# Merged Poisson Process Rate

If I have two process $$X_t$$ with rate λ and $$Y_t$$ with rate μ and I merge them I get an overall rate of λ + μ.

Is this the same logic that would follow if I have a process of $$2X_t + 3Y_t$$ to get 2λ + 3μ.I feel like it wouldn't be correct but I'm not sure what would be the correct answer then.

• $2X_t$ is not a Poisson process : if you have any confusion on this statement you may ask, since it can be confusing, especially if confused with the idea that this is a sum of Poisson processes. But the two Poisson processes being summed are dependent ($X_t$ and $X_t$ are of course dependent) so the sum isn't Poisson anymore. Having said that, the expression for the expectation is still correct. – Teresa Lisbon May 3 at 17:00

If by "rate $$\lambda$$" you mean a process $$X_t$$ such that $$\mathbb E[X_t] = \lambda t$$, then this is just linearity of expectation: $$\mathbb E[2 X_t + 3 Y_t] = 2 \mathbb E[X_t] + 3 \mathbb E[Y_t] = (2\lambda + 3 \mu) t$$
On the other hand, $$2 X_t + 3 Y_t$$ is certainly not a Poisson process (e.g. this never takes the value $$1$$), while if $$X_t$$ and $$Y_t$$ are independent Poisson processes, $$X_t + Y_t$$ is a Poisson process.