Aleah Avery

2022-11-23

Let ${X}_{1},{X}_{2},...,{X}_{n}$ be a random sample from a $N\left({\theta }_{1},{\theta }_{2}\right)$ distribution. Find the uniformly minimum variance unbiased estimator of $3{{\theta }_{2}}^{2}$.

Kristen Garza

Expert

Lets ${X}_{i}\sim N\left(\mu ,{\sigma }^{2}\right)$. We want to show $\left(\sum {X}_{i},\sum {X}_{i}^{2}\right)$ is complete for $\left(\mu ,{\sigma }^{2}\right)$.
It is enough to show $\left(\overline{X},S=\sum \left({X}_{i}-\overline{X}{\right)}^{2}\right)$ is complete. We know $\overline{X}$ and $S$ are independent and $\overline{X}\sim N\left(\mu ,\frac{{\sigma }^{2}}{n}\right)$, $S\sim Gamma\left(\frac{n-1}{2},2{\sigma }^{2}\right)$.
We should show if $\mathrm{\forall }\left(\mu ,{\sigma }^{2}\right)$
$E\left(g\left(\overline{X},S\right)\right)=0⇒P\left(g\left(\overline{X},S\right)=0\right)=1$
$0=E\left(g\left(\overline{X},S\right)\right)={\int }_{0}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}g\left(\overline{x},s\right)f\left(\overline{x}\right)f\left(s\right)d\overline{x}\phantom{\rule{thinmathspace}{0ex}}ds$
$=\frac{1}{\mathrm{\Gamma }\left(\frac{n-1}{2}\right)\left({\sigma }^{2}{\right)}^{\frac{n-1}{2}}}{\int }_{0}^{\mathrm{\infty }}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}g\left(\overline{x},s\right)f\left(\overline{x}\right){s}^{\frac{n-1}{2}-1}{e}^{-\frac{s}{{\sigma }^{2}}}d\overline{x}\phantom{\rule{thinmathspace}{0ex}}\right)ds$
$=\frac{1}{\mathrm{\Gamma }\left(\frac{n-1}{2}\right)\left({\sigma }^{2}{\right)}^{\frac{n-1}{2}}}{\int }_{0}^{\mathrm{\infty }}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}g\left(\overline{x},s\right)f\left(\overline{x}\right){s}^{\frac{n-1}{2}-1}d\overline{x}\phantom{\rule{thinmathspace}{0ex}}\right){e}^{-\frac{s}{{\sigma }^{2}}}ds$
$=\frac{1}{\mathrm{\Gamma }\left(\frac{n-1}{2}\right)\left({\sigma }^{2}{\right)}^{\frac{n-1}{2}}}{\int }_{0}^{\mathrm{\infty }}\left(h\left(s\right)\right){e}^{-\frac{s}{{\sigma }^{2}}}ds$
The above is a Laplace transform of $h\left(s\right)$, which implies $h\left(s\right)=0$, a.e.
So
$0={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}g\left(\overline{x},s\right)f\left(\overline{x}\right)d\overline{x}$
$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}g\left(\overline{x},s\right)\frac{1}{\sqrt{2\pi \frac{{\sigma }^{2}}{n}}}{e}^{-\frac{n}{2{\sigma }^{2}}\left(\overline{x}-\mu {\right)}^{2}}d\overline{x}$
$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\left(g\left(\overline{x},s\right)\frac{1}{\sqrt{2\pi \frac{{\sigma }^{2}}{n}}}{e}^{-\frac{n}{2{\sigma }^{2}}{\overline{x}}^{2}}{e}^{-\frac{n}{2{\sigma }^{2}}{\mu }^{2}}\right){e}^{\frac{n}{2{\sigma }^{2}}2\overline{x}\mu }d\overline{x}$
The above is a Two-sided Laplace transform.
So $g\left(\overline{x},s\right)=0$ a.e.

Do you have a similar question?