review_discrete

# Now we can write expression 2 as n r r 1n nn n

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Unformatted text preview: ￿ N − r ￿ E (X ) = ￿ N ￿ n = n At this point we have to use a result from combinatorial analysis from the beginning of the course (hw1 ex6). Here is the result: ￿n + m￿ r = ￿n￿￿m￿ 0 r + Or we can write it as: ￿n + m￿ r = ￿n￿￿ m ￿ r−1 1 r ￿ ￿n￿￿ m ￿ k=0 r−k k + ··· ￿n￿￿m￿ r 0 . . Therefore the summationof expression (1) above is equal to E (X ) = r ￿N ￿ n n−1 ￿ ￿r − 1￿￿ N − r ￿ n−y−1 y y =0 = r ￿N ￿ n ￿ N −1￿ n−1 ￿ N − 1￿ . Now we can write expression (1) as: = n−1 r (N − n)!n! (N − 1)! rn(n − 1)!(N − 1)! = N! (N − n)!(n − 1)! N (N − 1)!(n − 1)! n E (X ) = r . N ⇒ To ﬁnd the variance we start with EX (X − 1): EX (X − 1) = r(r − 1) ￿N ￿ n n ￿ x=2 n ￿ x=0 ￿r ￿￿N −r￿ n ￿ x n− ￿N ￿ x = x(x − 1) x(x − 1) n x=2 ￿N − r￿ r(r − 1) (r − 2)! = ￿N ￿ (r − x)!(x − 2)! n − x n EX (X − 1) = r (r − 1) ￿N ￿ n r! (r − x)!x! ￿N −r￿ n− ￿N x ￿ n ￿ ￿ r − 2 ￿N − r￿ x−2 x=2 n−x ￿ ￿r − 2￿￿ N − r ￿ y n−y−2 Using the same result as before, the summation of expression (2) is equal to EX (X − 1) = r (r − 1) ￿N ￿ n n−2 ￿ ￿r − 2￿￿ N − r ￿ y =0 y n−y−2 ⇒ Let y = x − 2 n−2 y =0 = n = (2) ￿N −2￿ n−2 r (r − 1) ￿N − 2￿ ￿N ￿ n−2 . Now we can write expression (2) as: = n r (r − 1)(N − n)!n! (N − 2)! r(r − 1)n(n − 1)(n − 2)!(N − 2)! = N! (N − n)!(n − 2)! N (N − 1)(N − 2)!(n − 2)! r (r − 1)n(n − 1) r(r − 1)n(n − 1) EX (X − 1) = ⇒ EX 2 − EX = N (N − 1) N (N − 1) r (r − 1)n(n − 1) n 2 EX = +r . N (N − 1) N 18 ⇒ ⇒ Therefore the variance is: σ 2 = EX 2 − µ2 = nr r(r − 1)n(n − 1) n n r (r − 1)n(n − 1)N + rnN (N − 1) − r2 n2 (N − 1) +r − (r )2 = N (N − 1) N N N 2 (N − 1) (r − 1)(n − 1)N + N (N − 1) − rn(N − 1) N rn − N r − N n + N + N 2 − N rn − N + rn = nr 2 (N − 1) N N 2 (N − 1) nr = = −N r − N n + N 2 + rn N (N − r) − n(N − r) nr(N − r)(N − n) = nr ⇒ σ2 = N 2 (N − 1) N 2 (N − 1) N 2 (N − 1) Example 12 We know that the probability mass function of a negative binomial random variable is: P ( X = x) = ￿ x − 1￿ r−1 pr (1 − p)x−r , x = r, r + 1, r + 2, · · · where X is the number of trials required until r successes occur. Therefore the expected value of X is: ∞ ￿ ￿ x − 1￿ pr (1 − p)x−r = ￿ x! (x − 1)! pr (1 − p)x−r = pr (1 − p)x−r (x − r)!(r − 1)! (x − r)!(r − 1)! = µ = E (X ) = x r−1 x=r ∞ ￿ x=r ∞ x x=r Factor outside of the summation r p ∞ ￿ x=r x! r pr+1 (1 − p)x−r = (x − r )!r! p ∞ ￿ ￿ x￿ r x=r r : p pr+1 (1 − p)x−r = Now let y = x + 1 : ∞ r ￿ ￿ y − 1￿ p y =r +1 r pr+1 (1 − p)y−r−1 . We observe now that Y is a negative binomial random variable that represents the number of trials required until r + 1 successes r occur. Therefore the summation is equal to 1 (it is the sum of all the probabilities). And we have that µ = E (X ) = p . To ﬁnd the variance we start with EX (X + 1) whic is equal to: ∞ ￿ ∞ ￿ = ￿ (x − 1)! (x + 1)! pr (1 − p)x−r = pr (1 − p)x−r (x − r )!(r − 1)! (x − r )!(r − 1)! = x(x + 1) x=r ∞ x(x + 1) x=r ￿ x − 1￿ pr (1 − p)x−r = EX (X + 1) = r−1 x=r Factor outside of the summation r (r + 1) p2 ∞ ￿ x=r (x + 1)! r (r + 1) pr+2 (1 − p)x−r = (x − r )!(r + 1)! p2 ∞ ￿ ￿x + 1￿ x=r r+1 r (r + 1) : p2 pr+2 (1 − p)x−r = Now let y = x + 2 : ∞ r(r + 1) ￿ ￿y − 1￿ p2 y =r +2 r+1 pr+2 (1 − p)y−r−2 We observe now that Y is a negative binomial random variable that represents the number of trials required until r + 2 successes r (r +1) occur. Therefore the summation is equal to 1 (it is the sum of all the probabilities). And we have that EX (X + 1) = p2 . We can ﬁnd EX 2 : EX (X + 1) = r (r + 1)...
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## This note was uploaded on 10/04/2012 for the course STATISTICS 100a taught by Professor Cristou during the Spring '10 term at UCLA.

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