*Review: The derivative and integral of some important functions one should remember.

Transcription

1 AMS 570 Professor Wei Zhu *Review: The derivaive and inegral of some imporan funcions one should remember. d ( k ) k k d d ( e ) e d d (ln ) d b k k b k k d ( ) ( b a ) a k a k b a b e d e e e a b a *The Chain Rule d g [ f ( )] g '[ f ( )] f '( ) d d For eample: e e d *The Produc Rule d [ g ( ) f ( )] g '( ) f ( ) g ( ) f '( ) d *Review: MGF is second funcion: The m.g.f. will also generae he momens Momen: s (populaion) momen: E( ) f ( ) d nd (populaion) momen: E( ) f ( ) d K h (populaion) momen: ( k ) k E f d

2 Take he K h derivaive of he M () wih respec o and he se = 0 we obain he K h momen of as follows: d M ( ) E ( ) d 0 d M E d... 0 k d k k M E d 0 Noe: The above general rules can be easil proven using calculus. Eample: When ~ N( ) we wan o verif he above equaions for k= & k=. d M e d (using he chain rule) So when =0 d M ( ) E ( ) d 0 d M () d d d [ M ( )] d d d d [( e ) ( )] ( e ) ( ) ( e ) And d M () d E 0 (using he resul of he Produc Rule) Considering Var( ) E( ) *** Join disribuion and independence Definiion. The join momen generaing funcion of wo random variables and is defined as

3 Theorem. Two random variables and are independen (if and onl if) Definiion. The covariance of wo random variables and is defined as []. Theorem. If wo random variables and are independen hen we have. (*Noe: However does no necessaril mean ha and are independen.) Eercise. Prove ha and are independen in wo approaches: () pdf () mgf Soluion: () The pdf approach: Le Then we have This is a - ransformaion beween ( ) and (W Z). Define he Jacobian J of he ransformaion as: Then he join disribuion of he new random variables is given b: Since and are independen we have: 3

4 Thus we find: ( ) So Z and W are independen. Furhermore we see ha and () The mgf approach: Now we p ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4

5 ( ) ( ) Q. () please derive he covariance beween ; () Are independen? [( ) ( )] = [] [ ] [ ] () are no independen. ou can do i in man was using (a) pdf (b) cdf/probabiliies or (c) mgf. For eample [ ] Q3. Le ~ N (34) please calculae P( 3) P( 3) P( ) P( Z 0) P(0 Z )

6 Q4. Sudens A and B plans o mee a he SAC Seawolf marke beween noon and pm omorrow. The one who arrives firs will wai for he oher for 30 minues and hen leave. Wha is he chance ha he wo friends will be able o mee a SAC during heir appoined ime period assuming ha each one will arrive a SAC independenl each a a random ime beween noon and pm? Soluion: 3/4 as illusraed b he following figure. Le and denoe he arrival ime (in hour where 0 indicaes noon and indicaes pm) of A and B respecivel. Our assumpion implies ha and each follows Uniform[0] disribuion and furhermore he are independen Thus he join pdf of and follows he uniform disribuion wih f() = if () is in he square region and f() = 0 ouside he square region. A and B will be able o mee iff as represened b he red area M bounded b he lines: and. Tha is: P(A & B will mee) = ( ) *Definiions: populaion correlaion & sample correlaion 6

7 7 Definiion: The populaion correlaion coefficien ρ is defined as: Definiion: Le ( ) ( n n) be a random sample from a given bivariae populaion hen he sample correlaion coefficien r is defined as: ( )( ) [ ( ) ][ ( ) ] *Definiion: Bivariae Normal Random Variable ( ) ~ ( ; ; ) BN where is he correlaion beween & The join p.d.f. of ( ) is ep f Eercise: Please derive he mgf of he bivariae normal disribuion. Q5. Le and be random variables wih join pdf ep f where. Then and are said o have he bivariae normal disribuion. The join momen generaing funcion for and is ep M. F h m g l f s of ; (b) ov h f o l f ρ = 0. (H ρ s h < o ul o > co l o co ff c b w.) (c) Find he disribuion of. (d) Find he condiional pdf of f( ) and f( )

8 Soluion: (a) The momen generaing funcion of can be given b [ ] Similarl he momen generaing funcion of can be given b [ ] Thus and are boh marginall normal disribued i.e. and. The pdf of is The pdf of is [ ] [ ] (b) If 0 hen M ( ) ep M ( 0) M (0 ) Therefore and are independen. If and are independen hen M ( ) M ( 0) M (0 ) ep ep Therefore 0 (c) ( ) M () E e E e 8

9 Recall ha M ( ) E e in M ( ) herefore we can obain M () Tha is M ( ) M ( ) ep ep ~ N( ) (d) The condiional disribuion of given = is given b b seing ( ) ( ) { Similarl we have he condiional disribuion of given = is } ( ) { ( ) } Therefore: ( ) ( ) Q6: and he value of random variable W depends on a coin (fair coin) flip: { find he disribuion of W. Is he join disribuion of Z and W a bivariae normal? Answer: 9

10 ( ) ( ) So. The join disribuion of Z and W is no normal. This can be shown b deriving he join mgf of Z and W and compare i wih he join mgf of bivariae normal. ( ( )) ( ) ( ) ( ) ( ) ( ) This is no he join mgf of bivariae normal. Alernaivel ou can also derive he following probabili: ( ) ( ) This shows ha W+Z can no possibili follow a univariae normal disribuion which in urns shows ha W and Z can no possible follow a bivariae normal disribuion. 0