I have seen that X ~ G(p) and we have X - 1 ~ NB(1,p) (*)
I do not understand the meaning of (*)
I definitely know that geometric dist helps to find the number of trials till the first success and and the negative binomial dist helps to find the number of trials till the rth success In this case we have negative binomial dist with r = 1. Why it would not be the geometric dist itself but have to be one less than the geometric (as (*) indicated)
Any help is appreciated!! Thanks for your help in advanced!!
Personally, I do not know what the (*) is attempting to address.
I would say that if we have $X\sim \text{NB}(1,p)$, then we also have that $X \sim \text{G}(p)$
Here is how the formula for the negative binomial turns into the formula for the geometric:
First we have that the general formula for the negative binomial, with $k$ successes in $n$ trials and we stop after the $k^{th}$ success, is: $$P(X=n) = \binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}p$$ In the special case of $k=1$, see that $$P(X=n)=\binom{n-1}{1-1}p^{1-1}(1-p)^{n-1}p$$ $$=\binom{n-1}{0}p^0(1-p)^{n-1}p$$ $$=1*1*(1-p)^{n-1}p$$ $$=(1-p)^{n-1}p$$ Which is the formula for the geometric distribution with probability of success $p$.
There are a few different variations of the Negative Binomial Distribution where the random variable $X$ means different things, so perhaps it was a different definition being used that the (*) was attempting to explain.