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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?

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    $\begingroup$ General formula: X and Y any random variables: E(X+Y)=E(X)+E(Y). Means are integrals and general rule - integral of sum is sum of integrals. (Integral may be Stieltjes, such as discrete r.v.) $\endgroup$ Jun 24, 2021 at 17:06

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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
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