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?