# Why the log likelihood is positive in some cases

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).

• $F(p, \Theta)$ is the cumulative distribution of the $p^{th}$ occurrence of $\Theta$? May 15 at 15:28
• 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. May 15 at 15:30
• 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 $Z$). May 15 at 15:38
• $\sigma_n$ is a constant. I should just write it as another symbol. May 15 at 20:39
• 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. May 15 at 21:25