hw3solutions

hw3solutions - Solutions to Homework Set No. 3 Probability...

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Solutions to Homework Set No. 3 – Probability Theory (235A), Fall 2011 1. Let X be an exponential r.v. with parameter λ , i.e., F X ( x ) = (1 - e - λx )1 [0 , ) ( x ). Define random variables Y = b X c := sup { n Z : n x } (“the integer part of X ”) , Z = { X } := X - b X c (“the fractional part of X ”) . (a) Compute the (1-dimensional) distributions of Y and Z (in the case of Y , since it’s a discrete random variable it is most convenient to describe the distribution by giving the individual probabilities P ( Y = n ) ,n = 0 , 1 , 2 ,... ; for Z one should compute either the distribution function or density function). Solution. b x c ≥ n x n for n Z . So, P ( Y n ) = e - λn and so P ( Y = n ) = P ( Y n ) - P ( Y n + 1) = e - λn (1 - e - λ ) Also, f Z ( t ) = 0 for t < 0 and t 1. For 0 t < 1, F Z ( t ) = P ( Z t ) = X n =0 P ( n X n + t ) = X n =0 ( e - λn - e - λ ( n + t ) ) = 1 - e - λt 1 - e - λ (b) Show that Y and Z are independent. (Hint: Check that P ( Y = n,Z t ) = P ( Y = n ) P ( Z t ) for all n and t .) Solution. P ( Y = n,Z t ) = P ( n X n + t ) = e - λn (1 - e - λt ) = P ( Y = n ) P ( Z t ) 2. (a) Let X,Y be independent r.v.’s. Define U = min( X,Y ), V = max( X,Y ). Find expressions for the distribution functions F U and F V in terms of the distribution functions of X and Y . Solution. By the independence, P ( U u ) = 1 - P ( X > u,Y > u ) = 1 - P ( X > u ) P ( Y > u ) = 1 - (1 - F X ( u ))(1 - F Y ( u )) 1
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Similarly, P ( V v ) = P ( X v ) P ( Y v ) = F X ( v ) F Y ( v ) (b) Assume that X Exp( λ ) ,Y Exp( μ ) (and are independent as before). Prove that
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hw3solutions - Solutions to Homework Set No. 3 Probability...

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