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Lets' say I have a Poisson distribution for which I use a maximum likelihood defined as:

$P_i=\frac{m_i^{n_i}}{e^{m_i}n_i!}$

where $m_i$ represents the model value of bin $i$ (real, $m_i$> 0) and $n_i$ its observed value (integer, number of counts in bin $i$). The cumulative likelihood for the whole data set is then:

$L=\prod\limits^{N}_{i=1} \frac{m_i^{n_i}}{e^{m_i}n_i!}$

whwre $N$ is the total number of bins.

Now, as I understand it, $P_i$ denotes the probability that the observation $n$ is drawn from model $m$.

(1) So a bigger likelihood value $L$ means a better fit between the model and the data (Is this correct?)

What I don't get is the necessity for a maximum likelihood ratio, defined as

(2) the ratio of probability of drawing $n_i$ points from model $m_i$, from that of drawing $n_i$ points from model $n_i$ (Is this correct?)

whose cumulative form is:

$LR= \prod\limits^{N}_{i=1} \frac{\frac{m_i^{n_i}}{e^{m_i}n_i!}}{\frac{n_i^{n_i}}{e^{n_i}n_i!}} = \prod\limits^{N}_{i=1} \left(\frac{m_i}{n_i}\right)^{n_i}e^{n_i-m_i}$

(3) Won't maximizing this value $LR$ give me the same results as maximizing $L$?

Finally:

(4) How can I calculate the p-value associated with a Poisson likelihood $L$ or a Poisson likelihood ratio $LR$ for a model($m$)-observation($n$) analysis?

Gabriel
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1 Answers1

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I'll try to take your questions in order.

  1. Your expression for $P_i$ is not the maximum likelihood, it is just the likelihood of the $i$th data point given your model. The likelihood of the entire data set is your expression $L$ (assuming all observations are independent). The larger $L$ is, the better the fit is.
  2. The likelihood ratio comes in when you want to compare two hypotheses. Denoting the true underlying probability for bin $i$ as $\lambda_i$, for the form you have written down these would be $H_0: \lambda_i=m_i$ versus $H_1:$ at least one $\lambda_i \ne m_i$. The likelihood ratio test statistic is defined as $$LR = \frac{\max_{H_0}L(\lambda)}{\max_{H_0\cup H_1} L(\lambda)},$$ where $L(\lambda)=\prod_i \frac{\lambda_i^{n_i}\exp(-\lambda_i)}{n_i!}$. The result happens to equal your expression for $LR$.
  3. Note that we are not maximizing LR: a separate maximization of the same function is performed in the numerator and the denominator under different assumptions, so your question does not really make sense.
  4. There are statistical theorems that say that under certain conditions the null-distribution of $-2\log LR$ is asymptotically $\chi^2$ with degrees of freedom equal to the difference in the number of parameters in the numerator and denominator. Since you don't specify where $m_i$ come from, it is difficult to tell. There are also situations when the distribution of the likelihood ratio test statistic can be obtained exactly based on the model (not here, though). The calculation of a p-value follows directly from the null-distribution of the test statistic.
Aniko
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