I am new to stochastic processes and just know three types of Poisson counting process
- Homogeneous
- Non-homogeneous
- Renewal
My issue is suppose if my mean of the Poisson process is a random variable which takes different values with different probabilities like $\lambda_i$ with probability $p_i$ which kind of process it should be .It does not fit into first category. For non-homogeneous it needs to be a function of time, but it is just a random variable with a certain probability distribution. And for a renewal process it needs to be a stochastic process having a distribution as well as dependence on time. Also please tell me how to simulate it as does the exponential inter-arrival time also valid for this, or should I use thinning as to make mean now $\lambda_i \times p_i$?
What you describe is not actually a Poisson distribution, but a mixture of Poisson distributions, also called an average of Poisson distributions or a convex combination of Poisson distributions.
Suppose, for example that $$ \Lambda= \begin{cases} 1 & \text{with probability } 1/2, \\ 3 & \text{with probability } 1/2., \end{cases} $$ and $$ X\mid \Lambda \sim \operatorname{Poisson}(\Lambda). $$ Then $$ \operatorname{E}(X) = \operatorname{E}(\operatorname{E}(X\mid \Lambda)) = \operatorname{E}(\Lambda) = 2, $$ and therefore if the distribution of $X$ is a Poisson distribution, then it would be a Poisson distribution with expected value $2$ and we would have (for example) $$ \Pr(X=5) = \frac{e^{-2} 2^5}{5!} \overset{\Large\text{?}} \approx 0.036. $$ But in fact we have \begin{align} \Pr(X=5) & = \operatorname{E}(\Pr(X=5\mid\Lambda)) = \operatorname{E}\left( \frac{e^{-\Lambda} \Lambda^5}{5!} \right) \\[10pt] & = \frac 1 2 \cdot \frac{e^{-1} 1^5}{5!} + \frac 1 2 \cdot \frac{e^{-3} 3^5}{5!} \approx 0.052. \end{align} Hence this mixture of Poisson distributions is not itself a Poisson distribution.
There is something called a "compound Poisson distribution", which is not a Poisson distribution, but rather is the distribution of a random variable of the form $\sum_{x=1}^X Y_x$ where $X$ has a Poisson distribution and $Y_1,Y_2,Y_3,\ldots$ are i.i.d. So this is an instance of the term "Poisson distribution" (but preceded by another word) being used to refer to something that is not a Poisson distribution. However, as far as I know, that is not done with mixtures of Poisson distributions.
To find the time until the first arrival, we need some concept of time. Thus suppose $\Lambda$ as described above is the rate, so that the number of arrivals before time $t$ is $\Lambda t.$ Let $T$ be the time until the first arrival. Then \begin{align} \Pr( T> t) & = \Pr(\text{the first arrival is after time } t) \\[10pt] & = \Pr(\text{the number of arrivals before time }t \text{ is } 0) \\[10pt] & = \Pr\left( \begin{array}{l} \Big(\Lambda=1\ \&\ \text{the number of such arrivals is }0 \Big) \\ \text{ or } \Big( \Lambda=0\ \&\ \text{the number of such arrivals is }0 \Big) \end{array} \right) \\[10pt] & = \frac 1 2 \cdot \frac{(1\cdot t)^0 e^{-1\cdot t}}{0!} + \frac 1 2 \cdot \frac{(3\cdot t)^0 e^{-3\cdot t}}{0!} = \frac{e^{-t} + e^{-3t} } 2. \end{align} The question is how to prove this is not an exponential function, i.e. we do not have $$ \text{for some } r>0, \text{ for all } t>0, \quad \frac{e^{-t} + e^{-3t}} 2 = e^{-r t}. $$ Maybe one of the quickest ways to show this is impossible is to compute the derivative of the function on the left and show that it is now just some constant multiplied by the function being differentiated. $$ \frac d {dt} \, \frac{e^{-t} + e^{-3t}} 2 \quad \overset{\Large{\text{?}}} = \quad \left(\text{some constant} \times \frac{e^{-t} + e^{-3t}} 2 \right) \text{ ?} $$