Question c We have E Z 1 see eg the mean of an exponential variable page 279 or

Question c we have e z 1 see eg the mean of an

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Question c) We have E ( Z ) = 1 (see e.g. the mean of an exponential variable page 279 or the distribution summary page 477 or page 480).
1 IMM - DTU 02405 Probability 2003-10-16 BFN/bfn Solution for review exercise 23 (chapter 4) in Pitman We introduce Y = M 3 such that Y has the exponential distribution with mean 2. Question a) E ( M ) = E ( Y + 3) = E ( Y ) + 3 = 5 V ar ( M ) = V ar ( Y + 3) = V ar ( Y ) = 4 where we have used standard rules for mean and variance see eg. page 249, and the result page 279 for the variance of the exponential distribution. Question b) We get the density f M ( m ) of the random variable M is f M ( m ) = 1 2 e 1 2 ( m 3) m > 3 . from the stated assumptions. We can apply the box page 304 to get f X ( x ) = f M ( m ) d x d m = 1 2 e 1 2 (log ( x ) 3) x = e 3 2 2 x x , x > e 3 where X = g ( M ) = e M . Alternatively F X ( x ) = P ( X x ) = P (log ( X ) log ( x ) = P (log ( X ) 3 log ( x ) 3) = P ( Y log ( x ) 3) = 1 e (log ( x ) 3) 2 = 1 e 3 2 x x > e 3 taking derivative we get f X ( x ) = d F X ( x ) d x == e 3 2 2 x x , x > e 3 Question c) We do the calculations in terms of the random variables Y i = M i 3, M i = log ( X i ). Here X i denotes the magnitude of the i ’th earthquake. From Example 3 page 317 we know that the minimum Z of the Y i ’s, Z = min ( Y 1 , Y 2 ) is exponentially distributed with mean 1. P ( M > 4) = P ( Z > 1) = e 1
1 IMM - DTU 02405 Probability 2003-11-1 BFN/bfn Solution for review exercise 25 (chapter 4) in Pitman Question a) We first note that the range of Y is 0 < Y 1 2 . P ( Y y ) = P U 1 2 P ( Y y | U 1 2 + P 1 2 < U P ( Y y | 1 2 < U = 2 P ( U y ) The density is 2 for 0 < y < 1 2 0 elsewhere. Question b) The standard uniform density f ( y ) = 1 for 0 < y < 1, 0 elsewhere. Question c) E ( Y ) = 1 2 0 2 = 1 4 , V ar ( Y ) = ( 1 2 0 ) 2 12 = 1 48
1 IMM - DTU 02405 Probability 2003-11-12 BFN/bfn Solution for review exercise 26 (chapter 4) in Pitman Question a) E ( W t ) = E ( Xe tY ) = E ( X ) E ( e tY ) by the independence of X and Y . We find E ( e tY ) from the definition of the mean. E ( e tY ) = 3 2 1 e ty · 2d y = 2 e t t e t 2 1 Inserting this result and E ( X ) = 2 we get E ( W t ) = 2 2 e t t e t 2 1 Alternatively we could derive the joint density of X and Y to f ( x, y ) = 2(2 x ) 3 e 2 x , 0 < x, 0 < y < 1 where we have used that X has Gamma (4,2) density, and apply the formula for E ( g ( X, Y )) page 349. Question b) Since X and Y are independent we find E ( W 2 t ) E ( W 2 t ) = E ( X 2 ) E ( e tY ) 2 where E ( X 2 ) = V ar ( X ) + ( E ( X )) 2 = 5, see eg. page 481. Next we derive E ( e tY ) 2 = e 2 t t ( e t 1 ) and apply the computational formula for the variance page 261 SD ( W t ) = 5 e 2 t t ( e t 1) 2 2 e t t e t 2 1 2 =
1 IMM - DTU 02405 Probability 2003-11-1 BFN/bfn Solution for review exercise 1 (chapter 5) in Pitman First apply the definition of conditional probability page 36 P Y 1 2 | Y X 2 = P ( Y 1 2 Y X 2 ) P ( Y X 2 ) The joint density of X and Y is the product of the marginal densities since X and Y are independent (page 349). We calculate the denominator using the formula for the probability of a set B page 349 P ( Y X 2 ) = 1 0 1 x 2 1 · 1 · d y d x = 1 0 (1 x 2 )d x = 1 1 3 = 2 3 and the numerator P Y 1 2 Y X 2 = P ( Y X 2 ) P Y < 1 2 Y X 2 Now for the last term P Y < 1 2 Y X 2 = 1 2 0 1 2 x 2 1 · d y d x = 1 2 0 ( 1 2 x 2 )d x = 1 2 1 2 1 3 1 2 1 2 = 1 3 2 Finally we get P Y 1 2 | Y X 2 = 2 3 1 3 2 2 3 = 1 2 4
1 IMM - DTU 02405 Probability 2003-10-17 BFN/bfn Solution for review exercise 20 (chapter 5) in Pitman Question a) This is example 3 page 317. A rederivation gives us P (

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