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Homework 9 Solution

Homework 9 Solution - EE 351K PROBABILITY& RANDOM...

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Unformatted text preview: EE 351K PROBABILITY & RANDOM PROCESSES FALL 2011 Instructor: Sujay Sanghavi [email protected] Homework 9 Solution Problem 1 The parameter of an exponential random variable has to be estimated from one sample. What is the ML estimator? Is it unbiased? Sol : Let θ be the parameter of exponential distribution. Then the likelihood of single sample X = x is f X ( x ; θ ) = θe- θx . We want to maximize this probability with respect to θ , then from d dθ θe- θx = e- θx − θxe- θx = 0 , we have ˆ θ ML = arg max θ p X ( x ; θ ) = 1 /x. So the estimator is ˆ Θ ML = 1 X . By definition, it is easy to get E [ ˆ Θ] = E [ 1 X ] = ∫ ∞ 1 x θe- θx dx. Also note that ∫ 1 x e- θx dx = ln x − θx 1! + θ 2 x 2 2 · 2! − θ 3 x 3 3 · 3! + ··· . So E [ ˆ Θ] ̸ = θ , which implies the estimator is biased. Problem 2 Alice models the time that she spends each week on homework as an exponentially distributed random variable with unknown parameter θ . Homework times in different weeks are independent. After spending 10, 14, 18, 8, and 20 hours in the first 5 weeks of the semester, what is her ML estimator of θ ? Sol : The likelihood can be expressed as f X ( x 1 ,x 2 ,x 3 ,x 4 ,x 5 ; θ ) = θe- 10 θ · θe- 14 θ · θe- 18 θ · θe- 8 θ · θe- 20 θ = θ 5 e- 70 θ , so ˆ θ ML = arg max θ p X ( x 1 ,x 2 ,x 3 ,x 4 ,x 5 ; θ ) = 1 14 . Problem 3 X is known to be a uniform random variable, with range [ − a,a ] . However, the parameter a ≥ is unknown, and has to be estimated from n samples. (a) What is the ML estimator?...
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Homework 9 Solution - EE 351K PROBABILITY& RANDOM...

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