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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.

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  • $\begingroup$ $2X_t$ 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. $\endgroup$ – Teresa Lisbon May 3 at 17:00
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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.

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