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Re: compelling evaluation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg93221] Re: compelling evaluation
  • From: Bill Rowe <readnews at sbcglobal.net>
  • Date: Fri, 31 Oct 2008 03:08:13 -0500 (EST)

On 10/30/08 at 2:01 AM, randolph.silvers at deakin.edu.au wrote:

>I have created a function from a PDF and so it is non-zero on the
>unit interval and 0 elsewhere. But when I try to integrate some
>function of that function, it is not evaluated. How can I get it to
>be evaluated?

>For example,

>project=UniformDistribution[{0,1}];
>f[y_]:=PDF[project,y]/(CDF[project,1] - CDF[project,0]);
>F[y_]:=Integral_0^y f[z]dz;

You seem to be making things needlessly complex here. By
definition for any distribution CDF[dist,a]-CDF[dist,b] must be
1 when a and b are chosen to be the distribution bounds. So,
defining f as

f[y_]:= PDF[UniformDistribution[{0,1}],y]

does exactly the same as your f[y] above.

As for your F[y] what you wrote above isn't Mathematica syntax.
I assume you mean F[y] to be the cumulative distribution
function. If so, F can be defined as

=46[y_]:=CDF[UniformDistribution[{0,1},y]

>Now, I define

>pi0[y_]:= F[y]
>pi1[y_]:= Integral_y^1 (1-q) f[q] dq

Again, you use Integral, something you have not defined. Perhaps
the problem is you are unfamiliar with the Mathematica syntax to
do integration? I think the syntax you want would be

p11[y_]:=Integrate[(1-q) f[q],{q, y, 1}]


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