Inequalities: Markov, Chebyshev and Jensen

All you know about a random quantity is that it is never negative and its mean is 10. How likely is it to reach 50 or more? It can't be very likely. If more than a fifth of the probability sat at 50 or above, those outcomes alone would push the mean past 10.

Markov's inequality makes this exact. For $X \ge 0$ and any $a > 0$,

The proof takes one line. Since $X \ge 0$, you always have $X \ge a \cdot \mathbf{1}\{X \ge a\}$: when $X \ge a$ the right side is $a$, and otherwise it is 0. Take expectations to get $E[X] \ge a\,P(X \ge a)$.

With only the mean known, the bound can't be improved. Put probability 0.2 at exactly 50 and 0.8 at 0. The mean is 10 and $P(X \ge 50) = 0.2$. Any mass placed above 50, or strictly between 0 and 50, uses up mean without adding to the tail.

Both conditions matter. A P&L that pays $50 with probability 0.99 and loses $3,950 with probability 0.01 has mean 10, yet it reaches 50 with probability 0.99. And the mean must be finite, which the lesson on Infinite Expectations showed you can't take for granted.

A typical interview version: the average of 100 nonnegative numbers is 10, so at most how many can be 50 or more? Markov gives $100 \times 10/50 = 20$, and twenty 50s with eighty 0s shows 20 is reachable. The same one-liner says a desk whose average order is 200 shares sees orders of 1,000 shares or more at most 20% of the time. The true fraction is usually far smaller, because Markov uses nothing but the mean.