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Unformatted text preview: Stat 443 - Forecasting Homework Assignment 1 Solutions Fall 2010 1. By the Cauchy-Schwarz inequality, for any two vectors, say x and y in an inner product space, | < x,y > | || x |||| y || where < , > is an inner product and || x || = < x,x > 1 / 2 . Furthermore, it states that equality holds if and only if x and y are linearly dependent. Since the expectation of a product of random variables is an inner product, we de ne < X,Y > = E ( XY ) for some random variables X and Y . Consider the random variables U E ( U ) and V E ( V ) . We have that < U E ( U ) ,V E ( V ) > = E ( ( U E ( U ) )( V E ( V ) ) ) = Cov ( U,V ) , || U E ( U ) || = < U E ( U ) ,U E ( U ) > 1 / 2 = E (( U E ( U )) 2 ) = V ar ( U ) , and || V E ( V ) || = V ar ( V ) Therefore, by the Cauchy-Schwarz inequality, | Cov ( U,V ) | = V ar ( U ) V ar ( V ) if and only if U E ( U ) and V E ( V ) are linearly dependent. Rearranging, we have that | ( U,V ) | = | Cov ( U,V ) | V ar ( U ) V ar ( V ) = 1 i U E ( U ) and V E ( V ) are linearly dependent, that is, i V E ( V ) = a ( U E ( U )) = V = aU + ( E ( V ) aE ( U )) for some a R . This implies that there exists some linear relationship between U and V i | ( U,V ) | = 1 . Note: Instead of using the Cauchy-Schwarz inequality, you can use the fact that the angle, between two vectors, say x and y is given by cos( ) = < x, y > || x |||| y || Alternative Proof: " = ": Suppose that there exists a linear relationship between U and V , i.e. V = a + bU , 1 where a,b R and b = 0 . Then, by the de nition of correlation, ( U,V ) = Cov ( U,V ) V ar ( U ) V ar ( V ) = Cov ( U,a + bU ) V ar...
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This note was uploaded on 12/12/2010 for the course ACTSC 445 taught by Professor Christianelemieux during the Fall '09 term at Waterloo.
- Fall '09