[Date Index]
[Thread Index]
[Author Index]
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
Prev by Date:
**Re: DSolve and K$31, i am puzzled**
Next by Date:
**Re: bode diagram**
Previous by thread:
**Integrate gives wrong answer**
Next by thread:
**Re: Integrate gives wrong answer**
| |