hw2solutions

hw2solutions - Solutions to Homework Set No. 2 Probability...

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Solutions to Homework Set No. 2 – Probability Theory (235A), Fall 2011 1. (a) Let X be a random variable with distribution function F X and piecewise continuous density function f X . Let [ a,b ] R be an interval (possibly infinite) such that P ( X [ a,b ]) = 1 , and let g : [ a,b ] R be a monotone (strictly) increasing and differentiable function. Prove that the random variable Y = g ( X ) (this is the function on Ω defined by Y ( ω ) = g ( X ( ω )), in other words the composition of the two functions g and X ) has density function f Y ( x ) = f X ( g - 1 ( x )) g 0 ( g - 1 ( x )) x ( g ( a ) ,g ( b )) , 0 otherwise. Solution. It follows that P ( Y [ g ( a ) ,g ( b )]) = 1 from P ( X [ a,b ]) = 1 and the mono- tonicity and continuity of g . Since g is strictly increasing and continuous, F Y ( x ) = P ( Y x ) = P ( X g - 1 ( x )) = F X ( g - 1 ( x )) and f Y ( x ) = d dy F Y ( x ) = d dy F X ( g - 1 ( x )) for x ( g ( a ) ,g ( b )). Thus, by the chain rule, f Y ( x ) = f X ( g - 1 ( x )) g 0 ( g - 1 ( x )) (b) If λ > 0, we say that a random variable has the exponential distribution with parameter λ if F X ( x ) = 0 x < 0 , 1 - e - λx x 0 , and denote this X Exp( λ ) . Find an algorithm to produce a random variable with Exp( λ ) distribution using a random number generator that produces uniform random numbers in 1
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(0 , 1). In other words, if U U (0 , 1), find a function g : (0 , 1) R such that the random variable X = g ( U ) has distribution Exp( λ ). Solution. The density function for X is f X = λe - λx . Use the formula in part (a), find g such that 1 g 0 ( g - 1 ( x )) = λe - λx . You should obtain g ( x ) = - ln(1 - x ) /λ. Now note that if U is uniform in (0 , 1) then so is 1 - U , so this g can be replaced by the slightly simpler g ( x ) = - 1 λ ln x . (c) We say that a non-negative random variable X 0 has the lack of memory property if it satisfies that P ( X t | X s ) = P ( X t - s ) for all 0 < s < t. Prove that exponential random variables have the lack of memory property. Solution. Let X Exp( λ ) . P ( X t | X s ) = P ( X t ) P ( X s ) = e - λt e - λs = e - λ ( t - s ) = P ( X t - s ) (d) Prove that any non-negative random variable that has the lack of memory property has the exponential distribution with some parameter λ > 0. (This is easier if one assumes that
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This note was uploaded on 11/13/2011 for the course STAT 235A taught by Professor Danromik during the Fall '11 term at UC Davis.

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hw2solutions - Solutions to Homework Set No. 2 Probability...

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