Bias – p. 2

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Subsampling Bias (Continued)

This page is for those who like math with more rigor.

The theorems given below provide bounds for the bias of a sample chosen according to Gy’s criterion.

Definition   Suppose a lot to be sampled and tested for an analyte is com­posed of N frag­ments. A sample from this lot is defined to be a random non­empty sub­set of the N frag­ments. In other words a sample is a random variable whose pos­sible values are non­empty sub­sets of the N frag­ments. (Note that the term random sample has a dif­fer­ent mean­ing.) A sample is correct if each frag­ment in the lot has the same prob­abil­ity of being included in the sample.
Notation   Index the frag­ments of the lot using the set L = {1, 2, …, N}. For any integer j ∈ L, let mj denote the mass of the j th frag­ment, Aj the mass of the critical com­po­nent (analyte) in the j th frag­ment, and aj the critical con­tent (mass frac­tion of analyte) in the j th frag­ment (aj = Aj / mj). The frag­ment masses, mj, are assumed to be known, but the masses of critical com­po­nent, Aj, and critical con­tents, aj, are un­known. In prob­lems where Aj and aj are allowed to vary, Aj will be treated as a func­tion of aj and mj, which are con­sidered more funda­mental (Aj = aj mj).

For any non­empty sub­set G ⊆ L, identify G with the collec­tion of frag­ments indexed by the elements of G. For example if G = {1, 2, 3}, then identify G with the collec­tion that con­sists of the 1st, 2nd, and 3rd frag­ments in the lot. Also, for any non­empty subset G ⊆ L, let

mG = 
∑  mj
 j ∈ G  
     
AG = 
∑  Aj
 j ∈ G  
     
aG = 
AG
mG

In particular mL denotes the total mass of the lot, AL denotes the mass of critical com­po­nent in the lot, and aL denotes the critical con­tent of the lot. Further­more, if S denotes a sample from the lot, then

mS denotes the mass of sample S,
AS denotes the mass of the critical component in sample S, and
aS denotes the critical content of sample S.

In this case mS, AS, and aS are numerical random variables.
Theorem A.1   Let S be a correct sample from lot L. Then

E(AS)
E(mS)
  = aL .

Proof: Since S is correct, there is a real number p with 0 < p ≤ 1, such that Pr[ j ∈ S ] = p for j = 1, 2, …, N. For any event F, let IF denote the random variable whose value is 1 if F occurs and 0 if F does not occur. So, for example, if j ∈ L, then I[ j ∈ S] equals 1 if frag­ment j belongs to sample S and it equals 0 other­wise. Then

mS  = 
N  
∑  I[ j ∈ S] × mj
 j = 1  
   and    AS  = 
N  
∑  I[ j ∈ S] × Aj
 j = 1  

For any event F, the expected value of IF equals the probability of F. So, if j ∈ L, then E(I[ j ∈ S]) = Pr[ j ∈ S ] = p. So,

E(mS) = 
N  
∑  E(I[ j ∈ S] × mj) = 
 j = 1  
N  
∑  p mj = p mL
 j = 1  
and
E(AS) = 
N  
∑  E(I[ j ∈ S] × Aj) = 
 j = 1  
N  
∑  p Aj = p AL
 j = 1  

So,

E(AS)
E(mS)
 = 
 p AL 
 p mL 
  = 
AL
mL
  = aL .


A stronger result can also be proved. It can be shown that E(AS) / E(mS) = aL for all pos­sible values of a1, a2, …, aN if and only if S is correct.

Note that the sampling bias is a bias in the mass fraction of analyte in the sample, which is defined by aS = AS / mS. So, a sample S is unbiased if and only if E(AS / mS) = aL. Unfortunately, the mean of the quotient, E(AS / mS), is not necessarily equal to the quotient of the means, E(AS) / E(mS). If one measured the total mass of analyte in a correct sample, AS, and divided it by the expected mass of the sample, E(mS), rather than the actual mass, the result would be unaffected by sampling bias; however, this is not typically done, and in most cases it would not be desirable anyhow, because the elimination of a rather small bias would not be worth the increase in variability that would occur.

One may consider the sampling bias to be negligible if it is a small fraction of the standard devia­tion of aS, and the following corollary to Theorem A.1 shows that this is true when­ever the rela­tive stand­ard devia­tion of the sample mass, mS, is small.

Corollary A.1.1   Assume S is a correct sample. Then

Bias(aS) = −Cov(aS, mS) / E(mS) = −σ(aS) × RSD(mS) × ρ(aS, mS)
and
| Bias(aS) | ≤ σ(aS) × RSD(mS)

where “RSD” denotes relative standard deviation (coefficient of variation).

Proof: First use the fact that aL = E(AS) / E(mS) to derive the follow­ing equations.

Bias(aS)  =  E(aS) − aL
 = 
E(aS) − 
E(AS)
E(mS)
 = 
E(aS) − 
E(aS mS)
E(mS)
 = 
E(aS) − 
E(aS) E(mS) + Cov(aS, mS)
E(mS)
 = 
− 
Cov(aS, mS)
E(mS)
 = 
− 
 ρ(aS, mS) × σ(aS) × σ(mS) 
E(mS)
 =  −σ(aS) × RSD(mS) × ρ(aS, mS)

Then, since | ρ(aS, mS) | ≤ 1, it follows that | Bias(aS) | ≤ σ(aS) × RSD(mS).

Note that a large value for RSD(mS) does not nec­es­sarily imply a large sampling bias, because aS and mS may be only weakly correlated.

Theorem A.2   Assume S is a sample chosen in a manner such that the mass of the sample always falls between (1 − δ) × M and (1 + δ) × M for specified values of M and δ. If Pr[ j ∈ S ] = M / mL for all j ∈ L, then
(1 − δ) × E(aS) ≤ aL ≤ (1 + δ) × E(aS) .


Given the premise of Theorem A.2, it can also be shown that if S is un­biased for all pos­sible values of a1, a2, …, aN, then for all j ∈ L,

(1 − δ) × 
M
mL
  ≤ Pr[ j ∈ S ] ≤  (1 + δ) × 
M
mL
  .

So, if the mass of the sample is not allowed to vary much, one can en­sure zero sampling bias only if all the frag­ments have nearly the same selec­tion probability.

    ▲     26 Jul 2005
Keith