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Let $ x_1, \ldots, x_m $ be i.i.d. samples drawn from a distribution $P$, and $ y_1, \ldots, y_n $ be i.i.d. samples drawn from a distribution $Q$. Assume that the samples $x_i$ and $y_j$ are independent of each other. Suppose $\bar{X}$ is the uniform minimum variance unbiased estimator (UMVUE) for $\mu_X = E_{X \sim P}[X]$ and $\bar{Y}$ is the UMVUE for $\mu_Y = E_{Y \sim Q}[Y]$.

What is the UMVUE for the parameter $\mu_X \mu_Y$? Is $\bar{X} \bar{Y}$ the UMVUE for $\mu_X \mu_Y$, or is there a better estimator?

amWhy
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milad jalali
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1 Answers1

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The UMVUE is indeed $\overline{X}\overline{Y}$. First, note that $\mathbb{E}[\overline{X}\overline{Y}] = \mathbb{E}[\overline{X}]\mathbb{E}[\overline{Y}] = \mu_X\mu_Y$ by independence, so $\overline{X}\overline{Y}$ is at least an unbiased estimator for $\mu_X\mu_Y$.

To show that $\overline{X}\overline{Y}$ is UMVUE, we will apply the following theorem repeatedly.

Theorem 1: Let $T$ be an unbiased estimator of some parameter $\theta$. Then $T$ will be a UMVUE if and only if $T$ is uncorrelated with all unbiased estimators of zero.

To that end, suppose that $f(X,Y)$ is an unbiased estimator of zero. Let us define the marginal estimators $$f_Y(Y) = \int f(X,Y)\ P(dX),\quad\text{and}\quad g_X(X) = \int \overline{Y}f(X,Y)\ Q(dY).$$ Then we have $$ \mathbb{E}_{P,Q}[\overline{X}\overline{Y}f(X,Y)] = \int\int\overline{X}\overline{Y}f(X,Y)\ P(dX)Q(dY) = \int \overline{X}g_X(X)\ P(dX) = \mathbb{E}_P[\overline{X}g_X(X)]. $$ It follows that $\overline{X}\overline{Y}$ will be UMVUE if we can show that $g_X(X)$ is an unbiased estimator of zero whenever $f(X,Y)$ is. Calculating the expectation, we have $$\mathbb{E}_P[g_X(X)] = \int g_X(X)\ P(dX) = \int\int \overline{Y}f(X,Y)\ P(dX)\,Q(dY) = \int \overline{Y}f_Y(Y)\ Q(dY) = \mathbb{E}_Q[\overline{Y}f_Y(Y)].$$ The last expectation is zero since $\overline{Y}$ is UMVUE by assumption, and $\mathbb{E}_{Q}[f_Y(Y)] = \mathbb{E}_{P,Q}[f(X,Y)] = 0$ so that $f_Y(Y)$ is an unbiased estimator of zero. It therefore follows that $\overline{X}\overline{Y}$ is a UMVUE for $\mu_X\mu_Y$.

EuYu
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