Re: compelling evaluation
- To: mathgroup at smc.vnet.net
- Subject: [mg93213] Re: [mg93175] compelling evaluation
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 31 Oct 2008 03:06:45 -0500 (EST)
- Reply-to: hanlonr at cox.net
When you post you should use actual Mathematica notation,
project = UniformDistribution[{0, 1}];
f[y_] = PDF[project, y]/
(CDF[project, 1] - CDF[project, 0])
Piecewise[{{1, 0 <= y <= 1}}]
The definition of f can be written more simply as
f[y_] = PDF[project, y]
Piecewise[{{1, 0 <= y <= 1}}]
F[y_] = Assuming[{0 <= y <= 1}, Integrate[f[z], {z, 0, y}]]
Piecewise[{{y, Inequality[0, Less,
y, LessEqual, 1]}}]
pi0[y_] = F[y]
Piecewise[{{y, Inequality[0, Less,
y, LessEqual, 1]}}]
This definition obviously doesn't do anything.
pi1[y_] = Assuming[{0 <= y <= 1},
Integrate[ (1 - q) f[q], {q, y, 1}]] //
Simplify
Piecewise[{{(1/2)*(y - 1)^2,
Inequality[0, LessEqual, y,
Less, 1]}}]
pi2[y_] = Assuming[{0 <= y <= 1},
Integrate[ q f[q], {q, y, 1}]]
Piecewise[{{(1/2)*(1 - y^2),
Inequality[0, LessEqual, y,
Less, 1]}}]
T[y_] = Assuming[{0 <= y <= 1},
Integrate[ (q - y) f[q], {q, y, 1}]] //
Simplify
Piecewise[{{(1/2)*(y - 1)^2,
Inequality[0, LessEqual, y,
Less, 1]}}]
D[T[y], y]
Piecewise[{{0, y < 0},
{y - 1, 0 < y < 1},
{0, y == 1 || y > 1}},
Indeterminate]
T[pistar]
Piecewise[{{(1/2)*(pistar - 1)^2,
Inequality[0, LessEqual,
pistar, Less, 1]}}]
Bob Hanlon
---- 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;
Now, I define
pi0[y_]:= F[y]
pi1[y_]:= Integral_y^1 (1-q) f[q] dq
pi2[y_]:= Integral_y^1 q f[q] dq
T[y_]:= Integral_y^1 (q-y) f[q] dq
When I enter a numerical value, each is correctly computed; when I
differentiate, it also looks correct. For example,
D[T[y],y] returns
Integral_y^1 -{1 0<=q<=1 dq
and, T[pistar] returns
Integral_pistar^1 (-pistar + q)({1 0 <=q <= 1) dq
How can I compel Mathematica to "know" or evaluate T[pistar] and
return the symbolic expression assuming that q is in the relevant
domain?
Then, D[T[y],y] would return -(1-y) and T[pistar] would return:
1/2 - pistar + pistar^2/2
--
Bob Hanlon