# Reasons for why slutsky matrix may be non symmetric

In demand system estimation, theoretically we require this matrix to be symmetric. This unfortunately is not the case most of the time.

What are some reasons for why symmetry of the slutsky matrix may fail empirically?

• Might be relevant: economics.brown.edu/sites/default/files/papers/… Commented Sep 5, 2023 at 17:36
• @MichaelGreinecker thank you for the resource! I am a bit iffy on the examples provided but the idea of price paths is something new to me. Very interesting.
– EconJohn
Commented Sep 6, 2023 at 4:41

The necessary and sufficient conditions on a demand system are homogeneity of degree 0, Slutsky symmetry and Slutsky negativity.

From a theoretical point of view, symmetry of the Slutsky matrix is due to the fact that the cross partial derivatives of the expenditure function are equal (Young's theorem). If $$e(p,u)$$ is the expenditure function, then the Hicksian demand for good $$i$$ is the derivative of $$e(p,u)$$ with respect to the price of good $$i$$, say $$p_i$$. So: $$\frac{\partial h_i(p,u)}{\partial p_j} = \frac{\partial^2 e(p,u)}{\partial p_j \partial p_i} = \frac{\partial^2 e(p,u)}{\partial p_i \partial p_j} = \frac{\partial h_j(p,u)}{\partial p_i}.$$ So symmetry gives the basic integrability condition.

Negativity of the Slutsky matrix, is equivalent to the requirement that the expenditure function is concave. As such we get the intuitive connections:

• symmetry $$\sim$$ integrability condition
• negativity $$\sim$$ concavity condition

Here I use the symbol $$\sim$$ to say that they are related (although) not necessarily one to one from a pure mathematical perspective.

There also seems also te be a strong connnection between the various revealed preference axioms and the Slutsky conditions. In particular, WARP (the weak axiom of revealed preference) appears to be tightly connected to the negativity of the Slutsky matrix (see (Samuelson,1938) for an exposition).

For example let $$\widetilde{x}$$ be the optimal bundle at price $$p$$, so $$\widetilde{x} = x(p, p \cdot \widetilde{x})$$ where $$x(p,m)$$ is the Marshallian demand at prices $$p$$ and income $$m$$. Consider a compensated price change to $$\widehat{p}$$ giving rise to a new demand $$\widehat{x} = x(\widehat{p}, \widehat{p} \cdot \widetilde{x})$$ (so the new budget can still afford the bundle $$\widetilde{x}$$. Then, $$\widehat{p} \cdot \widehat{x} = \widehat{p} \cdot \widetilde{x}$$ and WARP requires that: $$p \cdot \widetilde{x} < p \cdot \widehat{x}.$$ Then adding up and rewriting gives $$(\widehat{p} - p) \cdot (\widehat{x} - \widetilde{x}) < 0.$$ Now if we specify $$\widehat{p} = p + t v$$ for some vector $$v$$, we get: $$t v \cdot (x(p + v t, (p+vt)\cdot \widetilde{x}) - x(p, p \cdot \widetilde{x}) < 0.$$ dividing by $$t^2$$ and taking the limit for $$t \to 0$$ gives, $$v \cdot S \cdot v < 0.$$ where $$S$$ is the Slutsky matrix. As $$v$$ can be any vector, this means that $$S$$ needs to be negative definite.

Now, given that WARP does not exhaust all testable implications on a demand function, the remaining part should be related to the symmetry of the Slutsky matrix. In particular, Slutsky symmetry seems to be tightly connected to transitivity of the (revealed) preference relation. I refer to Hurwicz & Richter (1979) who link symmetry to the so called Ville axiom as the argument is not that straightforward.

Anyway, if we summarize it seems that:

• WARP $$\sim$$ negativity of Slutsky matrix
• 'transitivity´ of the (revealed) preferences $$\sim$$ symmetry of the Slutsky matrix

Of course this is not a theoretical one to one mapping, but it should give some behavioural intuition behind the two Slutsky requirements.

Together WARP and transitivity give rise to the Strong Axiom of Revealed Preference (SARP) (see Houthakker,1950). It is known that this exhausts all testable implications on a (homogeneous) demand functions (to be consistent with utility maximisation).

Interesting note: it is known that if there are only two goods then WARP is equivalent to SARP (see Rose,1958). It is also known that in the two goods case, Slutsky symmetry is always satisfied (follows from Engel aggregation and homogeneity of degree 0). So WARP (or negativity of the Slutsky matrix) is sufficient in the 2 goods case.