Estimating likelihood from the Residual Sum of Squares

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.

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