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Topic: defining averages over unknown PDF
Replies: 5   Last Post: Nov 14, 2013 1:49 AM

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 Itai Seggev Posts: 61 Registered: 7/29/13
Re: defining averages over unknown PDF
Posted: Nov 6, 2013 12:26 AM

On Mon, Nov 04, 2013 at 11:16:59PM -0500, Sune wrote:
> Dear all.
>
> I want to do some symbolic manipulations of an expression involving averages over a stochastic variable with an unknown density. Therefore, I figured I could define a function av, which would correspond to the average over this unknown parameter density function.
> I did as follows:
> av[y_ + z_, x_] := av[y, x] + av[z, x]?
> av[c_ y_, x_] := c av[y, x] /; FreeQ[c, x]
> av[c_, x_] := c /; FreeQ[c, x]
>
> So these are basic properties of the average over the distribution of X. Some things work okay, for example
> In[52]:= av[Exp[-x y], x]?
> Out[52]= av[E^(-x y), x]
> and
> In[79]:= D[av[-x y, x], x]?
> Out[79]= -y
> and
> In[80]:= D[av[-x y, x], y]?
> Out[80]= -av[x, x].
>
> However, the most vital part for my calculations does not work:
> In[81]:= D[av[Exp[-x y], x], y]?
> Out[81]= -E^(-x y) x
>
> It should have been av[-Exp[-x y] x,x].
>
> Any clues to what I'm doing wrong? I'm thinking that I need to specify some rules for differentiation, but I don't know how. But then I'm wondering how come it got the other expressions for differentiation right.

Ahh, the subtle treacheries of partial differentiation. Note that by your
definition,

In[71]:= av[Exp[-x y] + h, x] - av[Exp[-x y], x]

Out[71]= h

So that

In[72]:= Limit[(av[Exp[-x y] + h, x] - av[Exp[-x y], x])/h, h -> 0]

Out[72]= 1

So both your "correct" and "incorrect" answers are consistent with the chain
rule and and the above computation of partial derivatives. So why is D
computing the partial derivative in such a stupid way? Well, it isn't, at
least not directly. D correctly computes the partial derivative as

f'[x] * Derivative[1, 0][av][f[x], x] + Derivative[0, 1][av][f[x],x]

But now Derivative helpfully tries compute these partials using pure functions,
and then your definitions kick in, giving 1 and 0 for the partials. In
particular, your third definitions means av[#1,#2]& === #1, and you're doomed.

So you want to abort the automatic differentiation rules with your own custom
rule, which you can do with the following syntax:

av /: D[av[f_, x_], y_] /; x =!= y := av[D[f, y], x]

In[65]:= D[av[Exp[-x y], x], y]

Out[65]= -av[E^(-x y) x, x]

--
Itai Seggev
Mathematica Algorithms R&D
217-398-0700

Date Subject Author
11/6/13 Itai Seggev
11/6/13 Sune
11/12/13 Itai Seggev
11/14/13 Itai Seggev