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Re: Area Under Curve (Min Length Interval)
*To*: mathgroup at smc.vnet.net
*Subject*: [mg53902] Re: Area Under Curve (Min Length Interval)
*From*: "Ray Koopman" <koopman at sfu.ca>
*Date*: Wed, 2 Feb 2005 18:10:44 -0500 (EST)
*References*: <ctqdm2$sd3$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
The minimum-length interval has equal densities at its endpoints.
Needs["Statistics`"]
F[x_] := CDF[ChiSquareDistribution[5],x]
f[x_] := PDF[ChiSquareDistribution[5],x]
FindRoot[{F[b]-F[a]==.93,f[b]==f[a]},{{a,1},{b,10}}]
{F[b]-F[a],f[b]-f[a]}/.%
{a->0.37253, b->10.3441}
{0.93, 3.46945*^-18}
Bruce Colletti wrote:
> Re Mathematica 5.1.
>
> How would I compute the minimum length interval over which the area
under f(x) is given?
>
> For instance, as shown below, f(x) is the PDF of a chi-square
distributed random variable whose CDF is F[x]. Seeking the minimum
length 93%-interval [a,b], the code returns "Obtained solution does not
satisfy the following constraints within Tolerance -> 0.001..."
Fiddling with options has been futile.
>
> Any ideas? Thankx.
>
> Bruce
>
> F[x_] := CDF[ChiSquareDistribution[5], x]
>
> Minimize[{b - a, F[b] - F[a] == 0.93, b > a > 0}, {a, b}]
>
> NMinimize[{b - a, F[b] - F[a] == 0.93, b > a > 0}, {a, b}]
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