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Can anyone provide an intuitive explanation of why the Slutsky matrix right multiplied by the price vector yields a zero matrix?

I know this is true but I do not really understand why it is true. Can anyone help here?

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3 Answers 3

This is a general mathematical property of the second derivative/Hessian matrix of multivariate functions that are homogeneous of degree one.

The Expenditure function $E$ is homogeneous of degree one in prices. Why? If all prices change in the same proportion (which is how we check for the mathematical property of homogeneity), relative prices do not change. If relative prices do not change, the quantitative composition of the minimum-cost compensated consumption bundle in order to achieve a given utility does not change at all. Then, since all prices have increased by the same proportion, budget shares remain the same, and the Expenditure needed to achieve the same utility, rises in that same proportion: homogeneity of degree one.

By duality, the Hicksian demand vector is the gradient of the Expenditure function, $H = \nabla_p E$.

The Hicksian demand vector, gives us minimum-cost quantities demanded. Due to the homogeneity of degree one of the Expenditure function, the inner product of the Hicksian demand vector times the price vector equals the Expenditure function. This also should be intuitive: we just multiply each quantity demanded by the unit price that must be paid for it, and by summing these products we obtain the total Expenditure we must incur in order to acquire the minimum-cost bundle for given utility.

So we have (simplifying differentiation notation) $ E = H\cdot p$ while also $\frac {\partial}{\partial p}E = H$. Therefore also

$$\frac {\partial}{\partial p}(H\cdot p) = H \implies H + \frac {\partial H}{\partial p}\cdot p = H$$

and it must be the case that

$$\frac {\partial H}{\partial p}\cdot p = 0$$

So the Hicksian demand vector is homogeneous of degree zero in prices (mathematically, this is a consequence of Euler's theorem for homogeneous functions, i.e. that if a function is homogeneous with degree of homogeneity $k$, its gradient has degree of homogeneity $k-1$).

But the 1st derivative (Jacobian) of Hicksian demand (which is the Hessian-matrix of second derivatives of the Expenditure function) is the Slutsky matrix, $\frac {\partial^2 E}{\partial p^2}=\frac {\partial H}{\partial p} = S(p,w)$. So $S(w,p) \cdot p = 0$.

So the result stems from the homogeneity of degree one of the Expenditure function. Is there an intuitive explanation, analogously with the intuition behind the homogeneity of degree one of the Expenditure function? Well, the former comes directly from the latter, so it is difficult to come up with a "separate" intuitive argument. One could informally say that compensated quantities demanded are "independent" of (not affected by) price variation when relative prices remain the same. Then in geometric terms, this means that the vectors of rates of change of compensated quantities demanded (which is what each row of the Slutsky matrix contains), are orthogonal to the price vector.

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Wow. This is a fantastic answer. –  mathtastic Oct 1 at 11:34
@mathtastic Thanks for the kind words. –  Alecos Papadopoulos Oct 1 at 14:07

I don't know if you will regard this as an explanation, or rather as a proof.

What we have a better understanding from univariate calculus is the first order Taylor approximation, i.e. a function satisfying some regularity conditions can be well-approximated by a linear function at a point. Say $f:\mathbb{R}\rightarrow\mathbb{R}$, then around $p^\ast$ (i.e. when $\delta$ is small) $$ f(p^\ast+\delta)\simeq f(p^\ast)+\delta\times \left.\frac{df}{dp}\right|_{p=p^\ast} $$

Now, we can do something similar for multivariable functions. If $h_i:\mathbb{R}^n\rightarrow\mathbb{R}$, then $$ h_i(\mathbf{p^\ast+\delta})\simeq h_i(\mathbf{p^\ast})+\left.\frac{\partial h_i(\mathbf{p})}{\partial p_1}\delta_1\right|_{p=p^\ast}+\dots+\left.\frac{\partial h_i(\mathbf{p})}{\partial p_n} \delta_n\right|_{p=p^\ast} $$

Now, it should be apparent that when we multiply all prices by the same number, the Hicksian demand does not change. So let's say we increase prices from $\mathbf{p^\ast}$ to $\mathbf{p^\ast(1+\Delta)}$. So each price ${p_j^\ast}$ changes proportionally at the amount of $\Delta \times {p_j^\ast}$. We should see no change in the value of $h_i$ above if we replace $\delta$ with $\Delta \mathbf{p^\ast}$. Then it must be true that the additional terms including partial derivatives would sum to 0, which basically results in your $S(\mathbf{p},w)\cdot \mathbf{p}=0$

Put another way, since the Hicksian demand for any good is not responsive to a change in prices that keeps the relative prices the same, then if we look at the total of the individual affects of these price changes on a good, we should observe a 0 change.

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I suppose that you know already the answer given in Mas-Colell, this property has to do with no money illusion in fact (under some regularity conditions) it is if and only if with the property of the demand $x(\alpha p,\alpha w)=x(p,w) \forall \alpha > 0$. Now, it is trivial to show that this means that $D_px(p,w)p+D_wx(p,w)w=0$, under Walras' law $p'x(p,w)=w$ or equivalently $x(p,w)'p=w$, replacing it in the firs expression one gets $S(p,w)p=0$. We can go the other way using integration. In other words, I think that the robust intuition is that your consumer does not suffer from money illusion. Note that the consumer may not be rational but she will at least not be subject to this behavioral bias.

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