Show the Markov Property of a Poisson Process

Question

I wish to show the Markov Property of a Poisson Process, i.e. for any $0=t_0 < t_1 < \cdots < t_n$and non-decreasing natural numbers $k_i \geq 0, i=1,\ldots,n, n \geq 2$,

$$P(N(t_n)=k_n\mid N(t_1)=k_1,\ldots,N(t_{n-1})=k_{n-1}) = P(N(t_n)=k_n\mid N(t_{n-1})=k_{n-1})$$

I'll write up my books definition of a Poisson process below:
A stochastic process $(N(t))_{t \geq 0}$ is said to be a Poisson process if the following conditions hold:

(1) The process starts at zero: $N(0)=0$ a.s.

(2) The process has independent increments: for any $t_i, i=0,\ldots,n,$ and $n \geq 1$ such that $0=t_0<t_1<\cdots<t_n$ the increments $N(t_{i-1},t_i], i=1,\ldots,n$, are mutually independent.

(3) There exists a non-decreasing right-continuous function $ \mu: [0, \infty) \rightarrow [0, \infty)$ with $\mu(0)=0$ such that the increments $N(s,t]$ for $0<s<t<\infty$ have a Poisson distribution Pois$(\mu(s,t])$.

(4) With probability $1$, the sample paths $(N(t, \omega))_{t \geq 0}$ of the process $N$ are right-continuous for $t \geq 0$ and have limits from the left for $t>0$. We say that $N$ has càdlàg sample paths.

I tried the following, but can't really seem to get anywhere, and I know I'm making a mistake, since I somehow make $N(t_n)$ independent of $N(t_n-1)$, which is certainly not the case, only the increments are independent. Anyways, here it is:
For every $k_i$ exists a $c_i, i=1,\ldots,n$ such that $k_i = c_1 +\cdots+c_i$. The Markov Property now becomes:

\begin{align} & P(N(t_n) =c_1+\cdots+c_n\mid N(t_{n-1})=c_1+\cdots+c_{n-1}, \ldots, N(t_1)=c_1) \\[10pt] = {} & P(N(t_n)=c_1+\cdots+c_n\mid N(t_{n-1})=c_1+\cdots+c_{n-1}) \end{align}

The left side yields:

\begin{align} & P(N(t_n) =c_1+\cdots+c_n\mid N(t_{n-1})=c_1+\cdots+c_{n-1},\ldots,N(t_1)=c_1) \\[10pt] = {} & \frac{P(N(t_n) = c_1+\cdots+c_n, N(t_{n-1})=c_1+\cdots+c_{n-1}, \ldots, N(t_1) = c_1)}{P(N(t_{n-1})=c_1+\cdots+c_{n-1}, \ldots, N(t_1)=c_1)} \\[10pt] = {} & \frac{P(N(t_1) = c_1, N(t_1,t_2] = c_2,\ldots,N(t_{n-1},t_n]=c_n)}{P(N(t_1)=c_1, N(t_1,t_2] = c_2, \ldots, N(t_{n-2}, t_n-1]=c_{n-1})} \end{align}

Since these are independent by definition (3), this would cancel out, and bring the "obvious" answer $P(N(t_n)=c_n)$, since I somehow made a wrong assumption that they're all independent. I feel like my whole approach is probably wrong, but I don't know how to continue, and I can't seem to find my mistake. I could really use a guiding hand.

Thanks a lot in advance, any help is appreciated.

Answer 1

The usual strategy is to rewrite things in terms of the increments, so that you can take advantage of the independent increments property.

Hint: $$\begin{align}&P(N(t_n) = k_n \mid N(t_1)=k_1,\ldots,N(t_{n-1}) = k_{N-1}) \\ &= P(N(t_n) - N(t_{n-1}) = k_n - k_{n-1} \mid N(t_1)=k_1, N(t_2) - N(t_1)= k_2 - k_1,\ldots, N(t_{n-1}) - N(t_{n-2}) = k_{n-1} - k_{n-2}) \end{align}$$ and $$P(N(t_n)=k_n \mid N(t_{n-1}) = k_{n-1}) = P(N(t_n) - N(t_{n-1}) = k_n -k_{n-1} \mid N(t_{n-1}) = k_{n-1}).$$

Answer 2

Seems to me this follows from indpendence of increments, unless I missed something about what you're trying to show.

\begin{align} & P(N(t_n) =c_1+\cdots+c_n\mid N(t_{n-1})=c_1+\cdots+c_{n-1}, \ldots, N(t_1)=c_1) \\[10pt] = {} & P( N(t_n) - N(t_{n-1}) = c_n \mid N(t_{n-1})=c_1+\cdots+c_{n-1}, \ldots, N(t_1)=c_1) \end{align}

The reason these are equal is that the two events are conditionally the same event. I.e. if $A\cap C = B\cap C$ then $A$ and $B$ are conditionally the same event, given $C.$ That implies that $P(A\mid C) = P(B\mid C).$

The probability above is the same as $$ P(N(t_n)-N(t_{n-1})=c_n \mid N(0,t_1)=c_1, N(t_1,t_2] = c_2,\ldots,N(t_{n-2}, t_{n-1}]=c_{n-1}). $$ By independence of increments this is $P(N(t_n) -N(t_{n-1}) = c_n).$

In the same way, one can show that $$P(N(t_n) -N(t_{n-1}) = c_n)=P(N(t_n)=k_n\mid N(t_{n-1})=k_{n-1}).$$

Hence both of the two conditional probabilities whose equality you want to prove are equal to $P(N(t_n) -N(t_{n-1})= c_n).$