Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2001
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2001

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: a couple of gripes

  • To: mathgroup at smc.vnet.net
  • Subject: [mg29765] Re: [mg29751] a couple of gripes
  • From: BobHanlon at aol.com
  • Date: Sun, 8 Jul 2001 01:00:09 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 2001/7/7 2:37:22 AM, J.A.Solomon at city.ac.uk writes:

>When I evaluate 
>Integrate[f[x]*If[x==y,1,0],{x,-Infinity,Infinity}]
>I don't get 
>f[y].
>
>When I evaluate 
>PDF[NormalDistribution[x,0],x]
>I don't get 
>1.
>

$Version

"4.1 for Power Macintosh (November 2, 2000)"

If[x == y, 1, 0] is equivalent to DiracDelta[x - y]

Integrate[f[x]*DiracDelta[x-y], {x, -Infinity, Infinity}]

f[y]

Needs["Statistics`NormalDistribution`"];

The PDF of the normal distribution evaluated at the mean is

PDF[NormalDistribution[mu, sigma], mu]

1/(Sqrt[2*Pi]*sigma)

As the standard deviation, sigma, approaches zero, the PDF evaluated at the 
mean 
will become arbitrarily large, not 1.  Presumably, you are trying to evaluate 
the 
CDF at the mean+ as the standard deviation approaches zero.  

CDF[NormalDistribution[mu, 10^(-n)], mu+10^(1-n)]

(1/2)*(1 + Erf[5*Sqrt[2]])

%//N

1.

For arbitrarily large n this is the case with the standard deviation 
arbitrarily small 
and the CDF evaluated arbitrarily close to the mean.


Bob Hanlon
Chantilly, VA  USA


  • Prev by Date: Re: a couple of gripes
  • Next by Date: Re: Partitioning 1-D list into 2-D Matrix
  • Previous by thread: Re: a couple of gripes
  • Next by thread: Re: a couple of gripes