What is Moments, variance?

Example 22.16 (Uncorrelated but glued together)

Roll two fair dice, $X$ and $Y$ independent, and set $S = X + Y$, $D = X - Y$. By bilinearity of the covariance,

sum and difference are uncorrelated. Independent? Certainly not: $S = 12$ forces $D = 0$, while $\P(D = 0) = \frac16$ unconditionally. Correlation only tests the linear part of a dependence; here the dependence is carried by the constraint that $S$ and $D$ have the same parity, invisible to covariance. (For this pair, zero covariance needed $V(X) = V(Y)$: identical distributions, not independence, did the work.)

Example 22.17 (When Markov is exact)

Markov’s inequality is an equality precisely when nothing is wasted in the bound $a\,\mathbf 1_{X\geq a} \leq X$: the variable must take only the values $0$ and $a$. Concretely, if $\P(X = a) = \pi$ and $\P(X = 0) = 1 - \pi$, then $\E(X) = a\pi$ and

A realistic reading: in a population where average wealth is $100$ and wealth is either $0$ or $10^6$, the proportion of millionaires is exactly $10^{-4}$ — Markov’s bound, hit exactly by maximal inequality. Whenever $X$ spreads over intermediate values the bound is strict, often wildly so; but as the extreme case shows, no better inequality can be extracted from the mean alone.

Example 22.18 (Chebyshev is sharp — without further hypotheses)

Fix $\varepsilon > 0$, $q \in \intoc01$, and let $X$ take the values $\pm\varepsilon$ with probability $\frac q2$ each and $0$ with probability $1 - q$. Then $\E(X) = 0$, $V(X) = q\varepsilon^2$, and

equality in Chebyshev. So the inequality cannot be improved using only the variance — the $1/\varepsilon^2$ decay is the exact price of second-moment information. Faster decay requires stronger hypotheses: boundedness of the variable buys exponential concentration, as Exercise 22.7 previews and this chapter’s weekend problem develops systematically.