Let $X \sim Geom(p_1), Y \sim Geom(p_2)$ be two independent geometrically distributed random variables with probabilities of success $p_1, p_2$ respectively.
I want to find the mean of the sum of these distributions; $X+Y$.
I know that the mean of a geometric distribution is $\frac{1}{p}$, where $p$ is the probability of success.
In this answer: Sum of two Geometric Random Variables with different success probability, I see that:
$$\mathbf{P}(X+Y = n) = \frac{p_1p_2}{p_1 - p_2} \big( (1-p_2)^{n-1} - (1-p_1)^{n-1} \big), \qquad n = 2,3,\ldots $$
How do I then find the mean of this distribution?
How does the mean of this distribution relate to the sum of the means of the independent distributions? ($\frac{1}{p_1} + \frac{1}{p_2}$)
Intuitively, I'm not sure whether we'd expect the mean of the sum of the independent distributions to be roughly the same as the sum of the means of the independent distributions?