The random telegraph signal with parameter $\lambda$ is derived from a Poisson process
of arrival rate $\lambda$. In the Poisson process, the number of arrivals in $(t_1, t_2]$ is a Poisson random variable with parameter $\lambda(t_2-t_1)$. Consider counting the
Poisson arrivals in $(0,\infty)$ with a digital counter whose least significant bit (LSB)
is $0$ or $1$ with equal probability at time $t=0$. The random telegraph signal is just
a model for the value of the LSB, or rather for the value of $(-1)^{\text{LSB}}$ which
has value $+1$ or $-1$ according as the LSB is $0$ or $1$.
The random telegraph signal is a random process (collection of random variables)
$\{X_t \colon 0 \leq t < \infty\}$ where each $X_t \in \{+1,-1\}$ and $X_0$ is equally likely to be $+1$ and $-1$. For any $t_2 > t_1 \geq 0$, the number of arrivals in
$(t_1,t_2]$ is Poisson with parameter $\lambda(t_2-t_1)$ and thus is an
even number with probability $\exp(-\lambda(t_2-t_1))\cosh(\lambda(t_2-t_1))
= \frac 12 + \frac 12 \exp(-2\lambda(t_2-t_1))$.
It follows that the LSB is in the same state at time $t_2$ as it was at
time $t_1$, that is, $X_{t_2} = X_{t_1}$
with probability $\frac 12 + \frac 12\exp(-2\lambda(t_2-t_1))$ while $X_{t_2} = -X_{t_1}$
with probability $\frac 12 - \frac 12\exp(-2\lambda(t_2-t_1))$. It is
readily verified that $P\{X_t = +1\} = P\{X_t = -1\} = \frac 12$ and so the
process has mean function $0$ while the autocorrelation
function is
$$\begin{align}R_X(t_1,t_2) &= E[X_{t_1}X_{t_2}]\\
&= (+1)\left[\frac 12 + \exp(-2\lambda(t_2-t_1))\right]
+ (-1)\left[\frac 12 - \exp(-2\lambda(t_2-t_1))\right]\\
&= \exp(-2\lambda(t_2-t_1))\end{align}$$
which is a function of the time difference only. Thus
The random telegraph signal is a zero-mean WSS process with
autocorrelation function $R_X(\tau) = E[X_tX_{t+\tau}]= \exp(-2\lambda \tau)$
and its power spectral density is of the form of a Cauchy density
function.
It is worth emphasizing that if one takes a finite segment of an
actual realization or sample path of a random telegraph signal,
one does not get the nice and smooth
autocorrelation function described above but rather a piecewise linear
function that decays away slowly.
So, can we get any of this from a DFT? The answer is that we could
get an approximate answer, but we need to take care in setting up
the problem.
The random telegraph signal shifts back and forth between $+1$ and
$-1$ at random times. Thus, when we take $N$ samples spaced $T$
seconds apart of a realization of this signal, each sample value is
$+1$ and $-1$,
but since this is not a band-limited signal, the samples cannot be
used to reconstruct the realization at all. Note also that unless
$T$ is carefully chosen to be quite small so that we can be virtually
certain (i.e. with very high probability)
that there is at most one transition from $+1$ to $-1$ (or vice versa),
we will have missed a lot of the nuances (a.k.a transitions)
in the signal. But, if $T$ is very very small and $N$ is very very large,
our time-domain vector will have reasonably long runs of consecutive
$+1$s or consecutive $-1$s during which interval there are no
transitions at all.
Now, the autocorrelation function that we need
is the aperiodic autocorrelation function, and not the periodic
autocorrelation function, and so we need to zero-pad our signal
vector of length $N$ with $N-1$ or more $0$s (say $N$ for
concreteness) and compute the DFT of this vector of length $2N$.
An FFT algorithm could be used for this. Now $|X[k]|^2$ will
roughly resemble the Cauchy spectrum of the random telegraph
signal.
