For 0 c 1 2c 2 fy c p x 1 c p 1 c x 1

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Unformatted text preview: , at least up to the accuracy of the Gaussian approximation. (e) The value of n needed for p = 0.5 works for any p (the situation is similar to that in part (d)) so n needs to be selected so that 0.01 = 2.58 3.7 0.5(1−0.5) . n This yields n = ( 2.58 )2 (0.5)(1 − 0.5) ≈ 16, 641. 0.01 ML parameter estimation for continuous-type variables As discussed in Section 2.8 for discrete-type random variables, sometimes when we devise a probability model for some situation we have a reason to use a particular type of probability distribution, but there may be a parameter that has to be selected. A common approach is to collect some data and then estimate the parameter using the observed data. For example, suppose the parameter is θ, and suppose that an observation is given with pdf fθ that depends on θ. Section 2.8 suggests estimating θ by the value that maximizes the probability that the observed value is u. But for continuous-type observations, the probability of a specific observation u is zero, for any value of θ. However, if fθ is a continuous pdf f...
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