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If the risk-free asset has a volatility of $0$, therefore making its mean equal to the risk-free rate, $r_f$, does this mean that it has no probability distribution, and therefore there is no reason to model it parametrically (i.e. with $\mathcal{N}(\cdot)$ or other)?
How does this situation change when we drop the usual assumption of a constant $r_f$, since, empirically, central banks actually make it time-varying?
The standard way to think about this is that at time $t$ the riskless asset gives you known return of $r_{f,t}$ over a short time period. However, this rate may itself be time-varying and stochastic so that we don't know its futures values, say $r_{f,t+s}$. E .g. a common assumption is that the rate follows an Ornstein Uhlenbeck process (implying that the conditional distribution is normal).
In case you assume that $r_{f,t}$ is actually constant, say $c$, it still has a probability distribution. Here you need to define $r_{f,t}$ as a random variable that takes the value $c$ for all outcomes of the sample space. Naturally the distribution of $r_{f,t}$ is then such that all the probability mass lies at this single point: $P(r_{f,t}=c)=1$.
Just to add to the previous answer, one example of such asset (returning 'risk-free rate') is a money market (or bank) account, but it is only locally risk-free, with value accruing continuously at the risk-free rate prevailing in the market at every instant. It is risk-free only over a short period of time. In long term it is stochastic too. Its SDE is:
$$ dB_t =r_t B_t \; dt, \; B_0 =1, $$ or, equivalently, $$ B_t = \exp \left( \int_0^t r_u\; du \right), $$
where $r_t$ is a stochastic progressively measurable process with locally integrable paths, making $B_t$ a finite variation process with null quadratic variation. Intuitively, this means that it has a smaller degree of randomness with respect to the other risky assets. (In FX or equity option pricing, based on such intuition, one even assumes time-dependent deterministic interest rate.)
