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Unformatted text preview: § 8 Properties of distributions § 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 ∈ X (Ω) x P ( X = x ); (ii) (for continuous case) E [ X ] = Z ∞∞ 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 (Ω)). 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 ··· X x n g ( x 1 ,...,x n ) P ( X 1 = x 1 ,...,X n = x n ); (ii) (for continuous case) E [ Y ] = Z ∞ x 1 =∞ ··· Z ∞ x n =∞ g ( x 1 ,...,x n ) f ( x 1 ,...,x n ) dx n ··· 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 α,β and random variables X,Y , E [ αX + β Y ] = α E [ X ] + β E [ Y ] . 8.1.5 X ≥ ⇒ E [ X ] ≥ (trivial!) 8.1.6 X ≥ Y ⇒ E [ X ] ≥ E [ Y ] (since X Y ≥ .) 8.1.7  E X  ≤ E  X  (since  X  ≥ both X and X .) 47 8.1.8 If X = a on Ω, i.e. X is a nonrandom constant, then E [ X ] = a . (Reason — 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 ⇒ E [ XY ] = E [ X ] E [ Y ] (Reason — 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 ¶ Z y f Y ( y ) dy ¶ ; discrete case similar.) 8.1.10 Examples. (i) X ∼ Poisson ( λ ): E [ X ] = ∞ X k =0 k P ( X = k ) = ∞ X k =1 k λ k e λ k ! ¶ = λ ∞ X k =1 λ k 1 e λ ( k 1)! = λ ∞ X k =1 P ( X = k 1) = λ . (ii) X ∼ geometric with success probability p : E [ X ] = ∞ X k =0 k (1 p ) k p = p (1 p ) ∞ X k =0 d dp (1 p ) k = p (1 p ) d dp 1 p ¶ = 1 p p . (iii) X ∼ negative binomial (no. of failures before k th success with success probability p ): Write X = X 1 + ··· + X k , where each X i ∼ geometric with success probability p . Then E [ X ] = E [ X 1 ] + ··· + 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 ¶ M 1 k 1 ¶ / N + M N ¶ . Easier... write X = X 1 + ··· + X N , where X i = ( 1 , i th man has a woman to his right, , otherwise....
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 Fall '08
 SMSLee
 Probability

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