Re: Unbiased MLE estimators

On Jul 27, 11:43 am, Jack Tomsky
<jtom...@xxxxxxxxxxxxx> wrote:
HI All,

In a well known statistical software I saw the
MLE
estimators as the
following:

Uniform (a,b)

a: minimum value - ( b-a )/(n-1) where n is the
sample size.
b: maximun value + ( b-a )/(n-1)

is the bias for these actually ( b-a )/(n-1) ??

The expected value of Xmin is a + [(b-a)/(N+1)].
The bias is E(Xmin)-a = (b-a)/(N+1).

Similarly, the expected value of Xmax is b -
[(b-a)/(N+1)]. The bias is E(Xmax)-b = -(b-a)/(N+1).

Note that the term in the denominator is N+1, not
N-1.

Jack



Exponential(alpha)

alpha = (n-1)/n ) /mean instead of 1/mean ...
is
that correct ?

Is there a good book where I can find the MLE and
the
unbiased MLE
estimators for different distributions?

Thanks..- Hide quoted text -

- Show quoted text -

Thanks a lot for the reply. Do you have a reference
for these
equations ?



I worked it out myself. Here is the derivation.

Let y ~ U(0,1) and let x = a+(b-a)*y. Then x ~ U(a,b).

The transformation is linear (and monotonic).

Let's start with the expectation of ymin = min(y1, ..., yN).

The cdf of ymin is

Fmin(y) = P(ymin<y) = 1-P(all yi>y) = 1-(1-y)^N.

The density of ymin is the derivative.

fmin(y) = N(1-y)^(N-1)

The expected value of ymin is E(ymin) = Int[y*fmin(y) dy], where the integral limits are from y=0 to y=1.

Thus, E(ymin) = Int[y*N(1-y)^(N-1) dy].

Next, I use the beta function identity Int[y^(a-1)* (1-y)^(b-1) dy] = Beta (a,b) = Gamma(a)*Gamma(b)/*Gamma(a+b). Also, if N is an integer, Gamma(N) = (N-1)!.

In the integral expression for E(ymin), a=2 and b=N. We therefore have

E(ymin) = N*Gamma(2)*Gamma(N)/Gamma(N+2) = N*1*(N-1)!/(N+1)! = N!/(N+1)! = 1/(N+1).

Reverting back to the more general uniform, xmin = a+(b-a)*ymin.

E(xmin) = a+(b-a)*E(ymin) = a+(b-a)/(N+1).

The bias is E(xmin)-a = (b-a)/(N+1).

Analogous results follow for xmax by symmetry.

Jack
.

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