Learning Notes

Stochastic – Shot Noise

Definition of Problem

To model probability distribution P(X=n, t) of noises caused by random arrival of electrons in a vacum tube.

Annotation

λ = Probability constant of electron arrive over a period of time, intensity of Poisson Dist.

n = Total number of electrons arrived at time t

G = Generating function

Derivation

Let P(n, t) be the probability of a total of n electrons arrived at time t. The probability of an electron arriving over a period Δt can be modulated by a constant  λ:

\displaystyle P(n+1, t+\Delta t) - P(n, t) = \lambda \Delta t

Then the probability of having n arrival at t + Δt is the sum of probabilities: 1) Already has n arrival and no arrivals in Δt. 2) Already has n – 1 arrival and one arrivals in Δt. Formulated as follow:

\displaystyle P(n, t+ \Delta t) = P(n, t)(1 - \lambda \Delta t)  + P(n-1, t)(\lambda \Delta t)

Re-arrange the upper equation we get [Eq.1]

\displaystyle \frac{\partial P(n, t)}{\partial t} = \frac{P(n, t+\Delta t) - P(n, t)}{\Delta t} = \lambda \left[P(n- 1, t) - P(n, t)\right]

At this point, we would like to introduce a math skill Generating Function and denotes:

\displaystyle G(s, t) = \sum_n^{\infty} s^n P(n, t)

so that:

\displaystyle \frac{\partial G(s, t)}{\partial t} = \sum_n^{\infty} s^n \frac{\partial P(n, t)}{\partial t}

\displaystyle = \lambda \sum_{n=0}^{\infty} s^n \left[P(n-1, t) - P(n, t)\right]

\displaystyle = s \cdot \sum_{n=-1}^{\infty} s^n P(n, t)-\sum_{n=0}^{\infty} P(n, t)

\displaystyle = s \cdot \sum_{n=0}^{\infty} s^n P(n, t)-\sum_{n=0}^{\infty} P(n, t)
[P(-1, t) = 0 for all t since number of arrival must not be negative]

\displaystyle = \lambda(s - 1) \sum_{n=0}^{\infty} s^n P(n, t)

= \lambda(s-1) \cdot G(s, t)

We then get ourselves a p.d.e:

\displaystyle \frac{\partial G}{\partial t} = \lambda (s-1) G

separating variables,

\displaystyle \frac{\partial G}{G} = \lambda (s-1) \partial t

\displaystyle \ln{[G(s, t)]} = \lambda t(s-1) + C

With the initial condition of G(s, 0) being 1 because P(0, 0) = 1 and P(n, 0) = 0 for all n, the particular solution would be:

\displaystyle G(s, t) = e^{\lambda t(s-1)}G(s, 0)

Perform Taylor expansion at s = 0:

\displaystyle G(s, t) = G(0, t) + s\cdot\frac{G'(0, t)}{1!} + s^2 \cdot \frac{G''(0, t)}{2!} + \cdots

\displaystyle \sum_n^{\infty} s^n P(n, t) = e^{-\lambda t} + \frac{s}{1!}\cdot(\lambda t)e^{-\lambda t} + \frac{s^2}{2!} e^{-\lambda t}

By comparing terms in powers of s, we can deduce:

\displaystyle P(n, t) = \frac{(\lambda t )^n e^{-\lambda t}}{n!}

which is actually a Poisson Process.

See Also

Stochastic – Poisson Process with Python example

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