Poisson Processes

Assume characterisation (a) holds. The first inter-arrival time $T_1$ satisfies

$$P(T_1 > t) = P(N(t) = 0) = e^{-\lambda t},$$

so $T_1 \sim \mathrm{Exp}(\lambda)$.

For the second inter-arrival time, condition on $T_1 = s$:

$$P(T_2 > t \mid T_1 = s) = P\bigl(N(s + t) - N(s) = 0 \mid T_1 = s\bigr).$$

By the independent-increments property, $N(s+t) - N(s)$ is independent of events in $[0, s]$ (and hence of $T_1$), so

$$P(T_2 > t \mid T_1 = s) = P\bigl(N(s+t) - N(s) = 0\bigr) = e^{-\lambda t}.$$

Therefore $T_2 \sim \mathrm{Exp}(\lambda)$ and $T_2$ is independent of $T_1$. By induction, all inter-arrival times are iid $\mathrm{Exp}(\lambda)$.

Assume the inter-arrival times $T_1, T_2, \ldots$ are iid $\mathrm{Exp}(\lambda)$. The $n$-th arrival time is $S_n = T_1 + T_2 + \cdots + T_n$, which has the $\mathrm{Gamma}(n, \lambda)$ distribution with density

$$f_{S_n}(t) = \frac{\lambda^n\,t^{n-1}}{(n-1)!}\,e^{-\lambda t}, \qquad t \ge 0.$$

Now compute $P(N(t) = k)$ for a fixed $t > 0$. The event $\{N(t) = k\}$ is equivalent to $\{S_k \le t < S_{k+1}\}$, i.e., exactly $k$ arrivals occur in $[0, t]$. We have

$$P(N(t) \ge k) = P(S_k \le t) = \int_0^t \frac{\lambda^k\,s^{k-1}}{(k-1)!}\,e^{-\lambda s}\,ds.$$

By direct computation (or by recognising the incomplete gamma function identity),

$$P(N(t) = k) = P(N(t) \ge k) - P(N(t) \ge k+1) = \frac{(\lambda t)^k}{k!}\,e^{-\lambda t},$$

which is the $\mathrm{Poisson}(\lambda t)$ PMF.

Independent increments follow from the memoryless property of the exponential: given that an arrival occurred at time $s$, the residual time until the next arrival is again $\mathrm{Exp}(\lambda)$, independent of the past. Hence the process "restarts" after each arrival and after any deterministic time, yielding independent increments. $\blacksquare$

The survival function of $T \sim \mathrm{Exp}(\lambda)$ is $\bar{F}(t) = P(T > t) = e^{-\lambda t}$. By the definition of conditional probability,

$$P(T > t + s \mid T > t) = \frac{P(T > t + s)}{P(T > t)} = \frac{e^{-\lambda(t+s)}}{e^{-\lambda t}} = e^{-\lambda s} = P(T > s).$$

Let $g(t) = P(T > t)$ for a continuous non-negative random variable $T$ satisfying the memoryless property. Then

$$g(t + s) = P(T > t + s) = P(T > t + s \mid T > t)\,P(T > t) = P(T > s)\,P(T > t) = g(s)\,g(t).$$

So $g$ satisfies Cauchy's functional equation $g(t + s) = g(t)\,g(s)$ on $[0, \infty)$. Since $T$ is a continuous random variable, $g$ is monotone non-increasing (being a survival function) and right-continuous, which guarantees the only solution is $g(t) = e^{-\lambda t}$ for some constant $\lambda \ge 0$. Since $P(T < \infty) = 1$ (the distribution is proper), we need $\lambda > 0$. Therefore $T \sim \mathrm{Exp}(\lambda)$. $\blacksquare$

For any non-overlapping intervals $(s_1, t_1]$ and $(s_2, t_2]$, the increment $N(t_i) - N(s_i) = \sum_{k=1}^K [N_k(t_i) - N_k(s_i)]$. Since each component process has independent increments and the $K$ processes are mutually independent, the increments of $N$ over non-overlapping intervals are sums of independent random variables and are therefore independent.

The moment generating function of a $\mathrm{Poisson}(\mu)$ random variable is $M(z) = \exp\!\bigl(\mu(e^z - 1)\bigr)$. For a fixed interval of length $s$, the increment of $N_k$ is $\mathrm{Poisson}(\lambda_k s)$ with MGF $M_k(z) = \exp\!\bigl(\lambda_k s(e^z - 1)\bigr)$.

