Aleah Avery

Answered

2022-11-23

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

Answer & Explanation

Kristen Garza

Expert

2022-11-24Added 13 answers

Lets ${X}_{i}\sim N(\mu ,{\sigma}^{2})$. We want to show $(\sum {X}_{i},\sum {X}_{i}^{2})$ is complete for $(\mu ,{\sigma}^{2})$.

It is enough to show $(\overline{X},S=\sum ({X}_{i}-\overline{X}{)}^{2})$ is complete. We know $\overline{X}$ and $S$ are independent and $\overline{X}\sim N(\mu ,\frac{{\sigma}^{2}}{n})$, $S\sim Gamma(\frac{n-1}{2},2{\sigma}^{2})$.

We should show if $\mathrm{\forall}(\mu ,{\sigma}^{2})$

$E(g(\overline{X},S))=0\Rightarrow P(g(\overline{X},S)=0)=1$

$0=E(g(\overline{X},S))={\int}_{0}^{\mathrm{\infty}}{\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)f(\overline{x})f(s)d\overline{x}\phantom{\rule{thinmathspace}{0ex}}ds$

$=\frac{1}{\mathrm{\Gamma}(\frac{n-1}{2})({\sigma}^{2}{)}^{\frac{n-1}{2}}}{\int}_{0}^{\mathrm{\infty}}({\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)f(\overline{x}){s}^{\frac{n-1}{2}-1}{e}^{-\frac{s}{{\sigma}^{2}}}d\overline{x}\phantom{\rule{thinmathspace}{0ex}})ds$

$=\frac{1}{\mathrm{\Gamma}(\frac{n-1}{2})({\sigma}^{2}{)}^{\frac{n-1}{2}}}{\int}_{0}^{\mathrm{\infty}}({\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)f(\overline{x}){s}^{\frac{n-1}{2}-1}d\overline{x}\phantom{\rule{thinmathspace}{0ex}}){e}^{-\frac{s}{{\sigma}^{2}}}ds$

$=\frac{1}{\mathrm{\Gamma}(\frac{n-1}{2})({\sigma}^{2}{)}^{\frac{n-1}{2}}}{\int}_{0}^{\mathrm{\infty}}(h(s)){e}^{-\frac{s}{{\sigma}^{2}}}ds$

The above is a Laplace transform of $h(s)$, which implies $h(s)=0$, a.e.

So

$0={\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)f(\overline{x})d\overline{x}$

$={\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)\frac{1}{\sqrt{2\pi \frac{{\sigma}^{2}}{n}}}{e}^{-\frac{n}{2{\sigma}^{2}}(\overline{x}-\mu {)}^{2}}d\overline{x}$

$={\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}(g(\overline{x},s)\frac{1}{\sqrt{2\pi \frac{{\sigma}^{2}}{n}}}{e}^{-\frac{n}{2{\sigma}^{2}}{\overline{x}}^{2}}{e}^{-\frac{n}{2{\sigma}^{2}}{\mu}^{2}}){e}^{\frac{n}{2{\sigma}^{2}}2\overline{x}\mu}d\overline{x}$

The above is a Two-sided Laplace transform.

So $g(\overline{x},s)=0$ a.e.

It is enough to show $(\overline{X},S=\sum ({X}_{i}-\overline{X}{)}^{2})$ is complete. We know $\overline{X}$ and $S$ are independent and $\overline{X}\sim N(\mu ,\frac{{\sigma}^{2}}{n})$, $S\sim Gamma(\frac{n-1}{2},2{\sigma}^{2})$.

We should show if $\mathrm{\forall}(\mu ,{\sigma}^{2})$

$E(g(\overline{X},S))=0\Rightarrow P(g(\overline{X},S)=0)=1$

$0=E(g(\overline{X},S))={\int}_{0}^{\mathrm{\infty}}{\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)f(\overline{x})f(s)d\overline{x}\phantom{\rule{thinmathspace}{0ex}}ds$

$=\frac{1}{\mathrm{\Gamma}(\frac{n-1}{2})({\sigma}^{2}{)}^{\frac{n-1}{2}}}{\int}_{0}^{\mathrm{\infty}}({\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)f(\overline{x}){s}^{\frac{n-1}{2}-1}{e}^{-\frac{s}{{\sigma}^{2}}}d\overline{x}\phantom{\rule{thinmathspace}{0ex}})ds$

$=\frac{1}{\mathrm{\Gamma}(\frac{n-1}{2})({\sigma}^{2}{)}^{\frac{n-1}{2}}}{\int}_{0}^{\mathrm{\infty}}({\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)f(\overline{x}){s}^{\frac{n-1}{2}-1}d\overline{x}\phantom{\rule{thinmathspace}{0ex}}){e}^{-\frac{s}{{\sigma}^{2}}}ds$

$=\frac{1}{\mathrm{\Gamma}(\frac{n-1}{2})({\sigma}^{2}{)}^{\frac{n-1}{2}}}{\int}_{0}^{\mathrm{\infty}}(h(s)){e}^{-\frac{s}{{\sigma}^{2}}}ds$

The above is a Laplace transform of $h(s)$, which implies $h(s)=0$, a.e.

So

$0={\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)f(\overline{x})d\overline{x}$

$={\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}g(\overline{x},s)\frac{1}{\sqrt{2\pi \frac{{\sigma}^{2}}{n}}}{e}^{-\frac{n}{2{\sigma}^{2}}(\overline{x}-\mu {)}^{2}}d\overline{x}$

$={\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}(g(\overline{x},s)\frac{1}{\sqrt{2\pi \frac{{\sigma}^{2}}{n}}}{e}^{-\frac{n}{2{\sigma}^{2}}{\overline{x}}^{2}}{e}^{-\frac{n}{2{\sigma}^{2}}{\mu}^{2}}){e}^{\frac{n}{2{\sigma}^{2}}2\overline{x}\mu}d\overline{x}$

The above is a Two-sided Laplace transform.

So $g(\overline{x},s)=0$ a.e.

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