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### math-2011-note7

Course: ECON 205, Spring 2011
School: Korea University
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34 Theorem (Thm 13.3, p.291). The general quadratic form Q(x1 , , xk ) = i j ai j xi x j can be written as 1 a11 2 a12 1 a1k x1 2 1 a12 a22 1 a2k x2 2 2 x1 x2 xk . . . . , .. . . . . . . . . . 1 1 xk akk 2 a1 k 2 a2 k or xT Ax, where A is a symmetric matrix. 11.2 Continuous Functions Denition 24 (p.293). A function f : Rk Rm is continuous at x0 if whenever {xn }=1 n converges to x0 the...

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34 Theorem (Thm 13.3, p.291). The general quadratic form Q(x1 , , xk ) = i j ai j xi x j can be written as 1 a11 2 a12 1 a1k x1 2 1 a12 a22 1 a2k x2 2 2 x1 x2 xk . . . . , .. . . . . . . . . . 1 1 xk akk 2 a1 k 2 a2 k or xT Ax, where A is a symmetric matrix. 11.2 Continuous Functions Denition 24 (p.293). A function f : Rk Rm is continuous at x0 if whenever {xn }=1 n converges to x0 the sequence { f (xn )}=1 in Rm converges to f (x0 ). The function f is n said to be continuous if it is continuous at every point in its domain. Theorem 35 (Thm 13.5, p.295). Let f = ( f 1 , , f m) be a function from Rk to Rm . Then, f is continuous at x if and only if each of its component functions f i : Rk R1 is continuous at x. If f and g are two functions from Rk to Rm f , + g, f g, f g are dened by ( f + g)(x) = f (x) + g(x), ( f g)(x) = f (x) g(x), ( f g)(x) = f (x) g(x), respectively. Theorem 36 (Thm 13.4, p.294). Let f and g be functions from Rk to Rm . If f and g are continuous at x, then f + g, f g, f g are all continuous at x. 12 Calculus of Several Variables (ch.14) 12.1 Partial Derivative Denition 25 (p.300). Let F : Rk R. The partial derivative of f with respect to xi at x0 = (x0 , , x0 ) is dened as 1 k F (x0 , , x0 + h, , x0 ) F (x0 , , x0 , , x0 ) F 0 1 i 1 i k k (x ) = lim , h0 xi h if the limit exists. Example 3. F (x1 , x2 ) = 3x2 x2 + 4x1 x3 + 7x2 . 12 2 F F (x) =?, (x) =? x1 x2 Economic Example: Marginal Product Let Q = F (K , L) be a production function. Then the partial derivative F (K0 , L0 ) L 32
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Korea University - ECON - 205
12 Calculus of Several Variables (ch.14)12.1 Partial DerivativeDenition 25 (p.300). Let F : Rk R. The partial derivative of f with respect to xi atx0 = (x0 , , x0 ) is dened as1kF (x0 , , x0 + h, , x0 ) F (x0 , , x0 , , x0 )F 0ii11kk(x ) = l
Korea University - ECON - 205
Theorem 41 (Thm 30.5, p.828). Let f : R1 R1 be a C2 function. For any point a &lt;b R1 , there is a point c (a, b) such that f (b) = f (a) + f (a)(b a) + 1 f (c)(b 2a)2 .Proof. Dene2f (b) f (a) f (a)(b a) .(b a)2Suppose that f (x) = M for all x (a, b
Korea University - ECON - 205
Example 6. Suppose that we have a production function Q = kxa yb . Then,Q= akxa1 yb ,x 2Q= abkxa1 yb1 , x yQ= bkxa yb1y 2Q= abkxa1 yb1 y x14 Some Linear Algebra14.1 Deniteness of Quadratic Forms (16.2)Denition 27. Let A be an m m symmetric
Korea University - ECON - 205
Theorem 53 (Thm 11.2, p.243). Let A = (a1 , a2 , , am). If a1 , a2 , , am are linearlyindependent if and only if det A = 0.Proof. a) only if: If a1 , a2 , , am are linearly independent, then A is one-to-one andonto, hence has an inverse function. Call
Korea University - ECON - 205
Denition 36 (p.161). An m m matrix A = (ai j ) is called an upper-triangular matrixif ai j = 0 for i &gt; j. A is called a lower-triangular matrix if ai j = 0 for i &lt; j. A is calleda diagonal matrix if ai j = 0 for i = j.Theorem 56 (Fact 26.11, p.731). Th
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Statistics 351 (Fall 2007)Covariance Zero Does Not Imply IndependenceThe following exercise illustrates that two random variables can have covariance zero yet need notbe independent.Exercise. Consider the random variable X dened by P (X = 1) = 1/4, P
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Statistics 351 (Fall 2007)Fishers F -distribution and Statistical Hypothesis TestingFor Problem #4 on page 27 you are asked to manipulate Fishers F -distribution. The F -distributionarises in the following context. (See Section 10.9 in the Stat 251 tex
