I'm trying to deal with a random set M &#x2282;<!-- ⊂ --> <mrow class="MJX-TeXAtom-O

nidantasnu

nidantasnu

Answered question

2022-07-10

I'm trying to deal with a random set M ⊂ R k of which we can define its size as
| M | = ∫ M d w
and I need to find a proof for the following expression
E ( | M | ) = ∫ P ( w ∈ M ) d w
But I don't know even how to start so any hint is helpful.

Answer & Explanation

verzaadtwr

verzaadtwr

Beginner2022-07-11Added 17 answers

E ( | M | ) = ∫ Ω | M | d P ( M ) = ∫ Ω ∫ M d ω   d P ( M ) = ∫ Ω ∫ R k χ M ( ω ) d ω   d P ( M ) = T o n e l l i ∫ R k ∫ Ω χ M ( ω ) d P ( M )   d ω ( ∗ )
But now
χ M ( ω ) = 1 ⇔ ω ∈ M ⇔ χ A ω ( M ) = 1
where A ω := { M : ω ∈ M }
Thus
( ∗ ) = ∫ R k ∫ Ω χ A ω ( M ) d P ( M )   d ω = ∫ R k P ( [ M : ω ∈ M ] ) d ω
dikcijom2k

dikcijom2k

Beginner2022-07-12Added 6 answers

If P is the distribution of M (It requires extreme caution; you also need to define the sigma algebra), let P ( R k ) be the power set of R k ( P is a distribution over P ( R k )), you can write :
E [ | M | ]  = ∫ P ( R k ) ∫ M d ω   d P ( M )   = ∫ P ( R k ) ∫ R k 1 ( ω ∈ M ) d ω   d P ( M )    = ∫ R k ∫ P ( R k ) 1 ( ω ∈ M ) d P ( M ) d ω   = ∫ R k P ( ω ∈ M ) d ω 
I would make sure that you are permitted to interchange the two integrals there if I were you.

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