Suppose I've checked the identity below, how to implement it in Mathematica ?
(* {\[Alpha] \[Element] Reals, \[Beta] \[Element] Reals, \[Mu] \[Element] Reals, \[Sigma] > 0} *)
Integrate[CDF[NormalDistribution[0, 1], \[Alpha] + \[Beta] x] PDF[
NormalDistribution[\[Mu], \[Sigma]],
x], {x, -\[Infinity], \[Infinity]}] -> CDF[NormalDistribution[0, 1], (\[Alpha] +
\[Beta] \[Mu])/Sqrt[1 + \[Beta]^2 \[Sigma]^2]]开发者_Go百科
Most ways to do what you request would probably involve adding rules to built-in functions (such as Integrate
, CDF
, PDF
, etc), which may not be a good option. Here is a slightly softer way, using the Block
trick - based macro:
ClearAll[withIntegrationRule];
SetAttributes[withIntegrationRule, HoldAll];
withIntegrationRule[code_] :=
Block[{CDF, PDF, Integrate, NormalDistribution},
Integrate[
CDF[NormalDistribution[0, 1], \[Alpha]_ + \[Beta]_ x_] PDF[
NormalDistribution[\[Mu]_, \[Sigma]_], x_], {x_, -\[Infinity], \[Infinity]}] :=
CDF[NormalDistribution[0, 1], (\[Alpha] + \[Beta] \[Mu])/
Sqrt[1 + \[Beta]^2 \[Sigma]^2]];
code];
Here is how we can use it:
In[27]:=
withIntegrationRule[a=Integrate[CDF[NormalDistribution[0,1],\[Alpha]+\[Beta] x]
PDF[NormalDistribution[\[Mu],\[Sigma]],x],{x,-\[Infinity],\[Infinity]}]];
a
Out[28]= 1/2 Erfc[-((\[Alpha]+\[Beta] \[Mu])/(Sqrt[2] Sqrt[1+\[Beta]^2 \[Sigma]^2]))]
When our rule does not match, it will still work, automatically switching to the normal evaluation route:
In[36]:=
Block[{$Assumptions = \[Alpha]>0&&\[Beta]==0&&\[Mu]>0&&\[Sigma]>0},
withIntegrationRule[b=Integrate[CDF[NormalDistribution[0,1],\[Alpha]+\[Beta] x]
PDF[NormalDistribution[\[Mu],\[Sigma]],x],{x,0,\[Infinity]}]]]
Out[36]= 1/4 (1+Erf[\[Alpha]/Sqrt[2]]) (1+Erf[\[Mu]/(Sqrt[2] \[Sigma])])
where I set \[Alpha]
to 0
in assumptions to make the integration possible in a closed form.
Another alternative may be to implement your own special-purpose integrator.
精彩评论