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Unformatted text preview: Statistics 403 Problem Set 2 Due in lab on Friday, September 24th 1. We are given the following probability distribution, where p is an unspecified proba bility: x1 1 P ( X = x ) 0.3 p . 7 p (a) Find the value of p that maximizes the variance of this distribution. Solution: The expected value is (0 . 3 × 1) + ( p × 0) + ((0 . 7 p ) × 1) = 0 . 4 p. The direct way to get the variance is from the definition: . 3 · ( 1 (0 . 4 p )) 2 + p · (0 (0 . 4 p )) 2 + (0 . 7 p ) · (1 (0 . 4 p )) 2 If you expand this out and simplify you will get p 2 . 2 p + 0 . 84 You can also use the identity var( X ) = E ( X 2 ) ( EX ) 2 : E ( X 2 ) = ( 1) 2 · . 3 + 0 2 · p + 1 2 · (0 . 7 p ) = 1 p Thus the variance is 1 p (0 . 4 p ) 2 = p 2 . 2 p + 0 . 84 This is a concave parabola with a global maximum value at p = . 1. Since a negative number is not possible as a probability, we can conclude that there is no global maximum within the allowed range, which is the interval (0 , . 7). Thus we only need to check the endpoints of this interval, which gives us values 0 . 84 (when p = 0) and 0.21 (when p = 0 . 7). So 0.21 is the minimum possible variance. Note that the minimum variance occurs when the point x = 1 is eliminated from the sample space. So we see that the variance is minimized when we squeeze as much probability mass as possible into a small range. 1 (b) Find the value of p that minimizes the variance of this distribution. Solution: Following the work from part (a), 0 . 84 is the maximum possible vari ance. This occurs when when p = 0, which eliminates x = 0 from the sample space. Thus the variance is maximized when we stretch the probability mass overspace....
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This note was uploaded on 02/06/2012 for the course STAT 403 taught by Professor Kerbyshedden during the Winter '12 term at University of MichiganDearborn.
 Winter '12
 KerbyShedden
 Statistics, Probability, Variance

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