Does replacing a random variable in a sum with it expectation shift the mass of the expectation towards the mean?

Question

Assume any two positive, independent random variables $X$, $Y$ with pmf's $f_X$ and $f_Y$, as well as a third (degenerate) random variable $Z$ that is defined to be equal to the expectation of $Y$, i.e., $Z=E[Y]$.
Then the expected values of X+Y and X+Z are obviously the same by linearity of expectation, i.e., \begin{align} &\sum_{x= 0}^\infty x f_{X+Y}(x) =E[X+Y] = E[X+Z] = \sum_{x= 0}^\infty x f_{X+Z}(x) \end{align}

What I'm now unclear about is how the partial sums of the two expected values compare, specifically whether the following statement is true in general: \begin{align} \sum_{x= 0}^{c}x f_{X+Y}(x) \leq& \sum_{x= 0}^c x f_{X+Z}(x) \end{align} for all (integer) $c \geq E[X+Y]$.

Since the variance of $Z$ is set to 0, it stands to reason that the impact of extreme values on the expectation is at least usually lowered when replacing Y with Z and at least for $c$ high enough the statement should typically hold. What I'm unclear about is whether it really holds for all possible $X$, $Y$ and all $c\geq E[X+Y]$ (note that it does NOT hold for all $c < E[X+Y]$! For small $c$, the inequality has to be reversed). Note that the fact that the variance decreases when exchanging $Y$ for $Z$ alone is not enough for the statement to hold (i.e., consider extremely skewed $Z$), but similar counterexamples seem to fail as long as $Z$ is degenerate (i.e. all central moments of Z are 0).

As a sidenote, the equivalent statement for the cumulative probability function does NOT hold in general, i.e., we can construct distributions such that \begin{align} \sum_{x= 0}^{c} f_{X+Y}(x) >& \sum_{x= 0}^{c} f_X(x) \\ \end{align}

Edit: As a small addendum, if no one has a ready made solution at hand, I would also be interested in pointers to possibly relevant literature or keywords that might help me solving it myself. I'm sadly a bit out of my depth with this type of probability theory.

Answer 1

I've actually to find an answer myself. As posed, the statement does not hold. The following X and Y provide a counterexample: $P(X=11/10)= 0.9$, $P(X=10)=0.1$, $P(Y= 1)= 0.1$, $P(Y=21/10)= 0.9$

Then
\begin{align} &\sum_{x= 0}^{11}x f_{X+Y}(x) = 2.91 > 2.8 = \sum_{x= 0}^{11} x f_{X+Z}(x) \end{align}

Note that this counterexample makes use of the fact that both $X$ and $Y$ are quite skewed, importantly with both skews having opposing signs. So by replacing $Y$ with its expectation, the skewness of the sum strictly increases.