Re: Laplace Distribution
- To: mathgroup@smc.vnet.net
- Subject: [mg11962] Re: [mg11888] Laplace Distribution
- From: Bob Hanlon <BobHanlon@aol.com>
- Date: Fri, 10 Apr 1998 01:03:50 -0400
In a message dated 4/3/98 6:05:23 AM, tie@cscn.com wrote:
>I am working on Lapalace Distribution with pdf f(x) = (2*beta)^(-1) *
>exp(-(absx)/beta). variance of this distribution is 2*(beta^2). so if
>we take beta=1/sqrt(2), we have laplace dist'n with unit variance. how
>can i write a Mathematica function for this density and how can i plot
>the theoretical q-q plot of laplace against Normal.
>
>note. it has to integrate to one.
Needs["Statistics`ContinuousDistributions`"];
Needs["Statistics`DescriptiveStatistics`"];
dist = LaplaceDistribution[m, b]; PDF[dist, x]
E^(-((-m + x)*Sign[-m + x])/(2*b^2))
This PDF is in error.
D[CDF[dist, x] /. Sign[x - m] -> 1, x]
1/(2*b*E^((-m + x)/b))
D[CDF[dist, x] /. Sign[x - m] -> -1, x]
E^((-m + x)/b)/(2*b)
The PDF should be
Exp[-((Sign[x - m]*(x - m))/b)]/(2*b)
Since you are interested in the case of zero mean and unit variance,
then the PDF is
Exp[-Sign[x - m]*(x - m)/b]/(2*b) /. {m -> 0, b -> 1/Sqrt[2]}
1/(Sqrt[2]*E^(Sqrt[2]*x*Sign[x]))
dist01 = LaplaceDistribution[0, 1/Sqrt[2]]; f[x_] :=
Exp[-Sign[x]*Sqrt[2]*x]/Sqrt[2];
Integrating the PDF over the domain should give unity:
Integrate[f[x] /. Sign[x] -> -1, {x, -Infinity, 0}] +
Integrate[f[x] /. Sign[x] -> 1, {x, 0, Infinity}]
1
Plot[{f[x], CDF[dist01, x]}, {x, -3, 3}];
The mean is
Integrate[x*f[x] /. Sign[x] -> -1, {x, -Infinity, 0}] +
Integrate[x*f[x] /. Sign[x] -> 1, {x, 0, Infinity}]
0
With a zero mean, the variance is equal to the mean square
Integrate[x^2*f[x] /. Sign[x] -> -1, {x, -Infinity, 0}] +
Integrate[x^2*f[x] /. Sign[x] -> 1, {x, 0, Infinity}]
1
Quantile is the inverse of the CDF.
Quantile[dist01, q]
-((Log[1 - (-1 + 2*q)*Sign[-1 + 2*q]]*Sign[-1 + 2*q])/Sqrt[2])
Plot[Quantile[dist01, q], {q, 0.1, 0.9}];
Plot[Quantile[dist01, CDF[dist01, x]], {x, -5, 5}];
Plot[CDF[dist01, Quantile[dist01, q]], {q, 0, 1}];
distNorm = NormalDistribution[0, 1]; PDF[distNorm, x]
1/(E^(x^2/2)*Sqrt[2*Pi])
Quantile[distNorm, q]
Sqrt[2]*InverseErf[0, -1 + 2*q]
I do not know what "the theoretical q-q plot of laplace against Normal"
means; however, it may be something like one of the following
Plot[CDF[distNorm, Quantile[dist01, q]], {q, 0, 1}];
Plot[CDF[dist01, Quantile[distNorm, q]], {q, 0, 1}];
Bob Hanlon