Let f : <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struck">R </mrow>

Taniyah Estrada

Taniyah Estrada

Answered question

2022-06-05

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let f : R R be a continuous bounded function, ( X k ) k a sequence of i.i.d random variables such that
x R , f ( x ) = R f ( x + y ) d P X 1 ( y ) .
Let x R , Y k = f ( x + q = 1 k X q ) , F k = σ ( X 1 , . . . , X k ) .

Verify that:
(1) E [ ( Y 1 f ( x ) ) 2 ] E [ ( Y 2 Y 1 ) 2 ]
and deduce that for every y supp P X 1 := { u , r > 0 , P X 1 ( [ y r , y + r ] ) > 0 } , f ( x + y ) = f ( x ) .

One way to prove the first question is to note that
R ( f ( x + y ) f ( x ) ) 2 d P X 1 ( y ) = R ( R ( f ( x + y + u ) f ( x + u ) ) d P X 2 ( u ) ) 2 d P X 1 ( y ) R ( R ( f ( x + y + u ) f ( x + u ) ) 2 d P X 2 ( u ) ) d P X 1 ( y ) E [ ( Y 2 Y 1 ) 2 ]
For part 2 we can deduce that for P X 1 -almost every y R , f ( x + y ) = f ( x ) ,, how to conclude using continuity that f ( x + y ) = f ( x ) for every y supp P X 1 ?
How to deduce part 2 ?

Answer & Explanation

Blaze Frank

Blaze Frank

Beginner2022-06-06Added 18 answers

You have already found that f ( x + y ) = f ( x ) for P X 1 -a.e. y R . Now, let y s u p p P X 1 . Then P X 1 ( y 1 / n , y + 1 / n ) > 0 for every n N . Therefore, there must exist y n ( y 1 / n , y + 1 / n ) such that f ( x + y n ) = f ( x ). Obviously y n y and since f is continuous we have
<math xmlns="http://www.w3.org/1998/Math/MathML" "> f ( x + y ) = lim n f ( x + y n ) = f ( x ) .

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