Expectation of the "truncated" partial sum of random variables

Question

Suppose $X_i, i = 1,2,..., N$ are i.i.d. nonnegative random variables. Given any constant $C > 0$, can I prove the following?

$\mathbb{E}[\sum_{n=1}^N(C - \sum_{k=1}^nX_i)^+] - \sum_{n=1}^N(C - \sum_{k=1}^n\mathbb{E}[X_i])^+$ is $\mathcal{O}(\sqrt{N})$, where $(a)^+ = \max\{a, 0\}$.

Answer 1

$$\mathbb{E}[\sum_{n=1}^N(C - \sum_{k=1}^nX_i)^+] - \sum_{n=1}^N(C - \sum_{k=1}^n\mathbb{E}[X_i])^+ = \sum_{n=1}^N\mathbb{E}(C - \sum_{k=1}^nX_i)^+] - \sum_{n=1}^N(C - \sum_{k=1}^n\mathbb{E}[X_i])^+ = \sum_{n=1}^N[\mathbb{E}(C - \sum_{k=1}^nX_i)^+ - (C - \sum_{k=1}^n\mathbb{E}[X_i])^+]$$

To prove that it is $O(\sqrt{N})$ it is sufficient to prove, that $\mathbb{E}(C - \sum_{k=1}^nX_i)^+ - (C - \sum_{k=1}^n\mathbb{E}[X_i])^+$ is $O(\frac{1}{\sqrt{n}})$. Moreover, it is sufficient to prove that for $\mathbb{E}(C - \sum_{k=1}^nX_i)^+$ as $(C - \sum_{k=1}^n\mathbb{E}[X_i])^+ = 0$ for large $n$.

$\mathbb{E}\sqrt{n}(C - \sum_{k=1}^nX_i)^+ \leq C\sqrt{n}P(\sum_{k=1}^nX_i < C) \leq C\sqrt{n}P(|\sum_{k=1}^nX_i - nEX_1| > |C - n EX_1|) \leq C\sqrt{n}\frac{nDX_1}{|C - nEX_1|^2}$

by Chebyshev inequality.

And that value is indeed bounded.