Re: Cumulative distribution of Gauss

• To: mathgroup at smc.vnet.net
• Subject: [mg18762] Re: [mg18639] Cumulative distribution of Gauss
• From: "Will Cooper" <wcooper1 at san.rr.com>
• Date: Tue, 20 Jul 1999 01:33:25 -0400
• Organization: SouthWest Cable of San Diego, CA
• References: <7mp30t\$l8p@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Yes, there is! See the paper by M.K. Simon referenced below where he shows
how the CDF of a Gaussian, i.e.

Gaussian Q function , Q(x) =1/(2*pi)* Integral(exp(-x^2/2)) dx, x, infinity

can be represented by:

Q(x) = 1/pi * Integral(exp(-x^2/(2*(sin(Theta))^2)) dTheta, x, pi/2

Ref: Proceedings of the IEEE, Vol 86, No9, Sep 1998 " A Unified Approach to
the Performance Analysis of Digital Communications of Generalized Fading
Channels", (Eqns.1 & 2, pp1863-1864).

Hope this helps,

Will Cooper.

Tomas Garza wrote in message <7mp30t\$l8p at smc.vnet.net>...
> D. Sallier (sallier_daniel at wanadoo.fr) wrote:
>
>> Is there any analytical expression of the cumulative distribution
>> of Gauss ?
>> If not, can someone provided me with a fast executing algorithm. I am
>> currently using pre-tabulated data. It is fine but its ends up to
>> be rather
>> time consuming for highly iterative calculations.
>
>Why not try the add-on Statistics package? It looks as if the algorithm for
>Gauss cumulative distribution is fast enough. For example, I obtain 10,000
>values in less than 5 seconds:
>
>In[1]:=
>Needs["Statistics`ContinuousDistributions`"]
>In[2]:=
>dist = NormalDistribution[0, 1]
>Out[2]=
>NormalDistribution[0, 1]
>In[3]:=
>Table[CDF[dist, Random[Real, {-3, 3}]], {10000}]; // Timing
>Out[3]=
>{4.89 Second, Null}
>
>

```

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