# Poisson Twister Generator by Cumulative Frequency Technology
This makes it possible to simplify it without considering the events with an infinitesimal small probability of a higher order \(o(\tau)\):

\[P_{1}(\tau)=\lambda\cdot\tau. \tag{5}\]

Equation (5) allows for reaching the determination of probability of the event absence in the time interval \(\tau\):

\[P_{0}(\tau)=1-P_{1}(\tau)=1-\lambda\cdot\tau. \tag{6}\]

Combining together Equations (2) and (6), the probability of the event absence at the time moment \(t\) in the general interval \([0,t+\tau]\) could be obtained as follows:

\[P_{0}(t+\tau)=P_{0}(t)\cdot P_{0}(\tau)=P_{0}(t)(1-\lambda\tau). \tag{7}\]

This expression (7) leads to the definition of the derivative with respect to probability:

\[\frac{dP_{0}(t)}{d\tau}=\lim_{\tau\to 0}\frac{P_{0}(t+\tau)-P_{0}(t)}{\tau}=- \lambda P_{0}(t). \tag{8}\]

The solving of the differential Equation (8) determines the probability of the absence of events at the time moment \(t\), in which the constant \(c=1\) is derived from the initial condition \(P_{0}(0)=1\). The result is the following:

\[P_{0}(t)=ce^{-\lambda t}=e^{-\lambda t}. \tag{9}\]

Equation (9) with allowance for Equation (2) and Constraints (5) and (6) admits calculating the probability of a single event \(P_{1}([0,t+\tau])\) as follows:

\[P_{1}([0,t+\tau])=P_{0}(t)P_{1}(\tau)+P_{1}(t)P_{0}(\tau)=e^{-\lambda t} \lambda\tau+P_{1}(t)(1-\lambda\tau). \tag{10}\]

By analogy with Transformations (7) and (8), Equation (10) leads to the following differential equation:

\[\frac{dP_{1}(t)}{d\tau}=\lambda e^{-\lambda t}-\lambda P_{1}(t). \tag{11}\]The solving of Equation (11) determines \(P_{1}(t)\) with the initial condition \(P_{1}(0)=0\):

\[P_{\eta-k}((t,t+\tau])=P_{\eta-k}((0,\tau]).\]0

Performing successively the Transformations (10)-(12) for all variables \(\eta\in[0,\infty]\), the distribution of Poisson probabilities \(P_{\eta}(\lambda t)\) with respect to quantity \(\eta\) of random events with intensity \(\lambda\) at the time moment \(t\) could be obtained as follows:

\[P_{\eta-k}((t,t+\tau])=P_{\eta-k}((0,\tau]).\]1

Since the set of \(\eta\in\mathrm{H}=[0,\infty]\) of random events in Kolmogorov axiomatics contains \(\sigma\)-algebra, the probability measure \(P_{\eta}(\alpha=\lambda t)\) (13) uniquely determines the cumulative function \(F_{\mathrm{H}}(\eta,\alpha=\lambda t)\) of the distribution probabilities:

\[P_{\eta-k}((t,t+\tau])=P_{\eta-k}((0,\tau]).\]2

Using a definition of number \(e^{x}=\sum\limits_{k=0}^{\infty}x^{k}/k!\) for Equation (14), it follows that \(F_{\mathrm{H}}(\eta,\lambda)\in[0:1]\). It should be noted here that a probability space guarantees uniqueness of the determination of the inverse probability distribution function. If any value \(h\) of the cumulative distribution function \(F_{\mathrm{H}}(\eta_{h},\alpha=\lambda t)=h\) is given, then a value of the stochastic variable \(\eta_{h}\) can be obtained as the inverse transformation \(\eta_{h}=F_{\mathrm{H}}^{-1}(h)\). Therefore, by specifying the complete uniform random values \(F_{\mathrm{H}}^{-1}(\eta,\alpha=\lambda t)\in[0:1]\), it allows uniquely obtaining the stochastic values \(\eta\) of this distribution. This main mathematical model contains the bases for constructing the generators of random variables from given functions of their distribution. Let us use this statement for developing the generator of Poisson stochastic variables. For this it is necessary to get an absolutely complete and uniform generator, which has no repetitions and skipping of random variables. Let us use here the twister generator _nsDeonYuliTwist32D_[24; 25; 26; 27], which is of such properties. In discrete probability space, the number \(N\) of events is fixed.