I have a log likelihood that looks like the following.

$$ \log(p(Z|\Theta)) = -\sum_{p = 1}^{N} \left[ L \log\left(\pi\left(A F(p, \Theta) + \sigma^2\right) \right) + \frac{\sum_{l = 1}^{L} Z_l(p) }{A F(p, \Theta) + \sigma^2} \right] $$

It comes from an exponential distribution. The observations are $Z_l(p)$, and $A F(p, \Theta) + \sigma^2$ is the model of the expectation of Z(p).

However, the scale factor $A$ and additive factor $\sigma^2$ are estimated empirically or from prior knowledge before computing this likelihood.

Some real world measurements $Z$ do give me positive log-likelihood values. $Z_l$ is a discrete measurement with $1\times N$ dimension, and I have $L$ of them.

I want to have a quantitative measure of how good my data fits the measurements. That is why I thought of looking at the likelihood of all the different real world measurements. So, a measurement giving a log-likelihood value close to $0$ (but negative) is more fitted to my model than a measurement that gives a log-likelihood value far from $0$ (but negative).

However, now that I also have positive values for log-likelihood, I am confused how to say if the measurements fit my model.

Sometimes, by visual inspection, I find that a log likelihood of $-2000$ reconstruct my data better than a log-likelihood of 50 for example. To emphasize again, this is from a visual inspection.

Any input is appreciated.

My suspect: I think $A$ and $\sigma^2$ are the reason for which probably it is more than $0$. Am I right? If yes, how can I say something about my estimation (not just with visual inspection, but also from likelihood).

  • $\begingroup$ $F(p, \Theta)$ is the cumulative distribution of the $p^{th}$ occurrence of $\Theta$? $\endgroup$
    – TimWescott
    May 15 at 15:28
  • $\begingroup$ You are summing over $p$, but you have $\sigma_n^2$ in there, as well as functions of -- apparently -- $p$. It may do to edit your question for clarity. $\endgroup$
    – TimWescott
    May 15 at 15:30
  • $\begingroup$ Is $Z$ continuous? If so, then $p(Z|\Theta)$ is a probability density function, and can go higher than 1 -- if $Z$ is continuous, then to be a "fair" probability density function, the condition $\int_{-\infty}^{\infty} p(Z|\Theta) dz = 1$ must be satisfied (assuming that you replace my bounds of integration with whatever is the proper space for all possible values of $Z$). $\endgroup$
    – TimWescott
    May 15 at 15:38
  • $\begingroup$ $\sigma_n$ is a constant. I should just write it as another symbol. $\endgroup$
    – CfourPiO
    May 15 at 20:39
  • 1
    $\begingroup$ It would be nice if you would edit your question with this information -- it bears directly on the meaning of the question and as such belongs there, not in the comments. $\endgroup$
    – TimWescott
    May 15 at 21:25


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