Suppose we have a population data set X which is partitioned into two subsets A and B, with population variance 3 and 4, respectively. Is it true that the population variance of A is at least 3 (i.e. min{3,4})?

hogwartsxhoe5t

hogwartsxhoe5t

Answered question

2022-10-30

Minimum of the variance of a data set given the variances of subsets
Suppose we have a population data set X which is partitioned into two subsets A and B, with population variance 3 and 4, respectively. Is it true that the population variance of A is at least 3 (i.e. min{3,4})?

Answer & Explanation

dippoliticsxu

dippoliticsxu

Beginner2022-10-31Added 9 answers

Step 1
Let n 1 = | A | , n 2 = | B | , and n = | X | = n 1 + n 2 . Throughout I will use a, b, and x to denote a generic element in the set A, B, and X, respectively. The summations will then be understood as summing over all elements in the corresponding set. Also, a ¯ denotes the arithmetic mean of all elements in A, and similarly for b ¯ and x ¯ .
Proposition: 1 n ( x x ¯ ) 2 min { 1 n 1 ( a a ¯ ) 2 , 1 n 2 ( b b ¯ ) 2 }
Proof: WLOG, assume 1 n 2 ( b b ¯ ) 2 1 n 1 ( a a ¯ ) 2 .
Then we need to show that 1 n ( x x ¯ ) 2 1 n 1 ( a a ¯ ) 2 . First, we prove the following lemma:
Lemma: ( x x ¯ ) 2 ( a a ¯ ) 2 + ( b b ¯ ) 2 .
Indeed, applying the Cauchy-Schwarz inequality to the n-vectors ( a ¯ , a ¯ , . . . , a ¯ , b ¯ , b ¯ , . . . , b ¯ ), where the first n 1 components are a ¯ and the last n 2 components are b ¯ , and (1,1,...,1), we get
( n 1 a ¯ + n 2 b ¯ ) 2 ( n 1 a ¯ 2 + n 2 b ¯ 2 ) n ,
or
n 1 a ¯ 2 + n 2 b ¯ 2 n ( n 1 a ¯ + n 2 b ¯ n ) 2 = n x ¯ 2 ,
or
n x ¯ 2 n 1 a ¯ 2 n 2 b ¯ 2 0 ,
so
n x ¯ 2 n 1 a ¯ 2 n 2 b ¯ 2 2 ( n x ¯ 2 n 1 a ¯ 2 n 2 b ¯ 2 ) .
Using n x ¯ 2 = x ¯ x (and similarly for n 1 a ¯ 2 and n 2 b ¯ 2 and rearranging the terms, we get
2 x ¯ x + n x ¯ 2 2 a ¯ a + n 1 a ¯ 2 2 b ¯ b + n 2 b ¯ 2 .
Since x 2 = a 2 + b 2 , we obtain
x 2 2 x ¯ x + n x ¯ 2 a 2 2 a ¯ a + n 1 a ¯ 2 + b 2 2 b ¯ b + n 2 b ¯ 2 .
Hence, ( x x ¯ ) 2 ( a a ¯ ) 2 + ( b b ¯ ) 2
Step 2
Now, with the Lemma and the assumption 1 n 2 ( b b ¯ ) 2 1 n 1 ( a a ¯ ) 2 in mind, we get
n 1 ( x x ¯ ) 2 n 1 ( a a ¯ ) 2 + n 1 ( b b ¯ ) 2
and therefore
1 n ( x x ¯ ) 2 1 n 1 ( a a ¯ ) 2 .

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