Minimum of the variance of a data set given the variances of subsetsSuppose 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})?

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dippoliticsxu · Accepted Answer

Step 1Let       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 2Now, 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    .

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})?

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