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HW2SOLUTIONS

HW2SOLUTIONS - ENGRD 270 Summer 2007 PROBLEM SET 2...

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ENGRD 270 Summer 2007 PROBLEM SET 2 – Solutions Prepared by Shirshendu. Comments in italics are by Nikolay. Part 1 (1).1 First we sort the data as x (1) x (2) . . . x (10) . If x 11 x (1) , then the level 0.2 trimmed mean will be ( x (1) + · · · + x (9) ) / 9, which is constant as a function of x 11 . Sim- ilarly, for x 11 x (10) , the level 0.2 trimmed mean will be ( x (2) + · · · + x (10) ) / 9, which is again constant as a function of x 11 . For x (1) < x 11 < x (10) , the level 0.2 trimmed mean will be ( x (2) + · · · + x (9) + x 11 ) / 9, which is linear in x 11 . Hence the graph of the trimmed mean against x 11 will look like the following figure. (1).2 The trimmed mean is more robust than the sample mean. If there are a few obser- vations which are very small or very large compared to the rest of the data, then the sample mean will be highly influnced by that, whereas the trimmed mean will not be affected that much. For example, we see in the following figure, the trimmed mean remains constant when one observation is very small or very large. But the sample mean always decreases if one observation decreases keeping others fixed and it increases when one observation increases keeping others fixed. (1).2 For α = . 99, the sample trimmed mean is the sample median (for this data set but not always). 1

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0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 x (1) x (10) x 11 Level 0.2 sample trimmed mean 2
(2) All the set identities are valid. (2).1 Suppose x ( AB ) 0 . Then x is not in both of A and B , i.e. either x is not in A or x is not in B . So either x A 0 or x B 0 . Thus x A 0 S B 0 . On the other hand, suppose x A 0 S B 0 . Then either x A 0 or x B 0 . If x A 0 , x is not in A and so x cannot be in AB . If x B 0 , then x is not in B and so x cannot be in AB . Hence x ( AB ) 0 . So we conclude ( AB ) 0 = A 0 S B 0 . (2).2 Suppose x A T ( B S C ). Then x A and x B S C . As x B S C , either x B or x C . If x B , then x AB and if x C , then x AC . So, either x AB or x AC . Hence x ( AB ) S ( AC ). On the other hand, suppose x ( AB ) S ( AC ). Then either x AB or x AC . If x AB , then x A and x B , and in that case we can say x A T ( B S C ) as B ( B S C ). Similarly in the other case, if x AC , then x A and x C , and we can say that x A T ( B S C ), as C ( B S C ). So we conclude A T ( B S C ) = ( AB ) S ( AC ). (2).3 Suppose x A S ( BC ). Then either x A or x BC . If x A , then x ( A S B ) and x ( A S C ), as A is subset of both A S B and A S C , and hence in this case x ( A S B ) T ( A S C ). In the other case, if x BC , then x B and x C . As B i s a subset of A S B and C is a subset of A S C , we see that x A S B and x A S C . Hence in this case also x ( A S B ) T ( A S C ). On the other hand, suppose x ( A S B ) T ( A S C ). Then x A S B and x A S C . Now if x A , we see x A S ( BC ), as A is a subset of A S ( BC ). Lastly if x A c , then x A S B will imply x B and x A S C will imply x C . Hence, x BC A S ( BC ).

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HW2SOLUTIONS - ENGRD 270 Summer 2007 PROBLEM SET 2...

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