MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Mathematical expectation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg84137] Re: [mg84113] Mathematical expectation
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Tue, 11 Dec 2007 06:09:24 -0500 (EST)
  • References: <2929821.1197341645609.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

Add another pattern:

Clear[Esp]
Esp[X[_]] := Esp[X]
Esp[X[_]^y_] := Esp[X^y]
Esp[Esp[x_]] := Esp[x]
Esp[Esp[x_]^y_] := Esp[x]^y
Esp[x_ + y_] := Esp[x] + Esp[y]
Esp[x_ y_] := Esp[x] Esp[y] /; FreeQ[x, y] && FreeQ[y, x]
Esp[c_] := c /; FreeQ[c, X]
Esp[c_ x_] := c Esp[x] /; NumberQ[c]
Esp[expr_] /; expr =!= Expand@expr := Esp@Expand@expr

Esp[(X[1] + X[2])^2]

2 Esp[X]^2 + 2 Esp[X^2]

or, more efficient if Expand could take a lot of time...

Clear[Esp]
Esp[X[_]] := Esp[X]
Esp[X[_]^y_] := Esp[X^y]
Esp[Esp[x_]] := Esp[x]
Esp[Esp[x_]^y_] := Esp[x]^y
Esp[x_ + y_] := Esp[x] + Esp[y]
Esp[x_ y_] := Esp[x] Esp[y] /; FreeQ[x, y] && FreeQ[y, x]
Esp[c_] := c /; FreeQ[c, X]
Esp[c_ x_] := c Esp[x] /; NumberQ[c]
Esp[expr_] :=
  Module[{expanded = Expand@expr}, Esp@expanded /; expanded =!= expr]

Esp[(X[1] + X[2])^2]

2 Esp[X]^2 + 2 Esp[X^2]

Bobby

On Mon, 10 Dec 2007 19:40:00 -0600, Benoît Frénay  
<benoit.frenay at uclouvain.be> wrote:

> Hello,
>
> I'm new to Mathematica and I'm trying to define a generic mathematical
> expectation operator (called Esp) for i.i.d variables X[i] following the
> same distribution as the variable X (i.e. Esp[X[i]] == Esp[X]).  My code  
> is
>
> Esp[X[_]] := Esp[X]
> Esp[X[_]^y_] := Esp[X^y]
> Esp[Esp[x_]] := Esp[x]
> Esp[Esp[x_]^y_] := Esp[x]^y
> Esp[x_ + y_] := Esp[x] + Esp[y]
> Esp[x_ y_] := Esp[x] Esp[y] /; FreeQ[x, y] && FreeQ[y, x]
> Esp[c_] := c /; FreeQ[c, X]
> Esp[c_ x_] := c Esp[x] /; NumberQ[c]
>
> But when I feed
>  >>> In[14]:= Esp[(X[1] + X[2])^2]
> I obtain
>  >>> Out[14]= Esp[(X[1] + X[2])^2]
> In fact, I obtain the correct answer only if I ask Mathematica to expand
> the inner term:
> In[16]:= Esp[Expand[(X[1] + X[2])^2]]
> Out[16]= 2 Esp[X]^2 + 2 Esp[X^2]
>
> Where am I wrong?
>
> Thanks,
>
> Benoît Frénay
>
>
>



-- 
DrMajorBob at bigfoot.com


  • Prev by Date: Re: From 3D to 2D
  • Next by Date: Re: Prefix Forms on the BasicMath Palette
  • Previous by thread: Re: Mathematical expectation
  • Next by thread: Implementing own data functions