# Does increasing 2nd moment about "median" imply increasing variance?

This is a continuation of: Variance of X vs Variance of a binary function of X

In that thread, I posted a partial answer, which was accepted. However my answer has a gap. I'm now asking for help to fix the gap.

Let $$X$$ be a (not necessarily continuous) random variable in $$[0, 1]$$, with this additional property: Suppose there exists $$m \in (0, 1)$$ s.t. $$P(X < m) = P(X > m)$$. Define random variable $$Y$$ by "pushing" the probability "mass" on each side of $$m$$ to the limit values:

• If $$X < m$$ then $$Y = 0$$.

• If $$X > m$$ then $$Y = 1$$.

• If $$X = m$$ then $$Y = m$$.

Question: Is $$Var(X) \le Var(Y)$$? (If true, this would complete my proof in the old thread.)

Thoughts: I'm conjecturing that $$Var(X) \le Var(Y)$$ should be true, based on some "moment of inertia" considerations. By construction, the 2nd moment about $$m$$ has increased, i.e. $$E[(Y-m)^2] \ge E[(X-m)^2]$$. If $$m = E[X] = E[Y]$$ then the conjecture follows at once. The problem is that $$m, E[X], E[Y]$$ can be 3 different values, and I am unable to track/bound them enough to prove the conjecture.

Incidentally, there are 2 cases to this problem. I would like an answer that covers both cases:

(1) $$\forall x \in [0, 1]: P(X=x) = 0$$. In this case, $$m$$ always exists and is the median, and $$P(X=m)=0, P(Xm) = 1/2$$.

(2) $$\exists x \in [0,1]: P(X=x) > 0$$. In this case, $$m$$ might not exist (e.g. $$X = \{0, 0.5, 1\}$$ with probabilities $$\{0.2, 0.4, 0.4\}$$ respectively). This question presupposes that $$m$$ exists, in which case $$P(X=m) = p$$ might be positive or zero. (The $$p>0$$ subcase is what I need to complete my old proof.)

For ease, redefine $$X = X-m$$ and $$Y = Y-m$$ (this doesn't change variances). Let $$x = P(X=0)$$. Then $$X = 0$$ with prob $$x$$, $$X > 0$$ with prob $$\frac{1-x}{2}$$, and $$X < 0$$ with prob $$\frac{1-x}{2}$$. We first change $$X$$ to be $$0$$ with prob $$x$$, $$X = 1-m$$ with prob $$\frac{1-x}{2}$$, and $$X < 0$$ has the same distribution as before. The change in variance is $$\left[(1-m)^2(\frac{1-x}{2})+\int_{X < 0} X^2 - \left((\frac{1-x}{2})(1-m)+\int_{X < 0} X\right)^2\right] - \left[\int_{X > 0} X^2+\int_{X < 0} X^2 - \left(\int_{X > 0} X + \int_{X < 0} X\right)^2\right].$$ Since $$\int_{X < 0} X < 0$$ and $$\int_{X > 0} X \le (1-m)(\frac{1-x}{2})$$, the term $$-2(\frac{1-x}{2})(1-m)\int_{X < 0} X + 2\left(\int_{X > 0} X\right)\int_{X < 0} X$$ is non-negative, so it suffices to show $$\int_{X > 0} X^2 - \left(\int_{X > 0} X\right)^2 \le (1-m)^2(\frac{1-x}{2})-(1-m)^2(\frac{1-x}{2})^2.$$ Note $$\int_{X > 0} X^2 - \left(\int_{X > 0} X\right)^2 \le (1-m)\int_{X > 0} X - \left(\int_{X > 0} X\right)^2.$$ The quadratic $$(1-m)y-y^2$$ is increasing for $$y \le \frac{1-m}{2}$$, so since $$\int_{X > 0} X \le (1-m)(\frac{1-x}{2})$$, we get the desired inequality. To go from this changed $$X$$ to $$Y$$, just do the analogous change for $$X < 0$$ and run through the same proof.
• First, thanks very much for taking the time to answer my question from a year ago, and for which nobody else even voted. :) I really appreciate it. AFAICT there are two minor typos: (A) in the 2nd line it should be $x=P(X=0)$, and (B) in the 4th line after the big math expression the LHS should have $(\int_{X>0} X)^2$ instead of $(\int_{X>0}X)$. If you agree, and fix these, I will happily and gratefully accept this answer. Sep 5, 2019 at 1:34
• I have a follow-up question though: how did you find this proof? I don't recall what I tried from 1+ year ago, but I remember trying various manipulations vaguely similar to these. I probably even tried doing one side first... although I certainly don't recall anything similar to "the quadratic $(1-m)y- y^2$ is increasing for $y\le \frac{1-m}{2}$." Is there some geometric or other intuition that led you down this path? I would have been very surprised if my conjecture were false, but I just couldn't find the algebraic manipulations to chain together a proof... Sep 5, 2019 at 1:39
• @antkam I spent more time on this question than I would like to admit, trying various algebraic manipulations (without geometric insight or anything like that). I knew for a fact that the result had to be true. One source of motivation was that case (1) you had mentioned in your question became $E(X^2)-E(X)^2 \le \frac{1}{4}$, which is proved via $E(X^2) \le E(X)$ and analyzing the quadratic. For some reason, I only was able to obtain a proof after "doing one side first". Not much else I can say. PS: I've seen your other questions and liked them a lot so I did some searching and found this :) Sep 5, 2019 at 1:58
• "I spent more time on this question than I would like to admit" -- haha, now I felt better! :) Yeah it just had to be true, and I still feel there "should" be a simpler proof than this, but this certainly works. Come to think of it, shifting to $X-m$ helps the algebra quite a bit. Perhaps I was misled by thinking: since there is an intuitive argument based on 2nd moment, there has to be a simple argument too, when in fact the proofs of various "intuitive" properties about the 2nd moment themselves as complicated as what you found... Sep 5, 2019 at 2:12