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