Could anyone help prove the following claim? For Z a random variable, choose X , Y

cricafh

cricafh

Answered question

2022-05-24

Could anyone help prove the following claim?
For Z a random variable, choose X , Y appropriately in Holder's inequality to show that
f ( t ) = log ( E | Z t | )
is a convex function on the interval of t where E | X t | < .
I'm confused on the part of choosing appropriate X , Y in Holder's inequality. I tried using only X and Z and rearranging for the log ( E | Z t | ) term but it feels headed in the wrong direction.

Answer & Explanation

Annabella Velez

Annabella Velez

Beginner2022-05-25Added 7 answers

Hint. You want to prove
f ( λ t + ( 1 λ ) s ) λ f ( t ) + ( 1 λ ) f ( s )
for λ [ 0 , 1 ], s , t I (the interval in question). That is
log E | Z | λ t + ( 1 λ ) s λ log E | Z | t + ( 1 λ ) log E | Z | s
Using the properties of the logarithm, this translates to
log E | Z | λ t + ( 1 λ ) s log ( E | Z | t ) λ ( E | Z | s ) 1 λ
or - as log is increasing -
E | Z | λ t + ( 1 λ ) s ( E | Z | t ) λ ( E | Z | s ) 1 λ
Can you see Hölder now?
Another hint: Look at Hölder
E | X Y | ( E | X | p ) 1 / p ( E | Y | q ) 1 / q
and compare this with the right hand side of the last equation: Let 1 / p = λ, then 1 λ = 1 / q.

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