How are the pairs of two independent pure-birth processes a Markov process?

A pure-birth Process is a generalization of a homogeneous Poission process. Whereas in the Poisson process the holding times between jumps are iid exponentially distributed random variables with parameter $\lambda$, in a pure-birth process the parameters may vary between jumps, i.e. the waiting between states $n$ and $n+1$ is exponentially distributed with parameter $\lambda_n$.

Let $X_1$, $X_1$ be two independent pure-birth processes with rates $\lambda^1_i, \lambda^2_i$ respectivley. By pairs of $X_1$ and $X_2$ I mean the process $$X(t) := (X_1(t), X_2(t))$$

I'd like to prove that $X$ is a contininuous-time Markov process that starts in state $(0,0)$ and may jump from state $(n,m)$ to either $(n+1,m)$ or $(n,m+1)$ with rates $\lambda^1_n$ and $\lambda^2_m$ respectively, depending on which process $X_1$,$X_2$ the next jumps occurs.

While this might seem very intuitive I'm having trouble relating the properties of $X_1$ and $X_2$ to $X$.

Can you point to a treatment I can apply my problem as a special case of?

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Got something from the answer below? – Did Jul 29 '12 at 14:01
Yes, thank you very much! Thinking about it, of course it makes sense that it holds for Markov processes in general. I was able to find the same statement in Kallenberg's Foundation of Modern Probabilistics as an exercise. However i didn't fully get the notion of a infintesimal generator acting on functions. Is this another way to describe it instead of a matrix? And "acting" is just matrix multiplication? – Haderlump Aug 2 '12 at 12:42
Indeed, to identify functions defined on the state space to vectors (of infinite size) is to identify the action of $L$ on functions to the multiplication (on the left) by a square-matrix (of infinite size). – Did Aug 2 '12 at 13:34

Every vector process whose coordinates are independent continuous-time Markov processes is a continuous-time Markov process. Here, $(X(t))$ is a continuous-time Markov process on the state space $S\times S$, where $S=\{0,1,2,\ldots\}$ is the common state space of the independent processes $(X_1(t))$ and $(X_2(t))$.
While in $(n,m)$, the process $(X(t))$ jumps at rate $\lambda_{n,m}=\lambda^1_n+\lambda^2_m$, that is, after an exponential time of mean $1/\lambda_{n,m}$. When this jump occurs, $(X(t))$ jumps to $(n+1,m)$ or to $(n,m+1)$ with respective probabilities $\lambda^1_n/\lambda_{n,m}$ and $\lambda^2_m/\lambda_{n,m}$ and independently of the time of the jump.
The infinitesimal generators $L_1$ and $L_2$ of $(X^1(t))$ and $(X^2(t))$ act on functions $v$ defined on $S$ by $$L_iv(k)=\lambda^i_k\cdot(v(k+1)-v(k)),\qquad k\in S.$$ By definition, the infinitesimal generator $L$ of the independent product process $(X(t))$ acts on functions $u$ defined on $S\times S$ by $$Lu(n,m)=L_1u_{\cdot,m}(n)+L_2u_{n,\cdot}(m),\qquad (n,m)\in S\times S,$$ where the functions $u_{n,\cdot}$ and $u_{\cdot,m}$ are defined on $S$ by $u_{n,\cdot}(m)=u_{\cdot,m}(n)=u(n,m)$. In the present case, one gets, for every $(n,m)$ in $S\times S$ $$Lu(n,m)=\lambda^1_n\cdot(u(n+1,m)-u(n,m))+\lambda^2_m\cdot(u(n,m+1)-u(n,m)).$$