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 $P(X=m)=0, P(X<m) = P(X>m) = 1/2$.

(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.)


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – antkam
    Sep 5, 2019 at 1:34
  • $\begingroup$ 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... $\endgroup$
    – antkam
    Sep 5, 2019 at 1:39
  • $\begingroup$ @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 :) $\endgroup$ Sep 5, 2019 at 1:58
  • $\begingroup$ "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... $\endgroup$
    – antkam
    Sep 5, 2019 at 2:12
  • $\begingroup$ Also, great to hear you enjoy my other questions! Since math is mostly my hobby (i.e. not my job) some of my questions are really about elegance rather than result. E.g. I can answer both (a) math.stackexchange.com/questions/3140502/… & (b) math.stackexchange.com/questions/2963520/… by, respectively, (a) Monte-Carlo, & (b) careful counting. But in both I thought there might be an elegant answer (in case (a) even if it's only qualitative, not quantitative). $\endgroup$
    – antkam
    Sep 5, 2019 at 2:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .