For non-negative data the sample mean is not smaller than its standard error.(1) Let X 1 , X 2 , … , X n be a random sample from a population with non-negative values. Then show that X ¯ ≥ S / n , where S 2 = [ ∑ i = 1 n ( X i − X ¯ ) 2 ] / ( n − 1 ) .I have not seen this inequality stated before. It is not difficult to prove directly, but may be implied by some more general result I am overlooking.(2) Also use this inequality to find a counterexample, showing that the sample mean and variance are not independent for exponential (or beta) data; perhaps use n = 4 for simplicity. [of course, X ¯ and S are independent for normal data.]

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For non-negative data the sample mean is not smaller than its standard error.(1) Let       X    1    ,      X    2    ,  …  ,      X    n   be a random sample from a population with non-negative values. Then show that             X      ¯        ≥  S      /        n    , where       S    2    =  [      ∑          i      =      1        n    (      X    i    −            X      ¯            )    2    ]      /    (  n  −  1  )  .I have not seen this inequality stated before. It is not difficult to prove directly, but may be implied by some more general result I am overlooking.(2) Also use this inequality to find a counterexample, showing that the sample mean and variance are not independent for exponential (or beta) data; perhaps use   n  =  4 for simplicity. [of course,             X      ¯       and S are independent for normal data.]

Camden Stanton · Accepted Answer

Step 1For the counterexample: picking basically arbitrary numbers, if   n  =  4 and   S  ≥  2 then       X    ¯    ≥  1. So do you have  P  (  S  ≥  2  ∧      X    ¯    &amp;lt;  1  )  =  P  (  S  ≥  2  )  P  (      X    ¯    &amp;lt;  1  )  ?Step 2If not then you have your counterexample. Note that the left side is zero, so it&#039;s enough for both factors on the right side to be positive.Qualitatively speaking, this argument says that       X    ¯   and S are dependent because if S is large then       X    ¯   must also be large.

Nicholas Hunter · Accepted Answer

Step 1Answer to (1).Note that   (  n  −  1  )      S    2    =  ∑      X    i    2    −      1    n    (  ∑      X    i        )    2   follows from the definition of       S    2    .Also, for nonnegative data,   ∑      X    i    2    ≤  (  ∑      X    i        )    2   because the RHS contains nonnegative cross-product terms that the LHS does not.Step 2Then  n  (  n  −  1  )      S    2    =  n  ∑      X    i    2    −  (  ∑      X    i        )    2    ≤  n  (  ∑      X    i        )    2    −  (  ∑      X    i        )    2    =  (  n  −  1  )      n    2                      X        ¯              2    ,which implies (1).

(1) Let X_1,X_2,…,X_n be a random sample from a population with non-negative values. Then show that overlineX >= S/sqrtn, where S^2=[sum _{i=1}^n (X_i-overlineX)^2]/(n-1).

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