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Statistics 351 (Fall 2007)The Gamma FunctionSuppose that p &gt; 0, and deneup1 eu du.(p) :=0We call (p) the Gamma function and it appears in many of the formul of density functionsfor continuous random variables such as the Gamma distribution, Beta di
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Statistics 351 Fall 2007Independence of X and S 2Theorem. Suppose that X1 , . . . , Xn are independent N (0, 1) random variables. If1X=nnXii=11and S =n1n2(Xi X )2i=1denote the sample mean and sample variance, respectively, then X and S 2 a
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Statistics 351 (Fall 2007)The Jacobian for Polar CoordinatesExample. Determine the Jacobian for the change-of-variables from cartesian coordinatesto polar coordinates.Solution. The traditional letters to use arex = r cos and y = r sin .However, to a
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Statistics 351 (Fall 2007)Review of Linear AlgebraSuppose that A is the symmetric matrix1 1 0A = 1 2 1 .013Determine the eigenvalues and eigenvectors of A.Recall that a real number is an eigenvalue of A if Av = v for some vector v = 0. We callv a
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Statistics 351 Midterm #1 October 10, 2007This exam has 4 problems and 6 numbered pages.You have 50 minutes to complete this exam. Please read all instructions carefully, and checkyour answers. Show all work neatly and in order, and clearly indicate yo
University of Regina - STAT - 351
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University of Regina - STAT - 351
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University of Regina - STAT - 351
Statistics 351 Midterm #2 November 16, 2007This exam is worth 50 points.There are 5 problems on 5 numbered pages.You have 50 minutes to complete this exam. Please read all instructions carefully, and checkyour answers. Show all work neatly and in orde
University of Regina - STAT - 351
Statistics 351 (Fall 2007)The Density and Characteristic Function Denitions of Multivariate NormalitySuppose that the random vector X = (X, Y ) has a multivariate normal distribution with meanvector and covariance matrix given by=xyand =2xx y.
University of Regina - STAT - 351
Stat 351 Fall 2007Assignment #1 Solutions2. (a) If X Unif[0, 2], then FX (x) = x for 0 x 2, and if Y Exp(3), then FY (y ) = 1 ey/32for y &gt; 0. Since X and Y are independent, we conclude thatFX,Y (x, y ) = FX (x) FY (y ) =x1 ey/32for 0 x 2 and y &gt;
University of Regina - STAT - 351
Statistics 351 Fall 2007 Midterm #1 Solutions1. (a) By denition,x yfX (x) =2e exdy = 2ey= 2e2x , x &gt; 0,exandxyyx yfY (y ) =2e eydx = 2ex= 2ey 1 ey ,e0y &gt; 0.01. (b) Since fX,Y (x, y ) = fX (x) fY (y ), we immediately conclude that
University of Regina - STAT - 351
Statistics 351 Fall 2008 Midterm #1 Solutions1. (a) By denition,yefX (x) == ex ,ydy = exx &gt; 0.xNote that X Exp(1) so that E(X ) = 1.1. (b) By denition,yey dx = yey ,fY (y ) =y &gt; 0.0Note that Y (2, 1).1. (c) By denition,fY |X =x (y ) =
University of Regina - STAT - 351
Statistics 351 Fall 2008 Midterm #1 Solutions1. (a) By denition,xefY (y ) == ey ,xdx = eyy &gt; 0.yNote that Y Exp(1) so that E(Y ) = 1.1. (b) By denition,xex dy = xex ,fX (x) =x &gt; 0.0Note that X (2, 1).1. (c) By denition,fY |X =x (y ) =
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Statistics 351 Fall 2007 Midterm #2 Solutions1. (a) Let1 111B=and b =20so thatBX + b =1 111X1X2+20=X1 X2 2X1 + X2Y1Y2== Y.By Theorem V.3.1, we conclude that Y N (B + b, B B ) where1 111E(Y) = B + b =12+10=00andcov(Y) =
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University of Regina - STAT - 351
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University of Regina - STAT - 351
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University of Regina - STAT - 351
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University of Regina - STAT - 351
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University of Regina - STAT - 351
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