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I seem to have a bit of confusion about this particular distribution , and I would appreciate if people could help me get past it.

My question are as follows.

Let $X$ be a discrete random variable.

$1)$The probability mass function of the negative binomial distribution is defined to be :$P(X=N)=C_{N-1}^{k-1}p^k(1-p)^{N-k}$ where $k$ is the number of successes , $N$ the number of trials and $p$ the probability of success.

Unlike the binomial distribution whose $pmf$ is referred to as : $P(X=k)$, we refer to the $pmf$ of the negative binomial distrubution as: $P(X=N)$.

This bugs me , is this due to the fact that k is held constant while the number of trials is changing ?

$2)$The expected value $E[X]$ of a negative binomial distribution is :

$E[x]=\frac{k}{p}$

I know this can be derived by by some algebraeic approach knowing that the $pmf$ in a negative binomial distribution sums to 1.

However can this be proved using the linearity of expected value , assuming each trial is a Bernoulli distribution?

3)What's confusing me the most is that the expected value $E[X]$ turned out to be independent of the number of trials.

Take for example:

A basketball player A scores his 3rd 3 points shot after 10 trials.

A basketball player B scores his 3rd 3 points shot after 100 trials.

The expected value $E[x]$ would be the same in both cases according to the formula?

I would be grateful to anyone who could clarify these concepts.

Thanks in advance.

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1) That is just a matter of convention. Just don't allow it to confuse you.

2) You can write $X=X_1+\cdots X_k$ where the $X_i$ are iid with $X_1\sim\mathsf{Geometric}(p)$.

Then linearity of expectation combined with symmetry leads to $\mathsf EX=k\mathsf EX_1=\frac{k}{p}$.

3) $X$ is the number of trials.

In an expression $\mathsf EX=\dots$ you will never find an $X$ at the RHS (which would make it random).

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