Re: MLE Problem

Ok, I found an article by Saxena and Alam (The Annals of Stats, 10(3),
1982, pp. 1012-1016), who state that there is no closed form expression
for the MLE estimate of the nc chi-square parameter. The give the
expression (X-p)^+ for the MLE estimate they use(where p = degrees of
freedom and X = value of nc chi-squared rv) and ^+ stands for
Max(X-p,0), i.e. the 'positive part of").

One thing I do not understand though is their statement that this
estimator "dominates" the MLE estimator found by another method. What
does it mean for one estimator "to dominate" another? Does it mean that
is unbiased and has lower variance than the other estimator?

TA,

Matt



Brenneman wrote:
David Jones wrote:
Brenneman wrote:
Hi,

I am having a bugger of a time with a problem that doesn't seem too
hard.

I want to find an MLE for the non-centrality parameter of a
non-central chi-squared distbtn.

Normally, this would be a bear, but in this case, the problem is not
too bad, because the distribution has only 2 degrees of freedom, so
my
pdf is:

(i) f(z) = z * exp[-(z^2 + c^2)/2] * I_0(c*z)
where c is my non-centrality param and I_0 is modified Bessel
function
of zero order.

The log function is:
(ii) Ln(f) = ln(z) - (z^2 + c^2)/2 + ln[ I_0(cz) ]

and using the fact that:
(iii) d/dx I_0(x) = I_1(x)

gives us the derivative of the log liklihood function (I'll call F)
as: (iv) F(z) = 1/z - z + c * I_1(cz)/I_0(cz)


? Why are you taking the derivative with respect to z, and not c?

David Jones

Because I did something _very_ stupid: sorry.
But, even if you take the derivative wrt c, the problem still does not
work out, since the derivative of the Log liklelihood function is now:
(v.a) 0 = -c + z * I_1(cz)/I_0(cz)
or
(v.b) c*I_0(cz) = z*I_1(cz)
The power series expansion still will not work using this eqn since the
first few terms in the p.s. expansion of I_0(cz) are:
(vii.a) 1 + (cz)^2/4 + (cz)^4/36 + ...
and that of I_1(cz) is:
(vii.b) (cz)/2 + (cz)^3/16 + (cz)^5/384 + ...
and subtituting the last two expressions into (v.b) still gives an eqn
for which I cannot see a soln.

Matt

.

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