Say I have an observed data set ($n_i$) and I want to obtain the best fit out of 10 data sets produced by a model dependent on a single parameter $a$ ($m_i(a)\;a=1..10$).
Suppose I use a Poisson likelihood distribution:
$P_i(a)=\frac{m_i(a)^{n_i}}{e^{m_i(a)}n_i!}\; ; \; a=1..10$
where $m_i(a)$ represents the model value of bin $i$ (for a given value of the parameter $a$) and $n_i$ its observed value (which remains constant always) The likelihood of the whole data set or cumulative likelihood (for each value of $a$) is then:
$L(a)=\prod\limits^{N}_{i=1} \frac{m_i(a)^{n_i}}{e^{m_i(a)}n_i!}\; ; \; a=1..10$
where $N$ is the total number of bins. Now, I want to pick the model data set that best fits my observed data set. Could I just choose the maximum value of $L(a)$, ie: the maximum likelihood:
$Max\_Likelihood = max\,[ L(a)\; ; \; a=1..10]$
and say that the value of parameter $a$ associated with that particular modeled data set is the best estimation of parameter $a$, given my observed data set?
Or should I use a cumulative likelihood ratio, defined as :
$LR(a)= \prod\limits^{N}_{i=1} \frac{\frac{m_i(a)^{n_i}}{e^{m_i(a)}n_i!}}{\frac{n_i^{n_i}}{e^{n_i}n_i!}} = \prod\limits^{N}_{i=1} \left(\frac{m_i(a)}{n_i}\right)^{n_i}e^{n_i-m_i(a)}\; ; \; a=1..10$
and keep the value of $a$ that gives the maximum value of $LR(a)$, ie the maximum likelihood ratio:
$Max\_Likelihood\_Ratio = max\,[ LR(a)\; ; \; a=1..10]$
Since the observed data set ($n_i$) remains the same for all the modeled data sets ($m_i(a);\;a=1..10$), won't maximizing this value $LR(a)$ give me the same result (ie: the same value of $a$) as maximizing $L(a)$?