Re: likelihood greater than 1, beginner

Rune Allnor <allnor@xxxxxxxxxxxx> wrote in message
<747c47d0-08cd-4574-b500-
bb5ef704fe00@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
On 13 Apr, 13:22, "leo nidas" <bleonida...@xxxxxxxx>
wrote:
I have the following data and then i evaluate (with two
ways) the maximum of the likelihood function. I think
everything is correct since both ways give the same
result
but since the likelihood is the multiplication of the
contribution of each point, that is a probability, is
it
possible to take values beyond 1? The result I am
getting
is 112.2253 with both ways.

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



.

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