I'm trying to deal with a random set M ⊂ R k of which we...

nidantasnu

nidantasnu

Answered

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

Expert

2022-07-11Added 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

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

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