Let $X \sim \operatorname{Exp}(\lambda)$ and $Y \sim \operatorname{Exp}(\mu)$ be independent. Consider the random vector $Z = (X,Y)$.
How can I now calculate the probability density function $f_Z$?
Since $Z$ is independent its probability density function must be the product of the corresponding marginal density functions. How can I proceed from here?
Edit: Not a duplicate since my question resolves around a random vector.
You are right, the density of $Z$ is the product of the marginals, so the three densities are as follows: \begin{align*} &f_X(x) = \lambda \exp(-\lambda x), \\ &f_Y(y) = \mu \exp(-\mu y), \\ &f_Z(x,y) = f_X(x)f_Y(y) = \lambda\mu \exp(-\lambda x-\mu y). \end{align*} Note, that $f_X$ and $f_Y$ are defined only on $\mathbb R_{\ge 0}$ (you may define them to be zero for negative $x,y$, if you want densities on all of $\mathbb R$) and that $f_Z$ is a density on $\mathbb R_{\ge 0}^2$ (so depending on two variables).