A problem about expectation of partial sum of random variables Let $X_1, X_2,\dots,X_n$ be $n$ i.i.d. geometric distributed random variables with successful probability $p$, that is $$P(X_i=k)=(1-p)^{k-1}p$$
Let $Y_i=X_i-\mathbb{E}(X_i) = X_i-\frac{1}{p}$, so $Y_1,\dots,Y_n$ are also i.i.d. random variables.
Let $S_k=Y_1+\dots+Y_k$, and $S_k^+=\max(0,S_k)$
I want to know a close form of $\mathbb{E}(S_k^+)$
 A: Since $\sum\limits_{i = 1}^k {{X_i}} $ has negative binomial distribution with parameters $k$ and $1-p$, we have 
$$
\mathbb{E}\left[ {S_k^ + } \right] = \sum\limits_{j = \left\lceil {\frac{k}
{p}} \right\rceil }^\infty  {\left( {j - \frac{k}
{p}} \right)\left( \begin{gathered}
  j + k - 1 \\ 
  j \\ 
\end{gathered}  \right){p^k}{{\left( {1 - p} \right)}^j}} 
$$
A: Let $X$ denote a discrete random variable with finite mean $\mu$.  From 
$$E[X-\mu] = 0 = \sum_i (u_i - \mu) p_X(u_i) = \sum_{i\colon u_i > \mu} (u_i - \mu) p_X(u_i) + \sum_{i\colon u_i < \mu} (u_i - \mu) p_X(u_i),$$
we get that
$$\sum_{i\colon u_i > \mu} (u_i - \mu)p_X(u_i) 
= \sum_{i\colon u_i < \mu} (\mu - u_i) p_X(u_i)
$$ 
and so if $Y = \max\{0, X-\mu\}$, we have that
$$\begin{align*}
E[Y] &= \sum_i \max\{0, u_i - \mu\} p_X(u_i)\\
&= \sum_{i\colon u_i > \mu} (u_i - \mu)p_X(u_i)\\
&= \sum_{i\colon u_i < \mu} (\mu - u_i) p_X(u_i).
\end{align*}$$
In the question under consideration, $X = \sum_{j=1}^k X_j$ is a
negative binomial random variable with parameters $(k,p)$ and mean
$\frac{k}{p}$.  Its pmf is given by
$$p_X(n) = \binom{n - 1}{k-1} p^k (1 - p)^{n-k}, ~~ n = k, k+1, k+2, \ldots
$$
Hence,
$$\begin{align*}
E[S_k^+] &= E\left[\max\left\{0, X-\frac{k}{p}\right\}\right]\\
&= \sum_{n = \lceil \frac{k}{p}\rceil}^\infty
 \left(n - \frac{k}{p}\right)\binom{n - 1}{k-1} p^k (1 - p)^{n-k}\\
&= \sum_{n = k}^{\lfloor \frac{k}{p}\rfloor}
 \left(\frac{k}{p}- n \right)\binom{n - 1}{k-1} p^k (1 - p)^{n-k}.
\end{align*}$$
If $k/p$ is an integer, one of the terms in each sum is $0$.
The first form of the answer has already been given by @Alen
but the second might be easier to evaluate numerically
since it is a finite sum.

Neither form gives the floor of the expected value as has been alleged
  by OP Fan Zhang in comments on the main question.

