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Fin500J Topic 2Fall 2010 Olin Business School1 Fin500J Mathematical Foundations in Finance Topic 2: Matrix Calculus Philip H. Dybvig Reference: Matrix Calculus, appendix from Introduction to Finite Element Methods book Slides designed by Yajun Wang

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Fin500J Topic 2Fall 2010 Olin Business School2 Outline The Derivatives of Vector Functions The Chain Rule for Vector Functions

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Fin500J Topic 2Fall 2010 Olin Business School3 1 The Derivatives of Vector Functions

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Fin500J Topic 2Fall 2010 Olin Business School4 1.1 Derivative of Vector with Respect to Vector

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Fin500J Topic 2Fall 2010 Olin Business School5 1.2 Derivative of a Scalar with Respect to Vector If y is a scalar 1.3 Derivative of Vector with Respect to Scalar It is also called the gradient of y with respect to a vector variable x, denoted by.

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Fin500J Topic 2Fall 2010 Olin Business School6 Example 1 Given and

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In Matlab Fin500J Topic 2Fall 2010 Olin Business School7 >> syms x1 x2 x3 real; >> y1=x1^2-x2; >> y2=x3^2+3*x2; >> J = jacobian([y1;y2], [x1 x2 x3]) J = [ 2*x1, -1, 0] [ 0, 3, 2*x3] Note: Matlab defines the derivatives as the transposes of those given in this lecture. >> J' ans = [ 2*x1, 0] [ -1, 3] [ 0, 2*x3] >> syms x1 x2 x3 real; >> y1=x1^2-x2; >> y2=x3^2+3*x2; >> J = jacobian([y1;y2], [x1 x2 x3]) J = [ 2*x1, -1, 0] [ 0, 3, 2*x3] Note: Matlab defines the derivatives as the transposes of those given in this lecture. >> J' ans = [ 2*x1, 0] [ -1, 3] [ 0, 2*x3]

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Fin500J Topic 2Fall 2010 Olin Business School8 Some useful vector derivative formulas Homework

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Fin500J Topic 2Fall 2010 Olin Business School9 Important Property of Quadratic Form x T Cx Proof: If C is symmetric,

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Fin500J Topic 2Fall 2010 Olin Business School10 2 The Chain Rule for Vector Functions Let where z is a function of y, which is in turn a function of x, we can write Each entry of this matrix may be expanded as

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Fin500J Topic 2Fall 2010 Olin Business School11 The Chain Rule for Vector Functions (Cont.) Then On transposing both sides, we finally obtain This is the chain rule for vectors (different from the conventional chain rule of calculus, the chain of matrices builds toward the left)

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Fin500J Topic 2Fall 2010 Olin Business School12 Example 2 x, y are as in Example 1 and z is a function of y defined as

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In Matlab Fin500J Topic 2Fall 2010 Olin Business School13 >> z1=y1^2-2*y2; >> z2=y2^2-y1; >> z3=y1^2+y2^2; >> z4=2*y1+y2; >> Jzx=jacobian([z1; z2; z3; z4],[x1 x2 x3]) Jzx = [ 4*(x1^2-x2)*x1, -2*x1^2+2*x2-6, -4*x3] [ -2*x1, 6*x3^2+18*x2+1, 4*(x3^2+3*x2)*x3] [ 4*(x1^2-x2)*x1, -2*x1^2+20*x2+6*x3^2, 4*(x3^2+3*x2)*x3] [ 4*x1, 1, 2*x3] >> Jzx’ ans = [ 4*(x1^2-x2)*x1, -2*x1, 4*(x1^2-x2)*x1, 4*x1] [ -2*x1^2+2*x2-6, 6*x3^2+18*x2+1, -2*x1^2+20*x2+6*x3^2, 1] [ -4*x3, 4*(x3^2+3*x2)*x3, 4*(x3^2+3*x2)*x3, 2*x3] >> z1=y1^2-2*y2; >> z2=y2^2-y1; >> z3=y1^2+y2^2; >> z4=2*y1+y2; >> Jzx=jacobian([z1; z2; z3; z4],[x1 x2 x3]) Jzx = [ 4*(x1^2-x2)*x1, -2*x1^2+2*x2-6, -4*x3] [ -2*x1, 6*x3^2+18*x2+1, 4*(x3^2+3*x2)*x3] [ 4*(x1^2-x2)*x1, -2*x1^2+20*x2+6*x3^2, 4*(x3^2+3*x2)*x3] [ 4*x1, 1, 2*x3] >> Jzx’ ans = [ 4*(x1^2-x2)*x1, -2*x1, 4*(x1^2-x2)*x1, 4*x1] [ -2*x1^2+2*x2-6, 6*x3^2+18*x2+1, -2*x1^2+20*x2+6*x3^2, 1] [ -4*x3, 4*(x3^2+3*x2)*x3, 4*(x3^2+3*x2)*x3, 2*x3]

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