So I'm struggling to find the joint distribution of an (X,Y) bivariate population where X is discrete 0, 1 and 2 with equal probabilities and the conditional distribution of Y given X = x is exponential with mean $\beta(x + 1)$ with $\beta > 0$.
I know that
$$f_{x,y}(x,y) =f(y \mid x)*f_{x}(x)$$
But in this case, I'm not sure how to compute that product
(The exercise is actually about computing an ML estimator for $\beta$, but I can't even begin without having the joint distribution since my sample is from the bivariate population)
Speaking about joint distribution when one of the RVs is discrete is not correct, but you could compute the pmf of it by taking the derivative of the cdf. If $x$ has a discrete distribution on $[0,1,2]$ with equal probabilities and $Y|X=x \sim Exp(\beta(x+1))$ then the cdf will be
$$P(X=x,Y\leq y)=\frac{1}{3}\left(1-e^{\left(\beta(1+x)\right)^{-1}y}\right)$$
So taking the derivative $$f(x,y)= \frac{1}{3}\left(\beta\left(1+x\right)\right)^{-1}exp\left(-\left(\beta\left(1+x\right)\right)^{-1}y\right)$$
If you want to find the MLE of $\beta$, first the likelihood will be (dropping out terms that do not depend on $\beta$):
$$\mathcal{L} \propto \beta^{-n}exp\left(-\frac{1}{\beta}\left(n_{0}\bar{y}_{0} + \frac{1}{2}n_{1}\bar{y}_{1} + \frac{1}{3}n_{2}\bar{y}_{2}\right)\right)$$
where $\bar{y}_{i}$ is the conditional mean of $y$ conditional on $x=i$ and $n_{i}$ the sample size. Log likelihood:
$$\log\mathcal{L} \propto -n\log \beta -\frac{1}{\beta}\left(n_{0}\bar{y}_{0} + \frac{1}{2}n_{1}\bar{y}_{1} + \frac{1}{3}n_{2}\bar{y}_{2}\right)$$
First order condition yields $$\frac{n}{\beta} = \frac{1}{\beta^{2}}\left(n_{0}\bar{y}_{0} + \frac{1}{2}n_{1}\bar{y}_{1} + \frac{1}{3}n_{2}\bar{y}_{2}\right)$$
And therefore
$$\beta^{MLE}=\frac{n_{0}\bar{y}_{0} + \frac{1}{2}n_{1}\bar{y}_{1} + \frac{1}{3}n_{2}\bar{y}_{2}}{n}$$