I'm start studying Bayesian statistics, but I've found that I'm having troubles with the likelihoods. Let's say that I have a vector of observations $y$ and I want to calculate how likely it is $y$ given a parameter $\theta$, that is, $P(y|\theta)$.

I wonder if there is a conceptual problem here, but the thing is that we do not have a probability density function for $y$. Rather, we have a quite simple and common expectation as follows:

$E(y)=k · exp(\theta \cdot x)$

In this linear model, $k$ is a constant and $x$ an independent variable. We do not solve this by taking logarithms and using linear regression, but by numerical methods, but this is out of the topic.

In any case, we estimate $\theta$ numerically by minimizing the residual sum of squares ($RSS$):

$RSS = [E(y)-y]^{2}$

I have the notion that minimizing $RSS$ maximizes likelihood (i.e. the estimated $\hat{\theta}$ is the MLE), but I cannot find a way to estimate that likelihood $L(\theta|y) = P(y|\theta)$ from these data, which will be necessary to solve Bayes.

In addition, if I were interested in solving a particular case, say $P(y|\theta>0.2)$, how one would proceed in this case? I just cannot imagine how to calculate $RSS$ in this scenario.

  • $\begingroup$ The given does not define a model, only a moment connection, hence there is no likelihood provided and no (easy) way to implement Bayes. $\endgroup$ – Xi'an Feb 7 at 15:34
  • $\begingroup$ I would say that $E(y) = k · exp (\theta x)$ is a model, isn't it? $\endgroup$ – elcortegano Feb 7 at 15:44
  • 1
    $\begingroup$ There is an infinity of models that suits this constraint. $\endgroup$ – Xi'an Feb 7 at 16:13

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