Bound an answer before calculating it

A bound can expose an impossible answer before you finish the derivation. Start with what the variable can do, then use the information the problem actually supplies.

A probability lies between zero and one. The mean of a variable confined to 2 through 8 must lie in that interval. Doubling every payoff doubles its mean; adding a fixed fee reduces profit by that fee on every path where it is charged.

These checks are cheap. They are not a proof that an answer inside the permitted range is correct.

A bound from a mean

Let X be nonnegative with expected value four. What can you say about the probability that X is at least ten?

Whenever X is at least ten, it contributes at least ten to its own value. Pointwise:

X ≥ 10 × I(X ≥ 10).

Taking expectations gives 4 ≥ 10 P(X ≥ 10), so the probability is at most 0.4. This is an instance of Markov's inequality, derived here from the indicator rather than quoted without conditions.

The nonnegative assumption is essential. Negative values could offset a large positive tail in the mean, breaking this argument.

A bound is not an estimate

The mean alone does not identify the tail probability. If X is always four, the probability of reaching ten is zero. If X is ten with probability 0.4 and zero otherwise, the mean remains four and the probability is 0.4. Both satisfy the supplied facts.

Therefore “the answer is 40%” overstates the conclusion. “The probability is no more than 40%, and the exact value is not identified” is the supported statement.

Use small cases

Suppose a counting formula claims that two draws from three categories always cover three categories. The two-draw case immediately exposes the error: at most two categories can appear. A result can fail a support check even when every arithmetic operation was performed correctly.

For a stopping policy, inspect zero continuation cost and very large cost. A rule that keeps paying an enormous fee to replace a bounded payoff likely violates its own objective. For a signal problem, inspect a perfectly uninformative detector: its posterior should equal the prior.

Practice prompt

A nonnegative count has mean 1.5. Give a valid upper bound on the chance it is at least five, and an example showing the bound can be attained.

Answer: the probability is at most 1.5/5 = 0.3. A count equal to five with probability 0.3 and zero otherwise attains it.

Use indicators to make these arguments explicit. The probability workbook adds worked practice and checks across several problem types.