Given i.i.d. data $x_1,...,x_n$ from a r.v. $X$ with density
$$f(x) = \begin{cases} \frac{1}{4} \theta_1 &\text{if } x=1 \\ \frac{1}{4}(1-\theta_1) &\text{if } x=2 \\ \frac{3}{4}\frac{1}{\theta_2}e^{-(x-3)/ \theta_2} &\text{if } x \geq 3 \ , \end{cases} $$
find the MLE of $\theta =(\theta_1, \theta_2)$.
$f(x)$ needs to be normalized, and this results in $1=\int_3^{\infty}\frac{1}{\theta_2}e^{-(x-3)/ \theta_2}dx$, which is only satisfied for $\theta_2=1$, Is this then also the MLE for $\theta_2$?
Edit: The density is defined with respect to the measure $\mu=\mu_0 1\{x < 3\}+\mu_1 1\{x \geq 3\}$ where $\mu_0$ is counting measure and $\mu_1$ is length measure.
Say you have $k_1$ observations where $x_i=1$ and $k_2$ observations with $x_i=2$ and the rest are $x_i\geq 3$. Then the likelihood will be $$L(\theta_1,\theta_2)=\left(\frac{\theta_1}{4}\right)^{k_1}+\left(1-\theta_1\right)^{k_2}\frac14^{k_2}+\prod_{I}\frac{3}{4}\frac{1}{\theta_2}e^{-(x_i-3)/ \theta_2}$$ where in the last sum we have $I$ including all the indices for which $x_i\geq 3$.
The maximization is simple since the discrete and the continuous parts "do not have anything in common".