By independence of the $K$ processes,

$$M_{N(t+s)-N(t)}(z) = \prod_{k=1}^{K} M_k(z) = \prod_{k=1}^{K} \exp\!\bigl(\lambda_k s(e^z - 1)\bigr) = \exp\!\left(s(e^z - 1)\sum_{k=1}^{K}\lambda_k\right).$$

This is the MGF of a $\mathrm{Poisson}\!\left(s\sum_k\lambda_k\right)$ random variable. Since the MGF uniquely determines the distribution, $N(t+s) - N(t) \sim \mathrm{Poisson}(\lambda s)$ with $\lambda = \sum_k \lambda_k$. $\blacksquare$

Given $N(t) = n$, each of the $n$ events is independently retained with probability $p$. Therefore

$$N_{\mathrm{kept}}(t) \mid N(t) = n \sim \mathrm{Binomial}(n, p).$$

By the law of total probability,

$$P\bigl(N_{\mathrm{kept}}(t) = k\bigr) = \sum_{n=k}^{\infty} P\bigl(N_{\mathrm{kept}}(t) = k \mid N(t) = n\bigr)\, P\bigl(N(t) = n\bigr).$$

Substituting the Binomial and Poisson PMFs:

$$= \sum_{n=k}^{\infty} \binom{n}{k} p^k (1-p)^{n-k}\, \frac{(\lambda t)^n}{n!}\,e^{-\lambda t}.$$

Rearranging, with $m = n - k$:

$$= \frac{(p\lambda t)^k}{k!}\,e^{-\lambda t} \sum_{m=0}^{\infty} \frac{[(1-p)\lambda t]^m}{m!} = \frac{(p\lambda t)^k}{k!}\,e^{-\lambda t}\, e^{(1-p)\lambda t} = \frac{(p\lambda t)^k}{k!}\,e^{-p\lambda t}.$$

This is the $\mathrm{Poisson}(p\lambda t)$ PMF.

An analogous calculation shows $N_{\mathrm{disc}}(t) \sim \mathrm{Poisson}\bigl((1-p)\lambda t\bigr)$. To verify independence, compute the joint PMF:

$$P\bigl(N_{\mathrm{kept}}(t) = j,\, N_{\mathrm{disc}}(t) = k\bigr) = P\bigl(N_{\mathrm{kept}}(t) = j,\, N(t) = j + k\bigr)$$

$$= \binom{j+k}{j} p^j (1-p)^k\, \frac{(\lambda t)^{j+k}}{(j+k)!}\,e^{-\lambda t} = \frac{(p\lambda t)^j}{j!}\,e^{-p\lambda t} \cdot \frac{((1-p)\lambda t)^k}{k!}\,e^{-(1-p)\lambda t}.$$

This factors as the product of the marginal PMFs, confirming independence. The independent-increments property of the thinned processes follows from that of the original process. $\blacksquare$

Given $N(t) = n$, the vector $(N_1(t), \ldots, N_K(t))$ has the $\mathrm{Multinomial}(n;\, p_1, \ldots, p_K)$ distribution:

$$P\bigl(N_1(t) = n_1, \ldots, N_K(t) = n_K \mid N(t) = n\bigr) = \frac{n!}{n_1!\cdots n_K!}\,p_1^{n_1}\cdots p_K^{n_K},$$

where $n_1 + \cdots + n_K = n$.

Summing over $n = n_1 + \cdots + n_K$:

$$P\bigl(N_1(t) = n_1, \ldots, N_K(t) = n_K\bigr) = \frac{n!}{n_1!\cdots n_K!}\,p_1^{n_1}\cdots p_K^{n_K} \cdot \frac{(\lambda t)^n}{n!}\,e^{-\lambda t}$$

$$= \prod_{k=1}^{K} \frac{(p_k \lambda t)^{n_k}}{n_k!}\, e^{-p_k \lambda t} = \prod_{k=1}^{K} P\bigl(X_k = n_k\bigr),$$

where $X_k \sim \mathrm{Poisson}(p_k \lambda t)$. The joint PMF factors into a product of marginals, establishing both the Poisson marginal distributions and mutual independence. $\blacksquare$