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