Yeah, you're right; x_ij does appear in p_ij. I had in mind that p_ij
was sitting there, already calculated.
Maybe a better way to express my concern is that I'm interested in the
change in p_ij due to a change in x_ij, not due to a change in
ln(x_ij).
Thanks.
Misha
On Fri, Mar 5, 2010 at 11:20 AM, Austin Nichols <austinnichols@gmail.com> wrote:
> Misha Spisok <misha.spisok@gmail.com> :
>
> p_ij is a function of x_ij, so it does appear, right?
>
> On Fri, Mar 5, 2010 at 1:43 AM, Misha Spisok <misha.spisok@gmail.com> wrote:
>> One benefit I see is that the variable, x_ij, doesn't appear in this
>> expression.
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