This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Â§ 8 Properties of distributions Â§ 8.1 Expectation 8.1.1 Definition. Let X be a random variable. The expectation , (or expected value , or mean ) of X is: (i) (for discrete case) E [ X ] = X x âˆˆ X (Î©) x P ( X = x ); (ii) (for continuous case) E [ X ] = Z âˆžâˆž xf ( x ) dx , where f is the pdf of X . 8.1.2 The expectation of X is what we would expect of its value if we are to take an observation of X . It is a weighted average of the whole range of values attainable by X (i.e. X (Î©)). More weight is given to points with higher mass/density function. 8.1.3 Proposition. Let X 1 ,...,X n be random variables. Define Y = g ( X 1 ,...,X n ), for some given realvalued function g . Then (i) (for discrete case) E [ Y ] = X x 1 Â·Â·Â· X x n g ( x 1 ,...,x n ) P ( X 1 = x 1 ,...,X n = x n ); (ii) (for continuous case) E [ Y ] = Z âˆž x 1 =âˆž Â·Â·Â· Z âˆž x n =âˆž g ( x 1 ,...,x n ) f ( x 1 ,...,x n ) dx n Â·Â·Â· dx 1 , where f is the joint pdf of ( X 1 ,...,X n ). This allows us to compute the expectation of any transformation of random variables using their joint distribution. 8.1.4 For constants Î±,Î² and random variables X,Y , E [ Î±X + Î² Y ] = Î± E [ X ] + Î² E [ Y ] . 8.1.5 X â‰¥ â‡’ E [ X ] â‰¥ (trivial!) 8.1.6 X â‰¥ Y â‡’ E [ X ] â‰¥ E [ Y ] (since X Y â‰¥ .) 8.1.7  E X  â‰¤ E  X  (since  X  â‰¥ both X and X .) 47 8.1.8 If X = a on Î©, i.e. X is a nonrandom constant, then E [ X ] = a . (Reason â€” for continuous case, Z xf ( x ) dx = Z af ( x ) dx = a Z f ( x ) dx = a ; discrete case similar.) 8.1.9 X,Y independent random variables â‡’ E [ XY ] = E [ X ] E [ Y ] (Reason â€” for continuous case, Z Z xy f ( x,y ) dxdy = Z Z xy f X ( x ) f Y ( y ) dxdy = Z xf X ( x ) dx Â¶ Z y f Y ( y ) dy Â¶ ; discrete case similar.) 8.1.10 Examples. (i) X âˆ¼ Poisson ( Î» ): E [ X ] = âˆž X k =0 k P ( X = k ) = âˆž X k =1 k Î» k e Î» k ! Â¶ = Î» âˆž X k =1 Î» k 1 e Î» ( k 1)! = Î» âˆž X k =1 P ( X = k 1) = Î» . (ii) X âˆ¼ geometric with success probability p : E [ X ] = âˆž X k =0 k (1 p ) k p = p (1 p ) âˆž X k =0 d dp (1 p ) k = p (1 p ) d dp 1 p Â¶ = 1 p p . (iii) X âˆ¼ negative binomial (no. of failures before k th success with success probability p ): Write X = X 1 + Â·Â·Â· + X k , where each X i âˆ¼ geometric with success probability p . Then E [ X ] = E [ X 1 ] + Â·Â·Â· + E [ X k ] = k (1 p ) p . (iv) N men and M women are seated randomly round a table. Let X = no. of men with a woman seated immediately to their right. Calculate E [ X ]. Solution : It is hard to find mass function of X explicitly. In fact, P ( X = k ) = N + M k N 1 k 1 Â¶ M 1 k 1 Â¶ / N + M N Â¶ . Easier... write X = X 1 + Â·Â·Â· + X N , where X i = ( 1 , i th man has a woman to his right, , otherwise....
View
Full
Document
This note was uploaded on 04/29/2010 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.
 Fall '08
 SMSLee
 Probability

Click to edit the document details