TITLE: How do you calculate the expected value of geometric distribution without diffrentiation?
QUESTION [12 upvotes]: Is there any way I can calculate the expected value of geometric distribution without diffrentiation? All other ways I saw here have diffrentiation in them.
Thanks in advance!

REPLY [4 votes]: The geometric distribution is memoryless so either you succeed in the initial attempt with probability $p$ or you start again with probability $1-p$ having made a failed attempt, 
if the succeeding on the first attempt counts as $1$ attempt: 
$$E[X]=p\times 1+(1-p)\times (1+E[X])$$ so $$p\times E[X]=1$$ so  $$E[X]=\frac{1}{p} \text{ attempts}$$
while if succeeding on the first attempt counts as $0$ failures:  
$$E[X]=p\times 0+(1-p)\times (1+E[X])$$ so $$p\times E[X]=1-p$$ so  $$E[X]=\frac{1-p}{p} \text{ failures}$$
and naturally $\frac1p = \frac{1-p}p +1$ since you stop at the first successful attempt