Kassandra Ross

Answered

2022-06-24

I`m trying to find an answer, but i have some problems, help.

Let ${\mathbb{P}}_{\theta}=U[0,\theta ]$.

For $h,{\theta}_{0}>0$ and $Z\sim \mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{1}{{\theta}_{0}}\right)$ I have to show that:

$\frac{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}-h/n}^{n}}{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}}^{n}}\stackrel{d,{\mathbb{P}}_{{\theta}_{0}}^{n}}{\u27f6}{e}^{\frac{h}{{\theta}_{0}}}{\mathbb{1}}_{\{Z\ge h\}}$

I already proved that for ${Z}_{n}=n({\theta}_{0}-max\{{X}_{1},\dots ,{X}_{n}\})$ with ${X}_{1},\dots {X}_{n}\sim {\mathbb{P}}_{{\theta}_{0}}$ holds ${Z}_{n}\stackrel{D}{\to}Z$ and this task seems like I have to prove that the pdf is converging too. I'm not sure which technical steps I need to show this and I'm not sure which kind of convergence is meant by $d,{\mathbb{P}}_{{\theta}_{0}}^{n}$.

Let ${\mathbb{P}}_{\theta}=U[0,\theta ]$.

For $h,{\theta}_{0}>0$ and $Z\sim \mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{1}{{\theta}_{0}}\right)$ I have to show that:

$\frac{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}-h/n}^{n}}{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}}^{n}}\stackrel{d,{\mathbb{P}}_{{\theta}_{0}}^{n}}{\u27f6}{e}^{\frac{h}{{\theta}_{0}}}{\mathbb{1}}_{\{Z\ge h\}}$

I already proved that for ${Z}_{n}=n({\theta}_{0}-max\{{X}_{1},\dots ,{X}_{n}\})$ with ${X}_{1},\dots {X}_{n}\sim {\mathbb{P}}_{{\theta}_{0}}$ holds ${Z}_{n}\stackrel{D}{\to}Z$ and this task seems like I have to prove that the pdf is converging too. I'm not sure which technical steps I need to show this and I'm not sure which kind of convergence is meant by $d,{\mathbb{P}}_{{\theta}_{0}}^{n}$.

Answer & Explanation

Govorei9b

Expert

2022-06-25Added 21 answers

The convergence follows with Radon-Nikodym, Slutzky and continuous mapping theorem.

$\begin{array}{rl}\frac{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}-h/n}^{n}}{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}}^{n}}& =\frac{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}-h/n}^{n}}{\mathrm{d}\lambda}{\left(\frac{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}}^{n}}{\mathrm{d}\lambda}\right)}^{-1}\\ & ={\left(\frac{{\theta}_{0}}{{\theta}_{0}-\frac{h}{n}}\right)}^{n}\frac{{1}_{\{0\le {X}_{i}\le {\theta}_{0}-h/n\text{}\mathrm{\forall}i\}}}{{1}_{\{0\le {X}_{i}\le {\theta}_{0}\text{}\mathrm{\forall}i\}}}\\ & ={\left(\frac{{\theta}_{0}}{{\theta}_{0}-\frac{h}{n}}\right)}^{n}{1}_{\{{Z}_{n}\ge h\}}\\ & \stackrel{\text{Slutzky}}{\to}\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{h}{{\theta}_{0}}\right){1}_{\{Z\ge h\}}\end{array}$

$\begin{array}{rl}\frac{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}-h/n}^{n}}{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}}^{n}}& =\frac{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}-h/n}^{n}}{\mathrm{d}\lambda}{\left(\frac{\mathrm{d}{\mathbb{P}}_{{\theta}_{0}}^{n}}{\mathrm{d}\lambda}\right)}^{-1}\\ & ={\left(\frac{{\theta}_{0}}{{\theta}_{0}-\frac{h}{n}}\right)}^{n}\frac{{1}_{\{0\le {X}_{i}\le {\theta}_{0}-h/n\text{}\mathrm{\forall}i\}}}{{1}_{\{0\le {X}_{i}\le {\theta}_{0}\text{}\mathrm{\forall}i\}}}\\ & ={\left(\frac{{\theta}_{0}}{{\theta}_{0}-\frac{h}{n}}\right)}^{n}{1}_{\{{Z}_{n}\ge h\}}\\ & \stackrel{\text{Slutzky}}{\to}\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{h}{{\theta}_{0}}\right){1}_{\{Z\ge h\}}\end{array}$

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