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MathGroup Archive 2005

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Re: Integrate gives wrong answer

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57102] Re: Integrate gives wrong answer
  • From: Maxim <ab_def at prontomail.com>
  • Date: Sun, 15 May 2005 03:04:15 -0400 (EDT)
  • References: <d615eu$nnb$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On Fri, 13 May 2005 03:03:58 +0000 (UTC), Brian Rogers <brifry at gmail.com>  
wrote:

> Hi, I have a problem on which Integrate misses badly. The code consits
> of:
>
> fmin[x_, p_]:=[p(1-(x-a+e)/(2e))^(p-1)]/2e
>
> which is the pdf of the minimum of p draws from a uniform distribution
> on [a-e,a+e], and
>
> Integrate[Abs[Min[mi,mc]-Min[mj,mc]]*fmin[mc,k]*fmin[mi,ni-k]*fmin[mj,nj-k],
> {mc,a-e,a+e},{mi,a-e,a+e},{mj,a-e,a+e},Assumptions->{k\el\_Integers,
> ni\el\_Integers,nj\el\_Integers,1<=k<=Min[ni,nj],0<a-e<a+e<1}]
>
> This produces the output:
>
> \!\(k\ \((\(a +
>       e\)\/ni - \(2\ e\)\/\(1 + ni\) + \(a + e\)\/\(k - ni - nj\) +
> \(2\ \
> e\)\/\(1 - k + ni + nj\))\)\)
>
> which is not symmetric betwenn ni and nj, as it should be.
>
> Any ideas?
> Thanks!
> Brian Rogers
> brifry at gmail.com
>

This can be done using the symmetries of the integral. First, it's  
independent of a, and second, the integral over mi < mj is the same as the  
integral over mj < mi with ni and nj interchanged; also we can assume that  
ni <= nj. The condition 0 < a - e < a + e < 1 is equivalent to 0 < e < 1/2  
&& e < a < 1 - e and we can choose a == 0:

In[1]:=
fmin[x_, p_] := p*(1 - (x - a + e)/(2*e))^(p - 1)/(2*e)

Integrate[Abs[Min[mi, mc] - Min[mj, mc]]*
       fmin[mc, k]*fmin[mi, ni - k]*fmin[mj, nj - k],
     {mc, a - e, a + e},
     Assumptions -> {a == 0, 1 <= k <= ni <= nj, 0 < e < 1/2,
       a - e < mi < mj < a + e}] //
   Simplify;

Integrate[%, {mi, a - e, a + e}, {mj, mi, a + e},
   Assumptions -> {a == 0, 1 <= k <= ni <= nj, 0 < e < 1/2}];

% + (% /. {ni -> nj, nj -> ni}) // FullSimplify

Out[4]=
2*e*(1/(1 + ni) + 1/(1 + nj) - 2/(1 - k + ni + nj))

which seems to agree with the results of the numerical integration.

Generally it's a good idea to specify the version of Mathematica you're  
using; the above will work in version 5.1 only. In 5.0 you can carry out  
the first integration step with the help of PiecewiseIntegrate from  
http://library.wolfram.com/infocenter/MathSource/5117/ .

However, Mathematica 5.1 still has problems with this type of integrals:

In[5]:=
Assuming[e > 0 && p > 0,
   Integrate[x*(e - 1 - x)^p, {x, -1 - e, -1 + e}]]

Out[5]=
0

which is incorrect. Integrating on 0 < e < 1 and 1 < e separately gives  
the correct answer.

In[6]:=
Integrate[e^(k - ni - nj)*(e - mi)^(-1 - k + ni)*
       (e - mj)^(-1 + nj)*(mi - mj) +
     e^(k - ni - nj)*(e - mi)^(-1 - k + ni)*(e - mj)^(-1 - k + nj)*
       ((e - mi)^(1 + k) - (e - mj)^k),
   {mi, -e, e},
   Assumptions -> {1 < k < ni < nj, 0 < e < 1/2, -e < mj < e}]

Out[6]=
ComplexInfinity

In fact the integral converges. A trickier issue is that if we perform the  
integration steps in a different order or for different values of a, the  
result of integration may contain expressions of the form Beta[z, -ni, ni  
- 1]. For integer ni>0 (which is what we need) this is correct only as  
Limit[Beta[z, -n, n - 1], n -> ni].

Maxim Rytin
m.r at inbox.ru


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