100000];

(*Normal Distribution*)
h2 = Plot[
PDF[NormalDistribution[], x],
{x, -6, 6},
PlotStyle -> Directive[Red, Thick]];

Attributes /@ {Greater, Sign}

Non - =E1*Noff is a list and is neither positive nor negative. Note that
Greater is not Listable but Sign is Listable.

Clear[S2];
S2[off_, =E1_] =
Sign[Non - =E1*Noff]*
Sqrt[2] (Non*Log[(1 + =E1)/=E1 (Non/(Non + Noff))] +
Noff*Log[(1 + =E1) (Noff/(Non + Noff))])^(1/2);

ListPlot[S2[off, =E1],
Frame -> True,
Axes -> False,
PlotLabel -> "Formula 17"]

hist2 = Histogram[S2[off, =E1],
"Log",
"ProbabilityDensity",
PlotLabel -> "Formula 17 LOG"]

hist22 = Histogram[S2[off, =E1],
Automatic,
"ProbabilityDensity",
PlotLabel -> "Formula 17"];

Show[hist22, h2]

ProbabilityScalePlot[S2[off, =E1],
"Normal",
PlotLabel -> "Formula 17"]

Bob Hanlon

On Wed, Jul 17, 2013 at 1:50 AM, William Duhe <wjduhe@loyno.edu> wrote:

> Bellow is a code which attempts to change the sign (+/-) of S with an if
> statement. S1 is shown and S2 is the attempt to include the if statement.
>
> You can see from the information yielded from S1 that the Gaussian fit is
> off by roughly a factor of 2 when it should in theory match. This is
> because it is overestimating positive events due to the lack of the sign
> change. The sign change should change with the quantity I have assigned t=

he
> label "Signal". When the Signal is positive S2 should be positive and whe=
n
> Signal is negative S2 should become negative.
>
>
> (*VARIABLES*)
>
> Non = RandomVariate[PoissonDistribution[\[Alpha]*off], 100000];
> Noff = RandomVariate[PoissonDistribution[off], 100000];
> off = 140; (*Expected Number of Off Counts*)
> \[Alpha] = .3; (*Ratio of On/Off Counts*)
>
> (*Normal Distribution*)
> h2 = Plot[
> Evaluate@
> Table[PDF[NormalDistribution[0, \[Sigma]], x], {\[Sigma],
> 1}], {x, -6, 6}, PlotStyle -> Red];
>
> (*Signal*)
> S[off_, \[Alpha]_] = Non - \[Alpha]*Noff;
> hist = Histogram[S[off, \[Alpha]], Automatic, "ProbabilityDensity",
> PlotLabel -> "Signal"]
>
>
>
> (*Formula WITHOUT IF STATEMENT*)
> S1[off_, \[Alpha]_] = .5 Sqrt[
> 2] (Non*Log[(1 + \[Alpha])/\[Alpha] (Non/(Non + Noff))] +
> Noff*Log[(1 + \[Alpha]) (Noff/(Non + Noff))])^(1/2);
> ListPlot[S1[off, \[Alpha]], PlotLabel -> "Formula 17",
> PlotRange -> {{0, 100000}, {0, 8}}]
> hist1 = Histogram[S1[off, \[Alpha]], "Log", "ProbabilityDensity",
> PlotLabel -> "Formula 17 LOG"]
> hist11 = Histogram[S1[off, \[Alpha]], Automatic, "ProbabilityDensity",
> PlotLabel -> "Formula 17"]
> Show[hist11, h2]
> ProbabilityScalePlot[S1[off, \[Alpha]], "Normal",
> PlotLabel -> "Formula 17"]
>
>
>
>
> (*Formula WITH IF STATEMENT*)
> S2[off_, \[Alpha]_] =
> If[Non - \[Alpha]*Noff > 0,
> Sqrt[2] (Non*Log[(1 + \[Alpha])/\[Alpha] (Non/(Non + Noff))] +
> Noff*Log[(1 + \[Alpha]) (Noff/(Non + Noff))])^(
> 1/2), -Sqrt[
> 2] (Non*Log[(1 + \[Alpha])/\[Alpha] (Non/(Non + Noff))] +
> Noff*Log[(1 + \[Alpha]) (Noff/(Non + Noff))])^(1/2)];
>
> ListPlot[S2[off, \[Alpha]], PlotLabel -> "Formula 17"]
> hist2 = Histogram[S2[off, \[Alpha]], "Log", "ProbabilityDensity",
> PlotLabel -> "Formula 17 LOG"]
> hist22 = Histogram[S2[off, \[Alpha]], Automatic, "ProbabilityDensity",
> PlotLabel -> "Formula 17"]
> Show[hist22, h2]
> ProbabilityScalePlot[S2[off, \[Alpha]], "Normal",
> PlotLabel -> "Formula 17"]
>
>