ECSE
hw16sols.pdf

# X 1 e ? x 1 1 ? x 1 ? mle 1 x 1 this makes sense

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x 1 = 0 =⇒ e - λ x 1 (1 - λ x 1 ) = 0 =⇒ λ MLE = 1 x 1 This makes sense; given just one sample, the sample mean is x 1 and we know the exponential has mean 1 λ . (b) Now we see N independent samples. The joint distribution (for all x i 0, 0 otherwise) is p x 1 , x 2 ,..., x N | λ = N Y i = 1 p x i | λ = λ N e - λ N i = 1 x i To get the MLE we again maximize this with respect to λ , i.e., d d λ p x 1 , x 2 ,..., x N | λ = d d λ λ N e - λ N i = 1 x i = N λ N - 1 e - λ N i = 1 x i - λ N N X i = 1 x i e - λ N j = 1 x j = 0 which means that λ N - 1 e - λ N i = 1 x i ˆ N - λ N X i = 1 x i ! = 0 or λ MLE = ( 1 N N i = 1 x i ) - 1 . Again, this makes sense; the parameter estimate is the inverse of the sample mean.
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• Spring '11
• Normal Distribution, Estimation theory, Bayesian probability, Prior probability

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