PSD of a Poisson process
Earlier we have started talking about the Poisson processes. In the few posts before the summer holidays we have driven our discussion towards waiting time paradox, which is an interesting phenomenon we encounter in our day-to-day lives. Here, on Physics of Risk we have an interest in colors of noise exhibited by variety of stochastic processes. Thus in the next few posts let us examine the power spectral density of the Poisson process.
Fourier transform of a Poisson point process
When discussing waiting time paradox we have provided a plot where the bus arrival events are shown as sudden bursts. Mathematically such signal can be expressed as:
\begin{equation} I(t) = \sum_k A(t-t_k) , \end{equation}
with \( A(s) \) being the burst profile. For now let us keep the mathematical form of the profile general. The Fourier transform of such signal is given by:
\begin{equation} \mathcal{F}\left\{ I(t) \right\} = \sqrt{\frac{2}{T}} \int_0^T I(t) e^{-2 \pi \mathrm{i} f t} \mathrm{d}t = \sqrt{\frac{2}{T}} \int_0^T \sum_k A(t-t_k) e^{-2 \pi \mathrm{i} f t} \mathrm{d}t . \end{equation}
Because of the linearity of the Fourier transform we can rewrite it as a sum of the Fourier transforms of the individual bursts:
\begin{equation} \mathcal{F}\left\{ I(t) \right\} = \sum_k \sqrt{\frac{2}{T}} \int_0^T A(t-t_k) e^{-2 \pi \mathrm{i} f t} \mathrm{d}t = \sum_k e^{-2 \pi \mathrm{i} f t_k} \mathcal{F}\left\{A(t) \right\} . \end{equation}
If \( A(t) \) is the Dirac delta function, then:
\begin{equation} \mathcal{F}\left\{ I(t) \right\} = \sqrt{\frac{2}{T}} \sum_k e^{-2 \pi \mathrm{i} f t_k} . \end{equation}
The expression about is remarkably similar to the Discrete Fourier transform of a signal with unitary bursts. This is a purely algebraic reason to pick Dirac delta function as burst profile (to maintain consistency between continuous Fourier transform and its discrete counterpart).
Physical reasoning for the choice is that in physics nothing happens instantaneously. Events assumed to be instantaneous happen over extremely short (thus often irrelevant) periods of time. The change the events trigger is often fixed in magnitude, so the shorter event time the larger the density of the change introduced. \( I(t) \) should be seen as representing not the magnitude of the change, but the density. This is why in the first Poisson process app we have plotted \( S(t) \) (the count of the students who have arrived by the time \( t \)).
Power spectral density of a Poisson point process
To understand overall dynamics of the process let us consider the average (over distinct realizations of the process) power spectral density of the signal:
\begin{equation} S(f) = \lim_{T\rightarrow\infty} \left\langle \frac{2}{T} \left\lvert \sum_k e^{-2 \pi \mathrm{i} f t_k} \right\rvert^2 \right\rangle . \end{equation}
The modulus squared in the above can be rewritten as a multiplication of two sums: the sum in the expression above, and its complex conjugate. Thus we have:
\begin{equation} S(f) = \lim_{T\rightarrow\infty} \left\langle \frac{2}{T} \sum_k \sum_m e^{2 \pi \mathrm{i} f (t_m - t_k)} \right\rangle . \end{equation}
Let us separate the terms with identical indices \( m = k \), as their sum is trivially equal to \( N \) (number of bursts):
\begin{equation} S(f) = \lim_{T\rightarrow\infty} \left\langle \frac{2}{T} \left( N + \sum_k \sum_{m\neq k} e^{2 \pi \mathrm{i} f (t_m - t_k)} \right) \right\rangle . \end{equation}
The sum over \( m \neq k \) can be simplified by recalling that square of the modulus is always a real number. Then:
\begin{equation} \sum_k \sum_{m\neq k} e^{2 \pi \mathrm{i} f (t_m - t_k)} = 2 \mathrm{Re}\left[\sum_k \sum_{m > k} e^{2 \pi \mathrm{i} f (t_m - t_k)}\right] . \end{equation}
Recall that \( t_k \) is the time when \( k \)-th burst occurs, thus the difference between \( m \)-th occurrence time and \( k \)-th occurrence time is given by:
\begin{equation} t_m - t_k = \sum_{i=k}^{m-1} \tau_i . \end{equation}
If \( \tau_i \) are identically distributed, then:
\begin{equation} \left\langle \exp\left[ 2 \pi \mathrm{i} \left( t_m - t_k \right) \right] \right\rangle = \left\langle \exp\left[ 2 \pi \mathrm{i} \sum_{i=k}^{m-1} \tau_i \right] \right\rangle = \chi_\tau(f)^{m-1-k} . \end{equation}
In the above \( \chi_\tau(f) \) is the characteristic function of the inter-event time distribution. Thus we have that:
\begin{equation} S(f) = \lim_{T\rightarrow\infty} 2 \bar{\nu} \left( 1 + \frac{2}{N} \sum_k \sum_{m\neq k} \left\langle e^{2 \pi \mathrm{i} f (t_m - t_k)} \right\rangle \right) = 2 \bar{\nu} \left( 1 + \mathrm{Re} \left[ \frac{2}{1 - \chi_\tau(f)} \right] \right) . \end{equation}
In the above \( \bar{\nu} \) stands for the average number of bursts per unit time. As \( \tau \) are sampled from exponential distribution with rate \( \lambda \), we have that \( \bar{\nu} = \lambda \). Furthermore the characteristic function of exponential distribution is given by:
\begin{equation} \chi_\tau(f) = \frac{\lambda}{\lambda - 2 \pi \mathrm{i} f} , \end{equation}
thus the power spectral density simplifies to:
\begin{equation} S(f) = \lim_{T\rightarrow\infty} \left\langle 6 \bar{\nu} \right\rangle \approx 6 \frac{N}{T} . \end{equation}
It seems that the algebra above indicates that the Poisson process is a white noise process. And thinking from the perspective of the microscopic model for the Poisson process, this conclusion seems reasonable as distinct time intervals are uncorrelated: the probability that event occurs is constant and equal for every time interval.
Interactive app
Interactive app below samples time series from the Poisson process. To verify the analytical intuitions above, we perform spectral analysis on the discretized time series. The discretized time series are obtained by counting the number of events inside unit time windows (the top plot). In total \( 2^{20} \) (roughly one million) such time windows are observed each time "Generate" button is pressed. PSD of the sampled series is shown in the bottom plot as a red curve, the dark gray curve indicates theoretical prediction.
