# hw2sol - Probability with Applications-Spring 2008 Homework...

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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 ∈ { 0 , . . . , 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 ∈ { 0 , . . . , n } 2. Since f is a pdf, it must satisfy R + -∞ f ( x ) dx = 1. In our case, this integral is equal to R 0 -∞ f ( x ) dx + R + 0 f ( x ) dx = R 0 -∞ 0 dx + R + 0 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 -∞ 0 dt = 0. In the other case, x > 0. Here, R x -∞ f ( t ) dt is equal to R 0 -∞ f ( t ) dt + R x 0 f ( t ) dt = R 0 -∞ 0 dt + R x 0 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 > 0 0 x 0

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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 ) dx dy . First, to completely specify f ( x, y ), we use the fact that RR R 2 f ( x, y ) dx dy = 1. Outside of the region where both x and y are between 0 and 2, we know that f ( x, y ) = 0. Therefore, our double integral over the entire plane reduces to a double integral over the square region where both x and y are between 0 and 2. This is equal to the iterated integral R 2 0 ( R 2 0 f ( x, y ) dy ) dx or R 2 0 ( R 2 0 f ( x, y ) dx ) dy . Arbitrarily choosing the latter formulation, we have
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