Specifying frequency parameter in the absence of occurrences Let's say I have a process where the occurrences are independent, proportional to time. I made $n$ observations for which I only observed no occurrences. My goal is to define a frequency parameter and characterize the unknown value. By characterizing I mean being able to give a level of confidence on the value, given a threshold. Something like "given $n=100$ observations without occurrence, I can say with 95,76% "confidence" that the frequency parameter is below the given threshold of $0.04$".
Given the description of the process, I intuitively modelized my process as sequence of independent Poisson random variables. Under the Poisson law, the probability of observing $k$ events in an interval is given by the equation
$P(k{\text{ events in interval}})=e^{-\lambda }{\frac {\lambda ^{k}}{k!}}$
Apart from a guess that the parameter $\lambda$ for the Poisson law is somewhat low I haven't much information about it.
Given that $\lambda$ appears to be really low, I also considered a simpler approach with only two possibilities: the realisation of one event or not, at each time step. This approach appears to be a Bernoulli trial, which can be modelised as following :
$P(\text{event})= \lambda $
$P(\text{no event})= 1-\lambda $
I understand that there are then two distinct approaches: frequentist and Bayesian. I understand that the Bayesian approach requires more information on the prior distribution. 
1) In a Bayesian approach, would it make sense to take the given threshold mentioned above to determine the prior? How do I get from the prior to the posterior? 
In the frequentist approach we can define estimators $\hat{\lambda}$. My problem is that the Maximum Likelihood Estimator gives $\hat{\lambda}_{MLE}=0$ in both case as we have no occurrences.
2) Are there other estimators that would give a different result? How do I define and use the distribution of these estimators to build the wanted confidence level? Is this approach equivalent to making the assumption that $\lambda$ is below the given threshold and testing for it?
 A: From your initial specification, your model is $K_1,...,K_n \sim \text{IID Pois}(\lambda)$, which gives likelihood function:
$$L_\boldsymbol{k}(\lambda) \propto \prod_{i=1}^n \text{Pois}(k_i|\lambda) \propto \lambda^{\sum k_i} \exp(-n\lambda) \quad \quad\ \quad \text{for } \lambda \geqslant 0.$$
You have observed data values $k_1 = ... = k_n = 0$ so the likelihood function in this case is the monotonically decreasing function $L_\boldsymbol{k}(\lambda) \propto \exp(-n\lambda)$.  Your goal is to get an interval estimate for the parameter $\lambda$.
It is worth noting at the outset that it is possible to include $\lambda = 0$ as an allowable parameter in this model.  In this case the Poisson distribution degenerates down to a point-mass distribution on zero, which is a possibility in this case, since you have not observed any non-zero outcomes.  Thus, the above specification of the range of $\lambda$ as including zero is not a typographical mistake --- it is an extension of the standard Poisson model to include the possibility of a point-mass distribution at zero.

Bayesian analysis: This method is applied by giving a prior to your parameter and then determining the highest posterior density region as an appropriate interval estimate for the parameter.  For simplicity, we will take the "non-informative" Jeffrey's prior:
$$\pi(\lambda) \propto \frac{1}{\sqrt{\lambda}} \propto \text{Ga}(\lambda|\tfrac{1}{2},0).$$
This is an improper prior with infinite variance, so it is highly diffuse.  This gives the posterior:
$$\pi_n(\lambda) \propto \frac{\exp(-n\lambda)}{\sqrt{\lambda}} \propto \text{Ga}(\lambda|\tfrac{1}{2}, n).$$
Since the posterior is monotonically decreasing, the highest posterior density interval is an interval from zero up to some upper bound.  If we define the bound $\lambda_*$ as the solution to $\Gamma(\tfrac{1}{2}, n \lambda_*) = \alpha \sqrt{\pi}$ then the highest posterior density interval with coverage $1-\alpha$ is $0 \leqslant  \lambda \leqslant \lambda_*$.  That is, we have:
$$\mathbb{P}(0 \leqslant  \lambda \leqslant \lambda_*| \boldsymbol{k}) = 1- \alpha.$$
This gives you a useful solution to your inference problem.  As $n \rightarrow \infty$, if you keep observing zeros then the above form holds and you get $\lambda_* \downarrow 0$, which accords with intuition.  That is, if you keep observing no events in more and more data, your interval estimate for the true parameter will keep shrinking down to zero.

Classical analysis: Since the likelihood function is monotonically decreasing you have MLE $\hat{\lambda} = 0$ in this case, as you have recognised.  (Since we have included this parameter value in the allowable range for the model, the MLE exists in this case and is zero.)  It is unsurprising that this is the MLE, given that all the data values you observed were zero.
In order to get an interval estimate in classical statistics you form a confidence interval via a pivotal quantity.  Grosh (1989, p. 59) derives a confidence interval for the Poisson distribution using its relationship to the chi-squared distribution.  If we define the bound $\lambda_{**}$ as the solution to $\Gamma(1, n \lambda_*) = \alpha$ then this confidence interval (with minimum coverage $1-\alpha$) is $0 \leqslant \lambda \leqslant \lambda_{**}$.  We can see that this corresponds to the Bayesian HPDI in the case where we use an improper uniform prior.
