References

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Proof of Lemma 6:

The set of values of

as

ranges over the set of probability vectors satisfying the constraint is a convex polygon. Since the modulus function is convex the supremum is achieved at some extreme point. The extreme points consist of those two point distributions putting mass

and

on adjacent vertices of the polygon. If such a distribution puts mass on

and

then the same norm is achieved by putting the same masses on

and 1. It is then easy to check that the norm is the same for the masses

and

as for the masses

and

.

Proof of Lemma 7:

Our proof is via a symmetrization argument. If

is independent of

and has the same distribution as

then

is bounded by

, has mean 0 and variance

. The characteristic function of

is

. Thus (16) follows from

for all

.

The characteristic function

is real and has the expansion

where

lies between

and

. Denote by

the quadratic function on the right hand side of and by

the remainder. We will prove that

for all

; in view of the inequality

this would finish the proof.

If

then

. For

, since

we then have

Since

we see that this inequality holds for all

.

On the other hand the quadratic

is non-negative at

. If

then

almost surely. Hence

almost surely and

for all

. If

then

is non negative between its two roots

and

. Let

denote the event that

. Then the roots satisfy

and

It is now easily checked that

. Hence

for all

. Now use

to check that

for all

. Next

so that

. Thus we see that

implies ().

The case

is handled with the roles of

and

reversed.

It now follows from the lemma that

On the set

we have the Taylor expansion

where

for every

.