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Lecture4

# Lecture4 - X d M t t dt E X = = = Considering 2 2 2 Var X E...

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Feb 5 th Lecture * The derivative and integral of some important functions one should remember. 1 ( ) ( ) 1 (ln ) k k x x d x kx dx d e e dx d x dx x - = = = 1 1 1 1 1 ( ) ( ) 1 1 b k k k k a b x x b a a x b x dx x b a x a k k x b e dx e e e x a + + + = = = - = + + = = = - = * The Chain Rule [ ( )] '[ ( )] '( ) d g f x g f x f x dx = For example: 2 2 2 x x d e e x dx = * The Product Rule [ ( ) ( )] '( ) ( ) ( ) '( ) d g x f x g x f x g x f x dx = + * Extra: The m.g.f. will also generate the moments Moment: 1 st (population) moment ( ) ( ) E X x f x dx = 2 nd (population) moment 2 2 ( ) ( ) E X x f x dx = K th (population) moment ( ) ( ) k k E X x f x dx = Take the K th derivative of the ( ) X M t at t=0. , we get the K th moment of X

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2 2 2 ( ) ( ) 0 ( ) ( ) 0 ... ( ) ( ) 0 X X k k X k d M t E X t dt d M t E X t dt d M t E X t dt = = = = = = Example: When 2 ~ ( , ) X N μ σ , we want to verify the above equations for k=1 2 2 1 2 2 ( ) ( ) ( ) t t X d M t e t dt μ σ + = + (using the chain rule) So when t=0 ( ) ( ) 0 X d M t E X t dt = = = 2 2 2 2 2 2 2 2 1 2 2 1 1 2 2 2 2 2 ( ) [ ( )] [( ) ( )] ( ) ( ) ( ) X X t t t t t t d M t dt d d M t dt dt d e t dt e t e + + + = = + = + + (using the result of the Product Rule) And 2 2 2 2 2 ( )
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Unformatted text preview: ( ) X d M t t dt E X = = + = Considering 2 2 2 ( ) ( ) Var X E X = =-* Bivariate Normal Random Variable 2 2 ( , ) ~ ( , ; , ; ) x x y y X Y BN ρ where is the correlation between & X Y The joint p.d.f. of ( , ) X Y is 2 2 2 2 2 2 1 1 2 ( , ) exp 2(1 ) 2 1 x y x y x y x y xy f x y σ σ πσ σ =-+- -- * Extra Question (How to use the Z-table) Let ~ (3,4) X N We want to calculate (1 3) P X < < (1 3) 1 3 3 3 3 ( ) 2 2 2 ( 1 0) (0 1) 0.3413 P X X P P Z P Z < <---= < < =- < < = < < =...
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Lecture4 - X d M t t dt E X = = = Considering 2 2 2 Var X E...

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