Re: likelihood greater than 1, beginner

"leo nidas" <bleonidas25@xxxxxxxx> wrote in message
<ftss6u$8gv$1@xxxxxxxxxxxxxxxxxx>...
Rune Allnor <allnor@xxxxxxxxxxxx> wrote in message
It might be that you compute the Probability Density
Function, PDF. The PDF f(x) below can take any non-
negative
value as long a its integral P equals one:

inf
P = integral f(x) dx = 1
-inf

Rune



I am sorry Rune if I am still missing something and thank
you very much for replying my message.


The likelihood function L(&#946;,&#963;^2) is given by the following
right?:

L(&#946;,&#963;^2)=(2&#960;&#963;^2)^(-0.5*n)*exp(-1/(2&#963;^2)*((Y-X&#946;)'*(&#933;-&#935;&#946;)))

Solving dL/d&#946;=0 and dlogL/d&#963;^2=0 we obtain betahat=(X'X)^-
1*X'Y and &#963;hat^2=rss/n (mle's)

So the maximum of the likelihood L(&#946;,&#963;^2) will be at the
values of the maximum likelihood estimators. So the
maximum of the likelihood will be given from the formula
named "maxL" isn't it? If it is so, then why this value of
the likelihood is greater than one?

Thanks again Rune

As Rune explained, this is NOT a probability
that you are computing. It is not constrained
to be less than 1. Only the integral of this
function can be interpreted as a probability.

John
.

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