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MSE of sample standard deviation

In a previous post, we have discussed the MSE of sample variance. One could argue that we should compute the MSE of the sample standard deviation as sample s.d. has the same unit as the measurements.

Let Sn be the sample s.d. and σ be the true s.d., the goal is to compute

MSE(Sn)=E(Snσ)2=E(S2n)2σE(Sn)+σ2=2σ22σE(Sn).

To compute E(Sn), first, we consider Taylor’s expanding g(x)=x about x=σ2, we have g(x)=σ+12σ(xσ2)18σ3(xσ2)2+R(x), where R(x)=(18˜σ318σ3)(xσ2)2 for some ˜σ between x and σ.

Let κ=E(Xμ)4/σ4 be the kurtosis. It could be shown that E[n(S2nσ2)]2σ4(κ1) and nER(S2n)0 (and the proofs are beyond the discussion of this post. One could consider the fact that n(S2nσ2) converges to N(0,σ4(κ1))).

Thus, E(Sn)=Eg(S2n)=σ+12σE(S2nσ2)18σ3E(S2nσ2)2+o(n1). =σσ8(κ1n)+o(n1).

Therefore, MSE(Sn)=σ24(κ1n)+o(n1).

Similar calculation could be done for ˆσn=(XiˉX)2n,

MSE(ˆσn)=E[n1nSnσ]2=E[(112n+o(n1))Snσ]2=σ24(κ1n)+o(n1).

Unfortunately, MSE of Sn and ˆσn match at the first order. We will need to consider higher order expansion next time.

Randy Lai
Randy Lai
Data Scientist