References
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![](_26063_lemma626.gif)
Proof of Lemma 6:
The set of values of
![](_26063_displaymath1842.gif)
as
![](_26063_displaymath1384.gif)
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
![](_26063_displaymath1846.gif)
and
![](_26063_displaymath1848.gif)
on adjacent
vertices of the polygon. If such a distribution puts mass on
![](_26063_displaymath1850.gif)
and
![](_26063_displaymath1852.gif)
then the same norm is achieved by putting the same masses on
![](_26063_displaymath1854.gif)
and 1. It is then easy to check that the norm is the same for
the masses
![](_26063_displaymath1848.gif)
and
![](_26063_displaymath1846.gif)
as for the masses
![](_26063_displaymath1846.gif)
and
![](_26063_displaymath1848.gif)
.
Proof of Lemma 7:
Our proof is via a symmetrization argument. If
![](_26063_displaymath1864.gif)
is independent of
![](_26063_displaymath1866.gif)
and has the same distribution as
![](_26063_displaymath1866.gif)
then
![](_26063_displaymath1870.gif)
is bounded by
![](_26063_displaymath1872.gif)
, has mean 0 and variance
![](_26063_displaymath1874.gif)
.
The characteristic function of
![](_26063_displaymath1876.gif)
is
![](_26063_displaymath1878.gif)
.
Thus (16) follows from
![](_26063_displaymath1880.gif)
for all
![](_26063_displaymath1946.gif)
.
The characteristic function
![](_26063_displaymath1884.gif)
is real and has the expansion
![](_26063_eqnarray636.gif)
where
![](_26063_displaymath1886.gif)
lies between
![](_26063_displaymath1340.gif)
and
![](_26063_displaymath1890.gif)
.
Denote by
![](_26063_displaymath1892.gif)
the quadratic function on the right hand side of
and by
![](_26063_displaymath1894.gif)
the remainder.
We will prove that
![](_26063_equation640.gif)
for all
![](_26063_displaymath1946.gif)
; in view of the inequality
![](_26063_displaymath1898.gif)
this would finish the proof.
If
![](_26063_displaymath1900.gif)
then
![](_26063_displaymath1902.gif)
.
For
![](_26063_displaymath1904.gif)
, since
![](_26063_displaymath1906.gif)
we then have
![](_26063_displaymath1908.gif)
Since
![](_26063_displaymath1910.gif)
we see that this inequality holds for all
![](_26063_displaymath1946.gif)
.
On the other hand the quadratic
![](_26063_displaymath1930.gif)
is non-negative at
![](_26063_displaymath1916.gif)
. If
![](_26063_displaymath1918.gif)
then
![](_26063_displaymath1920.gif)
almost surely. Hence
![](_26063_displaymath1922.gif)
almost
surely and
![](_26063_displaymath1924.gif)
for all
![](_26063_displaymath1926.gif)
. If
![](_26063_displaymath1928.gif)
then
![](_26063_displaymath1930.gif)
is non negative between its two roots
![](_26063_displaymath1932.gif)
and
![](_26063_displaymath1934.gif)
. Let
![](_26063_displaymath1936.gif)
denote the event that
![](_26063_displaymath1938.gif)
. Then the roots satisfy
![](_26063_displaymath1940.gif)
and
![](_26063_displaymath1942.gif)
It is now easily checked that
![](_26063_displaymath1944.gif)
. Hence
![](_26063_equation656.gif)
for all
![](_26063_displaymath1946.gif)
. Now use
![](_26063_displaymath1948.gif)
to check that
![](_26063_equation658.gif)
for all
![](_26063_displaymath1950.gif)
. Next
![](_26063_displaymath1952.gif)
so that
![](_26063_displaymath1954.gif)
.
Thus we see that
![](_26063_displaymath1946.gif)
implies
(
).
The case
![](_26063_displaymath1958.gif)
is handled with the roles of
![](_26063_displaymath1960.gif)
and
![](_26063_displaymath1962.gif)
reversed.
It now follows from the lemma that
![](_26063_eqnarray661.gif)
On the set
![](_26063_displaymath1326.gif)
we have the Taylor expansion
![](_26063_displaymath1438.gif)
where
![](_26063_displaymath1440.gif)
for every
![](_26063_displaymath1322.gif)
.
Next: About this document
Up: Maximum likelihood estimation of
Previous: Discussion
Richard Lockhart
Thu Oct 26 23:26:04 PDT 1995