In Chapter8 of this great book, the authors build a Bayesian model and use Fig.1 to show the posterior distributions of the latent state $N_{t}$ and its credible intervals. The model is written as below:

\begin{aligned} \left[N,\beta,\sigma_p^2|y\right]\propto&\prod normal(y_{t}|N_{t},\sigma_{data,t}^2)\\ \times&\prod lognormal(N_{t}|log(g(\beta,N_{t-1},x_{t})),\sigma_{p}^2))\\ \times &normal(\beta_{0}|0.234,0.0185)\prod_{i=1}^3normal(\beta_{i}|0,100)\\ \times&inverse\ gamma(\sigma_{p}^2|0.01,0.01)\\ \times&normal(N_{1}|y_{1},\sigma_{data,t}^2) \end{aligned}

where: $N_{t}$ is the unobserved, true abundance of wildebeest in year t; $x_{t}$ is a measure of the annual rainfall; $\beta$ is the coefficients; $\sigma_{p}^2$ is the process variance; $\sigma_{data,t}^2$ is a variance for the total population based on the sample of paragraphs.

My questions are:

(1) In Fig.1, why do we compare "mean" observed number of animals with "median" of the marginal posterior distribution of the population size (see the Figure's caption)? Why not use "mean" of the marginal posterior distribution to be consistent with dots (i.e., mean)?

(2) The authors remind us that:

The fit of the model medians is not smooth but, rather, tends to jump a bit between time steps. ... if we assumed that observation variance was 0, such that the true state was exactly the same as the observed state, then the medians would go through every data point.

Then they emphasize that:

It may be counterintuitive that a model like this one that does hit every point, or close to it, is probably not a very good model. What this says is that process variance is high relative to observation variance. The predictive value of such a model will be low.

Why would the solid line go through every data point if the observation variance was 0?


1 Answer 1


Regarding the second question, it might be helpful to remember that there is an estimation procedure involved. If the variance of the measured data drops to zero, there is only one estimation result possible -- the one that is exactly the same as the measured one. This is a feature of any estimation procedure that accounts for the variance of observations. This however is absolutely no indication of whether it is a "good" model.

For the first question, I might need to edit/delete the next part of my answer: There might be some information missing or being assumed about the true and unobserved (unobservable!) population size. My guess would be, that the quantiles of the median are either smaller or better defined than the mean and its variance. It might even be, that the mean cannot easily be calculated in this particular case. However, for the estimation procedure (the original task) it makes no difference what the plot is showing.

  • $\begingroup$ I am a little confused in two places in your answer. (1) Why " the quantiles of the median are smaller than the mean and its variance" makes us choose median instead of mean? Should we choose "a smaller value"? (2) Why "the mean cannot easily be calculated" in some cases? I think we can calulate its value through MCMC estimations. $\endgroup$
    – T X
    Jun 28, 2019 at 14:19
  • $\begingroup$ Sorry for the delayed answer; hopefully I can clarify a bit. Regarding (1): it's speculative, I don't know. The idea is simply that the "median" might be favorable to the "mean". Both are valid and well-defined measures, and a priori it's not clear which might be better suited. Regarding (2), again: without more knowledge it's only speculation. For this particular case this idea is probably not applicable. What I was thinking is: the underlying distribution might have no mean; but with every estimation method one can extract a mean, but it is neither accurate nor meaningful (no pun intended). $\endgroup$
    – cherub
    Jul 10, 2019 at 14:27
  • $\begingroup$ Thanks so much! I think I get your point: sometimes "mean" is not meaningful, for example "average wage". One more confirmation, you said "the quantiles of the median are smaller than the mean and its variance", do you mean "the variance of median is smaller than the variance of mean" in some cases, so we might choose median because it has a smaller variance? $\endgroup$
    – T X
    Jul 11, 2019 at 2:07
  • $\begingroup$ To the last question: yes. But the important bit is to interpret "variance of mean/median" in terms of "accuracy" of said value. There are few words in statistics that do not have a specific meaning. For example "variance" is usually the second central moment of a distribution and only in special cases it has something to do with the uncertainty of the mean. $\endgroup$
    – cherub
    Jul 11, 2019 at 12:49

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