Consider a random variable $X = \sum\_{i=1}^{m} X\_i$, where each $X\_i$ is an indicator 
random variable that is $1$ with probability $k/m$ and $0$ otherwise. We are interested in the quantity $S\_X(m) = E[\min(X,k)]$. The motivation is that we have a bin of capacity $k$. At each step, a ball is thrown into the bin with probability
$k/m$. At the end of $m$ steps, we ask how many balls are in the bin. This quantity, in expectation, is precisely $S\_X(m)$. 


Now suppose we throw "fractional balls", i.e., instead of having $\{0,1\}$ random variables $X\_i$s, we have random variables $Y\_i$ that have support $[0,1]$.
We retain the same expectation, i.e., $E[Y\_i] = k/m = E[X\_i]$ and the $Y\_i$'s are iid. 
Let $S\_Y(m) = E[\min(Y,k)]$, where $Y = \sum\_{i=1}^{m} Y\_i.$ 


The question I am interested in is whether $S\_Y(m) \geq S\_X(m)$? 


I have an intuition for why this must be true: the variance of $X\_i$ is at least as much as that of $Y\_i$ --- this is because $E[X\_i^2] = E[X\_i]$, where as $E[Y\_i^2] \le E[Y\_i]$. Thus, one would expect a random variable with smaller variance (namely $Y = \sum\_i Y\_i$) to be more concentrated around the mean than a random variable with larger variance (namely $X = \sum\_i X\_i$), thus implying the result. Roughly speaking, one would expect the "wastage" of balls due to overflowing the capacity $k$ of the bin occurs lesser when we have fractional balls than integer balls. However this is not a definitive proof. Is there a simple proof for this?



Capped binomial random variables
The answer to your question is **positive** and, for example, follows immediately from Corollary 4 of



> 
> C. A. León and F. Perron (2003), [Extremal properties of sums of Bernoulli random variables](http://dx.doi.org/10.1016/S0167-7152%2803%2900037-3), *Statistics and Probability Letters*, vol. 62, 345–354.
> 
> 
> 


A slight specialization of the corollary states:



> 
> Let $\newcommand{\E}{\mathbb E} Y = Y\_1 + \cdots + Y\_n$ be a sum of iid random variables taking values in $[0,1]$ with mean $\E Y\_i = \mu$ and let $X \sim \mathrm{Bin}(n,\mu)$. For any convex function $g : [0,n] \to \mathbb R$,
>  $$
> \E g(Y) \leq \E g(X) \ .
> $$
> 
> 
> 


Your result follows by noting that $g(x,a) = -\min(x,a)$ is convex in $x$ for any $a \in \mathbb R$.


The paper appeals to other references, but an easy and direct proof can be constructed, so we may as well give a version of it here as several short lemmas. 


In what follows below, we assume $(Y\_i)$ are iid on $[0,1]$ with distribution function $F$ and mean $\mu$ and that $(X\_i)$ are iid Bernoulli random variables also with mean $\mu$. The function $g$ is assumed to be an arbitrary convex function defined on the appropriate domain.


**Lemma 1**: If $Y \sim F$ and $X \sim \mathrm{Ber}(\mu)$, then $\E g(Y) \leq \E g(X)$.


*Proof*: $Y = 1 \cdot Y + 0 \cdot (1-Y)$, so, by convexity, $\E g(Y) \leq g(1) \E Y + g(0) \E (1-Y) = \E g(X)$.


**Lemma 2**: $\E g(Y\_1 + Y\_2) \leq \E g(X\_1 + X\_2)$.


*Proof*: Assume wlog that $(Y\_1,Y\_2)$ is independent of $(X\_1,X\_2)$. If $g$ is convex, then so is $g(y+\cdot)$, hence
$$\E g(Y\_1 + Y\_2) = \int\_0^1 \E g(y + Y\_2) \ \mathrm dF \leq \int\_0^1 \E g(y + X\_2) \ \mathrm dF = \E g(Y\_1 + X\_2) \leq \E g(X\_1 + X\_2) \ . $$


**Corollary**: Let $Y = Y\_1 + \cdots + Y\_n$ and $X = X\_1 + \cdots X\_n$, where $X \sim \mathrm{Bin}(n,\mu)$ as in the problem. Then, $\E g(Y) \leq \E g(X)$.


*Proof*: Extend the previous lemma by induction.


The desired result now follows by taking the aforementioned choice for $g$ with $a = n \mu$.


