nidantasnu

Answered

2022-07-10

I'm trying to deal with a random set $M\subset {\mathbb{R}}^{k}$ of which we can define its size as

$|M|={\int}_{M}dw$

and I need to find a proof for the following expression

$E(|M|)=\int P(w\in M)dw$

But I don't know even how to start so any hint is helpful.

$|M|={\int}_{M}dw$

and I need to find a proof for the following expression

$E(|M|)=\int P(w\in M)dw$

But I don't know even how to start so any hint is helpful.

Answer & Explanation

verzaadtwr

Expert

2022-07-11Added 17 answers

$\begin{array}{r}E(|M|)={\int}_{\mathrm{\Omega}}|M|dP(M)={\int}_{\mathrm{\Omega}}{\int}_{M}d\omega \text{}dP(M)={\int}_{\mathrm{\Omega}}{\int}_{{\mathbb{R}}^{k}}{\chi}_{M}(\omega )d\omega \text{}dP(M)\\ \stackrel{Tonelli}{=}{\int}_{{\mathbb{R}}^{k}}{\int}_{\mathrm{\Omega}}{\chi}_{M}(\omega )dP(M)\text{}d\omega \phantom{\rule{1em}{0ex}}(\ast )\end{array}$

But now

${\chi}_{M}(\omega )=1\iff \omega \in M\iff {\chi}_{{A}_{\omega}}(M)=1$

where ${A}_{\omega}:=\{M:\omega \in M\}$

Thus

$(\ast )\phantom{\rule{1em}{0ex}}={\int}_{{\mathbb{R}}^{k}}{\int}_{\mathrm{\Omega}}{\chi}_{{A}_{\omega}}(M)dP(M)\text{}d\omega ={\int}_{{\mathbb{R}}^{k}}P([M:\omega \in M])d\omega $

But now

${\chi}_{M}(\omega )=1\iff \omega \in M\iff {\chi}_{{A}_{\omega}}(M)=1$

where ${A}_{\omega}:=\{M:\omega \in M\}$

Thus

$(\ast )\phantom{\rule{1em}{0ex}}={\int}_{{\mathbb{R}}^{k}}{\int}_{\mathrm{\Omega}}{\chi}_{{A}_{\omega}}(M)dP(M)\text{}d\omega ={\int}_{{\mathbb{R}}^{k}}P([M:\omega \in M])d\omega $

dikcijom2k

Expert

2022-07-12Added 6 answers

If $P$ is the distribution of $M$ (which you need to be super careful about, you want to define the sigma algebra too), let $\mathcal{P}({\mathbb{R}}^{k})$ be the power set of ${\mathbb{R}}^{k}$ ($P$ is a distribution over $\mathcal{P}({\mathbb{R}}^{k})$), you can write :

$\begin{array}{rl}\mathbb{E}[|M|]& ={\int}_{\mathcal{P}({\mathbb{R}}^{k})}{\int}_{M}d\omega \text{}dP(M)\\ & ={\int}_{\mathcal{P}({\mathbb{R}}^{k})}{\int}_{{\mathbb{R}}^{k}}\mathbf{1}(\omega \in M)d\omega \text{}dP(M)\\ \\ & ={\int}_{{\mathbb{R}}^{k}}{\int}_{\mathcal{P}({\mathbb{R}}^{k})}\mathbf{1}(\omega \in M)dP(M)d\omega \\ & ={\int}_{{\mathbb{R}}^{k}}P(\omega \in M)d\omega \end{array}$

If I were you, I would make sure that you are allowed to exchange the two integrals there.

$\begin{array}{rl}\mathbb{E}[|M|]& ={\int}_{\mathcal{P}({\mathbb{R}}^{k})}{\int}_{M}d\omega \text{}dP(M)\\ & ={\int}_{\mathcal{P}({\mathbb{R}}^{k})}{\int}_{{\mathbb{R}}^{k}}\mathbf{1}(\omega \in M)d\omega \text{}dP(M)\\ \\ & ={\int}_{{\mathbb{R}}^{k}}{\int}_{\mathcal{P}({\mathbb{R}}^{k})}\mathbf{1}(\omega \in M)dP(M)d\omega \\ & ={\int}_{{\mathbb{R}}^{k}}P(\omega \in M)d\omega \end{array}$

If I were you, I would make sure that you are allowed to exchange the two integrals there.

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