# What is the distribution of the risk-free asset?

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.)

I think this is a useful question. I'd like to add an illustrative example to the excellent answers above, if not for the OP then for others.

A tractable (albeit unrealistic) model for the short rate is the following $$dr(t) = \sigma dW(t), \quad r(0) = 0$$ where $$W$$ is a standard Brownian motion.

So, \begin{align} E[r(t)] &= 0\\ Var[r(t)] &= \sigma^2 t \end{align}

Now, the risk-free asset (the money market account) is $$\log B(t)/B(0) = \int_0^t r(u) du$$ So we can speak of the expectation of $$\log B(t)/B(0)$$, which is given by $$E \left[\log B(t)/B(0)\right] = 0$$ and also $$Var \left[\log B(t)/B(0)\right] = \frac{\sigma^2}{3}t^3$$

So, as said by ir7, it's locally risk-free but not 'globally' risk free.