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Lecture 7

# We used this theorem previously when we dealt with

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Unformatted text preview: of equations, , ,…, , ,…, ⋮ , ,…, (here, the j in is indexing the function; it is not the jth partial derivative of ). Assume the matrix of first partial derivatives has a non‐zero determinant. This determinant is called the Jacobian of this system. The Jacobian … ≡ ⋮ ⋱ ⋮ … If 0 about some point ∗ , ∗ , … , ∗ in the domain of the system, there exists a neighborhood of , ,…, and a series of functions, , with , ,…, , for 1, 2, … , . These functions are continuous and have continuous first partial derivatives. We used this theorem previously, when we dealt with 2‐variable constraints. It is particularly useful when characterizing the demands of a consumer or producer. In the case of the former, these demands can be expressed in terms of prices the consumer faces and her income. In the case of the later, the demands (these being demands for factors, i.e., inputs) are a function of the factor prices and product prices the producer faces. An Example Consider the problem of a consumer that wishes to maximize her utility , ,… subject to the budget constraint: ⋯ where is the price of good , is her consumption of good and is her income. Assuming our ‘nice’ properties on the objective function, our first‐order conditions for is problem are: , ,… for 1, 2, … , , and ⋯ . Here, is the Lagrange multiplier for this problem. We have K+1 variables (the K demands plus ) and K equations (the K first‐ order conditions for the consumptions) and the budget constraint. Note too we have K+1 ‘free parameters’ or exogenous variables, the K prices and income I. Using the implicit function theorem, we can then write: , ,… , . course, , ,… , and of We can do this, assuming the Jacobian doesn’t equal 0 at the critical point. The Jacobian for this problem is our friend, the bordered Hessian matrix....
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