Unbiased estimate of cross-product for unbiased vector

Question

Let $g$ be an unbiased estimate of a vector $G$. Can $g$ be used to find an unbiased estimate of the cross product $GG'$? I'm stuck because naively using $gg'$ is a biased estimator, with the (upwards) bias depending on the noise in $g$.

Answer 1

Since you are hunting for unbiasedness in such abstract terms, you should look into the theory of U-statistics. Meanwhile there is a simple way to control for the bias you correctly mention.
Some notation: Let $S$ be your sample (you do have a sample I assume). We assume itis random (if it is considered deterministic, its a different scenario). Since $g$ is un unbiased estimator function of $G$ (assumed fixed), it most probably will have the form

$$g= h(S) = G + f(S)u$$ where $h()$ and $f()$ denote some functions and $u$ denotes independent noise with mean conditional on $S$ equal to zero. Dimensions are conformable. Then we obtain $$E(g) = E\left[G + f(S)u\right] = G+ E\left(E\left [f(S)u|S\right]\right) = G+ E\left(f(S)E\left [u|S\right]\right) = G + 0 = G $$ where we have used the Law of Iterated expectations.

Now $$gg' = \left[G + f(S)u\right]\left[G + f(S)u\right]' = GG' + Gu'f(S)'+f(S)uG'+ f(S)uu'f(S)'$$ $$\Rightarrow E(gg') = GG' + E\Big (EGu'f(S)'|S\Big)+ E\Big (Ef(S)uG'|S\Big)+ E\Big(Ef(S)uu'f(S)'|S\Big)$$ $$=GG' + E\Big (GE(u'|S)f(S)'\Big)+ E\Big (f(S)E(u|S)G'\Big)+ E\Big(f(S)E(uu'|S)f(S)'\Big)$$ $$=GG' + 0 + 0 + E\Big(f(S)E(uu'|S)f(S)'\Big)$$

The last line suggests the bias-adjusted estimator for $GG'$, call it $$H(S) \equiv gg' - f(S)\widehat {E(uu'|S)}f(S)'$$ assuming you can obtain an estimate for $E(uu'|S)$ (and you usually can). Then we have $$ E\Big[H(S)\Big] = E(gg') - E\Big (f(S)\widehat {E(uu'|S)}f(S)'\Big)$$ $$= GG' + E\Big(f(S)E(uu'|S)f(S)'\Big) - E\Big(E\Big (f(S)\widehat {E(uu'|S)}f(S)'\Big)|S\Big)$$ $$= GG' + E\Big( f(S)E(uu'|S)f(S)'\Big) - E\Big(f(S)\Big (\widehat {E(uu'|S)}f(S)'\Big) $$ $$\Rightarrow E\Big[H(S)\Big] = GG'+ E\Big[f(S)\Big(E(uu'|S) - \widehat {E(uu'|S)}\Big)f(S)'\Big] $$

...which will control the bias, although not eliminate it.

If on the other hand the data are deterministic, call the case $S_d$, and the noise is assumed unconditionally zero-mean then following the same route we would obtain

$$E\Big[H(S_d)\Big] = GG'+ f(S_d)\Big( E(uu') - E(\widehat {E(uu')})\Big)f(S_d)'$$ and if the estimator of $E(uu')$ is unbiased, then $H(S_d) \equiv gg' - f(S_d)\widehat {E(uu')}f(S_d)'$ is an unbiased estimator of $GG'$.