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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 $\mathbb{E}(S_k^+)$

2 Answers 2

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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.

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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}} $

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    Oh, now I see it. Well, the aswer is yes since it's support is susbset of integers. However, ceiling in the sum comes from removing all negative members, not from taking the ceiling of RV2012-01-09