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.

  • $\begingroup$ Your language is ambiguous because the meaning of "weighted mass" is unclear. Unfortunately, the math notation doesn't help, because that first sum quite literally has a single term. Please clarify. $\endgroup$
    – whuber
    Nov 25, 2019 at 19:15
  • $\begingroup$ I tried to clarify the question. I'm not sure why you believe the sum only has a single term though, the sum runs from Exp[X+Y]+a (or the next highest integer) to infinity. Also note that the equality between the first and second line is simply the definition of the convolution. What I'm not sure about is the inequality between the second and the third line. $\endgroup$
    – Borlin
    Nov 25, 2019 at 22:25
  • 1
    $\begingroup$ You might find it useful to consider $\text{Var}(X+Y)$ vs $\text{Var}(X+E[Y])$; it's easy to see when the second is and is not smaller than the first. Perhaps it's not exactly what you're after, but it may be useful to consider, nonetheless. $\endgroup$
    – Glen_b
    Nov 25, 2019 at 22:51
  • $\begingroup$ The second is always smaller, considering Var(X+Y) = Var(X)+Var(Y) (because X and Y are independent) and Var(X+E[Y])= Var(X) (because E[Y] is a constant). Sadly, the variance being smaller on its own does not suffice to satisfy the inequality. $\endgroup$
    – Borlin
    Nov 25, 2019 at 22:58
  • $\begingroup$ Aren't you making some implicit assumptions that, say, $X$ and $Y$ are non-negative random variables? As with @Glen_b, I'm trying to understand what you might mean by "mass ... shifts closer to the expectation." I would, like him, have guessed you might have in mind some absolute central moment like the variance, but evidently that's not the case. Could you give us a definition or formula? $\endgroup$
    – whuber
    Nov 26, 2019 at 0:06

1 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: $P(X=11/10)= 0.9$, $P(X=10)=0.1$, $P(Y= 1)= 0.1$, $P(Y=21/10)= 0.9$

\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.

  • $\begingroup$ Now I'm kinda wondering whether the original statement holds if both $X$ and $Y$ are skewed in the same direction (i.e. their skewness has the same sign)? Is there a way to ask follow up questions or should I open a new question for that? $\endgroup$
    – Borlin
    Nov 28, 2019 at 0:06

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