Mean of the sum of two independent geometric distributions with different probabilities of success 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?
 A: Comment:
At least two versions of the geometric distribution are in common use, so you always have to say which version you are using. From $E(X) = 1/p,$ I deduce that you're using the version that counts the number of trials up to and including the first Success. The mean of your geometric random variable is $E(X_1) = 1/p.$ Also, $E(X_2) = 1/p.$ So by @herbsteinberg's comment (+1), $E(X_1+X_2) = 2/p.$
Illustration by simulation:
By contrast rgeom in R, generates geometric
observations that count the number of Failures before the first Success. So I add $1$ to my
geometric samples from R to match your problem.
Let $p = 1/3.$
Simulating a million sums of the two random variables, and finding the mean of
the huge sample of sums, for the case $p = 1/3,$ I get 2 or 3 digits of accuracy for $E(X_1+X_2) = 6.$
set.seed(2021)
x1 = rgeom(10^6, 1/3)+1
x2 = rgeom(10^6, 1/3)+1
mean(x1); mean(x2); mean(x1+x2)
[1] 3.003199  # aprx E(X1) = 1/p = 3 
[1] 2.999317  # aprx E(X2) = 3
[1] 6.002516  # aprx E(X1+X2) = 3 + 3 = 6

