0
$\begingroup$

Kamanzi-wa-Binyavanga, 2009, wrote the following paper, Calculating Cumulants of a Taylor Expansion of a Multivariate Function:

enter image description here

What I am confused about, is how precise the notation. I understand that we take each partial derivative with respect to some point ($r,s,t$ here) between $1$ and $p$, which might be the number of regressors in a statistical setting. What I don't understand is:

  1. Why we must sum over each $r,s,t$ individually?
  2. Why he sums over $z$ and $s$, not $r$ and $s$ in the double summation
  3. Why he actually says each of these summations is from 1 to $p_i$. Am I correct in saying, the third for example, should look like $\frac{1}{3!}\sum_{r=1}^{p_i}\sum_{s=1}^{p_i}\sum_{t=1}^{p_i}$.
  4. Lastly, this might seem simple, but if I wanted the Taylor expansion for the whole function $Y$, would I simply add $\sum_{i=1}^{q}$ before the term $c$?

Any help/clarity would be gratefully received. I'm trying to learn to use Einstein notation in this setting, but I need to understand exactly this expansion first.

$\endgroup$

1 Answer 1

2
$\begingroup$
  1. Because this is how Taylor expansion for a function of several variables looks like. For an overview you may consult respective wikipedia page or any of the numerous lecture notes available online. It is common to use more concise multi-index notation or express summations via vector/matrix products, latter perhaps being more relevant if you are studying Einstein notation

  2. It looks like a typo, since $z$ doesn't appear anywhere within that term

  3. That's correct. $p_i$ is just the number of arguments (independent variables) of the function $f_i$

  4. The whole variable $Y$ is a vector in $\mathbb{R}^{q}$ and each of its elements is explicitly defined as a function of certain $p_i$ variables (notation in the paper allows for them even to constitute nonoverlapping sets). If I understand correctly, in this setting you would compute expansion separately for each element (function). Aggregating them further doesn't make sense, in general

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.