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Let $X=(X_{1},...,X_{n})$ and $Y=(Y_{1},...,Y_{n})$ be independent simple samples from the distributions of $\mathcal{N}(m_{x},\sigma ^2)$ and $\mathcal{N}(m_{y},\sigma ^2)$, respectively. Which of the following two estimators: $T_{1}(X,Y) = \overline{X} \ \overline{Y}$, $T_{2} (X,Y)= \frac{1}{n} \sum \limits_{i=1}^{n} X_{i}Y_{i}$ is a better estimator for the parameter $t=m_{x}m_{y}$?
$(\overline{X} = \frac{1}{n}\sum \limits_{i=1}^{n}X_{i})$
. . .

I look forward to guidance. I know that we definitely need to calculate some expected values but I don't know what. How to start?

Gackt
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1 Answers1

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Both estimators are unbiased and consistent. So, in order to compare which estimator is better, we need a metric. I propose the variance of estimator.

According to CLT and because $X,Y$ are independant (hence, $cov(X,Y) = 0$ ), we have $$\sqrt{n}\left( \begin{pmatrix} \bar{X} \\ \bar{Y} \end{pmatrix} -\begin{pmatrix} m_x \\ m_y \end{pmatrix} \right) \xrightarrow{n \to +\infty} \mathcal{N}\left(0,\begin{pmatrix} \sigma^2 & 0 \\ 0 & \sigma^2 \end{pmatrix} \right) \tag{1}$$ From $(1)$ we can apply the Multivariate delta method, with $h(x,y) = xy$ $$\sqrt{n}(T_1(X,Y)- m_xm_y) \xrightarrow{n \to +\infty} \mathcal{N}(0,m_y^2\sigma^2+m_x^2\sigma^2) $$

And we know that $$\sqrt{n}(T_2(X,Y)- m_xm_y) \xrightarrow{n \to +\infty} \mathcal{N}(0,V(XY)) $$

We notice that $(m_x^2+m_y^2)\sigma^2 < V(XY) = E(X^2Y^2)=(\sigma^2+m_x^2)(\sigma^2+m_y^2) - m_x^2m_y^2$, we can conclude that $T_1$ is better than $T_2$

NN2
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