hw2sol - Probability with Applications—Spring 2008...

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Unformatted text preview: Probability with Applications—Spring 2008 Homework 2 Solutions for ISYE 2027, Section B (February 20, 2008) Solutions. 1. Suppose n = 5 and we want P ( X = 3). One way we can have X = 3 is by seeing the sequence (exactly, in order) HHTHT, which is the sample space element ( H,H,T,H,T ). By our independence assumption, P ( { ( H,H,T,H,T ) } ) is equal to P ( { H } × { H,T } 4 ) P ( { H,T } × { H } × { H,T } 3 ) ··· P ( { H,T } 4 × { T } ) which is the probability that the first toss is H, multiplied by the probability that the second toss is H, and so on, with the last term being the probability that the fifth toss is T. By our assumptions, this may be found to equal ( 2 3 ) 3 ( 1 3 ) 2 . Similarly, we will get ( 2 3 ) 3 ( 1 3 ) 2 as the probability of any sequence in which exactly three heads appear (as long as the sequence has length 5). Since the number of sequences with exactly three heads is (5 C 3), we have for n = 5 that P ( X = x ) = (5 Cx )( 2 3 ) x ( 1 3 ) 5- x for x ∈ { ,..., 5 } . For the general case where n ∈ { 1 , 2 , 3 ,... } , we reason more abstractly to obtain the pmf: f ( x ) = P ( X = x ) = ( nCx )( 2 3 ) x ( 1 3 ) n- x for x ∈ { ,...,n } 2. Since f is a pdf, it must satisfy R + ∞-∞ f ( x ) dx = 1. In our case, this integral is equal to R-∞ f ( x ) dx + R + ∞ f ( x ) dx = R-∞ dx + R + ∞ ce- λx dx = 0 + c (- 1 λ e- λx ) | x → + ∞ x =0 = c (0- (- 1 λ )) In the last step, note that- 1 λ e- λx approaches 0 as x → + ∞ since we are given that λ > 0. We conclude that c = λ . 3. The cdf is F ( x ) = P ( X ≤ x ), by definition, for all x ∈ R . Here X is continuous with pdf f , so F ( x ) = R x-∞ f ( t ) dt . Since f is defined in a piecewise manner, we should consider two cases separately. In one case, x ≤ 0. Here, R x-∞ f ( t ) dt = R x-∞ dt = 0. In the other case, x > 0. Here, R x-∞ f ( t ) dt is equal to R-∞ f ( t ) dt + R x f ( t ) dt = R-∞ dt + R x λe- λt dt = 0 + (- e- λt ) | t = x t =0 =- e- λx- (- 1) The two cases above account for all x ∈ R . We have F ( x ) = 1- e- λx x > x ≤ 4. Since we have two jointly distributed random variables with joint pdf f , what we want is P ( Y > 2 X ) = RR y> 2 x f ( x,y ) dxdy ....
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This note was uploaded on 04/18/2008 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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hw2sol - Probability with Applications—Spring 2008...

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