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This question came out of a discussion with a colleague from economics about price indices. Here is MattF's formulation of the question which differs somehow from the original problem.

Let $Y=({\mathbf R}^+)^3$, where an element $y=(r,v,w)$ is interpreted as the inflation rate $r$ of an item (price at time 1 / price at time 0) and the expenditures $v$ and $w$ in the two periods.

Let $\pi_1$ be the first projection from $Y$ or any other space.

A price index for $Y$ is a sequence of symmetric functions $P_n:Y^n\to {\mathbf R}^+$ with $P_1=\pi_1$. E.g.:

  • The Laspeyres index is $P_n((r_i,v_i,w_i)_{i\le n})=\sum v_ir_i /\sum v_i$.

  • The Fisher index is the geometric mean of two weighted arithmetic means, $$ P_n((r_i,v_i,w_i)_{i\le n})= \left(\sum v_i r_i/\sum v_i\right)^{1/2} \left(\sum w_i r_i/\sum w_i\right)^{1/2}.$$

We want to know if $P$ arises naturally from some type of aggregation.

Let us call $P$ consistent-in-aggregation (CIA) if there is a transformation $h:({\mathbf R}^+)^3\rightarrow {\mathbf R}^+\times H$ with $\pi_1\circ h=\pi_1$, $H \subset {\mathbf R}^k$, and an associative aggregator $\oplus:({\mathbf R}^+\times H)^2\rightarrow({\mathbf R}^+\times H)$, such that $P_n(y_1,\ldots,y_n)=\pi_1(h(y_1)\oplus\ldots\oplus h(y_n))$. Then:

  • The Laspeyres index is CIA with $h(r,v,w)=(r,v)$ and $(r,v)\oplus(s,u)=((vr+us)/(v+u),v+u)$.

  • Every price index is CIA for some transformation, using the axiom of choice. (The proof uses a Hamel basis $\lbrace e_y:y\in Y\rbrace$ of $\mathbb R$ considered as a $\mathbb Q$-vector space and $h(y)=(\pi_1(y), e_y)$. The sum of transformations is enough to reconstruct $y_1,\ldots,y_n$ up to permutation.)

  • Can we find a continuous transformation $h$ and continuous aggregator $\oplus$ which make the Fisher index CIA?

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  • $\begingroup$ @Yemon Choi, why did you revert the edit? $\endgroup$
    – user44143
    Oct 11, 2014 at 14:56
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    $\begingroup$ @MattF. It seemed overly drastic, and I believe Jochen should be free to choose his own wording. If you really want me to quibble over the formatting then we could be here forever. If you think the OP has mis-stated something then I feel it is more courteous to inform him via a comment. We should not be in the business of being sub-editors on MO unless we start waving sub-editing credentials $\endgroup$
    – Yemon Choi
    Oct 11, 2014 at 15:00
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    $\begingroup$ Jochen, I hope you'll consider adopting my edits, which may make the problem easier to appreciate. $\endgroup$
    – user44143
    Oct 11, 2014 at 18:24
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    $\begingroup$ @MattF. Whether your edit makes the problem easier is a matter of taste. Instead of discussing this I would like to get any ideas e.g. how to obtain any necessary conditions for CCIA. $\endgroup$ Oct 12, 2014 at 8:03
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    $\begingroup$ @MattF. Let us try with your formulation. Thanks for your efforts. $\endgroup$ Oct 14, 2014 at 6:30

2 Answers 2

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Let's think of it as making a Markov process out of a function of two Markov processes, by possibly increasing the dimension.

That is, let $h(r,v_1,v_2)=(r,r,v_1,r,v_2)\in\mathbb R^5$ and define $\oplus_{\text{Fisher}}$ by $$ (r,r_1,v_1,r_2,v_2)\oplus_{Fisher} (r',r'_1,v'_1,r'_2,v'_2):=(({\pi_1(\vec\alpha_1)\cdot\pi_1(\vec\alpha_2)})^{1/2},\vec\alpha_1,\vec\alpha_2)$$ where $$\vec\alpha_i=(r_i,v_i)\oplus_{\text{Laspeyres}} (r_i',v_i')\in\mathbb R^2,\quad i=1,2$$ with $\oplus_{\text{Laspeyres}}$ being the $\oplus$ that you defined for the Laspeyres index.

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    $\begingroup$ Looks very good, and I like very much the first sentence. $\endgroup$ Oct 14, 2014 at 14:50
  • $\begingroup$ Nice question (especially after @Mattf's edit finally went through) :) $\endgroup$ Oct 14, 2014 at 18:15
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Yes. Let $h(r,v,w)=(r,rv,v,rw,w)$ and let $(a,b,c,d,e)\oplus(a',b',c',d',e')=$ $$\big(\sqrt{\frac{b+b'}{c+c'}}\sqrt{\frac{d+d'}{e+e'}},\,b+b',\,c+c',\,d+d',\,e+e'\big).$$

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    $\begingroup$ Excellent. Although it looks slightly simpler than Bjorn Kjos-Hannssen's solution it is essentially the same. Unfortunately, I can's share the bounty and Bjorn was a little bit faster. $\endgroup$ Oct 14, 2014 at 16:17

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