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Distribution-free bounds on how far a variable strays, what they assume, and how loose they have to be.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will bound the probability that a non-negative variable exceeds a threshold from its mean alone, bound the probability that any variable strays a given number of standard deviations from its mean, and say what each inequality assumes. You will also run Chebyshev backwards to find how many independent measurements a stated guarantee costs.
A mean says where a distribution sits and a variance says how far it spreads. This lesson asks what those two numbers alone can guarantee about a distribution nobody has identified — which turns out to be surprisingly much, and never very sharp.
A bound is a ceiling no distribution of the stated kind can exceed; it is not an estimate of anything. A statement is distribution-free when it assumes nothing about the shape. A bound is tight when some distribution attains it — both inequalities here are tight, which is exactly why neither can be improved without assuming more.
Markov's inequality. For a non-negative variable and any $a > 0$,
$$P(X \ge a) \le \frac{E[X]}{a}.$$
The proof is one line: the expectation is at least the part of it contributed by the outcomes at or above $a$, and that part is at least $a\,P(X \ge a)$. Non-negativity is doing real work — without it a large positive tail could be paid for by a negative one, and the mean would say nothing.
Chebyshev's inequality. For any variable with finite variance,
$$P(|X - \mu| \ge k\sigma) \le \frac{1}{k^{2}}.$$
It is Markov applied to $(X - \mu)^{2}$, which is non-negative by construction and has expectation $\sigma^{2}$. That is the whole derivation, and it is why the bound squares.
Both are distribution-free: no bell, no symmetry, no continuity is assumed. The price is that they are loose — a normal variable has under $5\%$ beyond two standard deviations against Chebyshev's $25\%$ — and they cannot be improved, because for each $k$ there is a distribution attaining the bound exactly. Loose for almost everything and sharp for something is what a universal bound looks like.
Another way: story
You are told the average household in a town has two occupants, and nothing else. Can more than a fifth of households have ten or more? Markov says no: ten or more occupants, at an average of two, cannot happen more than a fifth of the time, whatever the distribution of household sizes is.
Another way: steps
The true tail beyond $k$ standard deviations, for three distributions, against the bound.
| $k$ | Chebyshev's bound | Normal | Uniform | Two-point extreme |
|---|---|---|---|---|
| 2 | 0.25 | 0.046 | 0 | 0.25 |
| 3 | 0.111 | 0.0027 | 0 | 0.111 |
| 4 | 0.0625 | 0.00006 | 0 | 0.0625 |
The last column is the distribution built to attain the bound: a variable sitting at $\mu$ with probability $1 - 1/k^{2}$ and at $\mu \pm k\sigma$ with the rest. It exists for every $k$, which is why no sharper universal bound is possible — and why the inequality is worth exactly as much as the ignorance it assumes.
Reading the bound as the probability. It is a ceiling. Quoting $25\%$ where the truth is $5\%$ is not a conservative estimate, it is a different claim.
Using Markov on a variable that can be negative. The inequality is false without non-negativity, and the failure is not subtle.
Forgetting to square. Chebyshev is $1/k^{2}$, not $1/k$.
Measuring the threshold in the wrong units. $k$ counts standard deviations, so a threshold given in the variable's units has to be divided by $\sigma$ first.
Households average $2$ occupants. Bound the share with $10$ or more.
Occupancy is non-negative.
$P(X \ge 10) \le 2/10 = 0.2$.
Mean over threshold.
True figure is far below, and this is all that the mean alone can support.
A ceiling, honestly reported.
The same households have standard deviation $1$. Bound the share with $10$ or more.
That is $8$ standard deviations above the mean.
$P(|X - 2| \ge 8) \le 1/64 \approx 0.016$.
One over the square.
Ten times tighter than Markov, from one extra number.
The variance buys the square.
The distance from the mean is $15$, which is $15/5 = 3$ standard deviations.
Convert to standard deviations first.
Chebyshev gives $1/3^{2}$.
One over the square.
So at most $1/9 \approx 0.111$ of the probability is outside.
A variable has standard deviation $9$ and an unknown distribution. Give Chebyshev's bound on the probability of being more than $k$ standard deviations from the mean, and the share that is therefore guaranteed to be within it.
| Bound on being outside | Guaranteed inside | |
|---|---|---|
| $k = 2$ | ||
| $k = 4$ | ||
| $k = 5$ |
A non-negative variable has mean $6$ and standard deviation $6$. Give Markov's bound on $P(X \ge 24)$ and Chebyshev's bound on $P(|X - 6| \ge 24)$.
Markov: p. Chebyshev: q.
Repair times are non-negative and average $9$ hours. What is the largest that the probability of a repair taking $45$ hours or more could be?
Answer:
Chebyshev's inequality gives $P(|X - \mu| \ge 2\sigma) \le 0.25$ for a variable with standard deviation $5$. What does that statement claim?
Independent measurements each have variance $3$. Their average has variance $3/n$. Using Chebyshev, give the smallest $n$ for which the average is within $1$ of the true mean with at least each guarantee.
| Smallest $n$ | |
|---|---|
| Within $1$ at least $75\%$ of the time | |
| Within $1$ at least $90\%$ of the time | |
| Within $1$ at least $96\%$ of the time |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A variable has mean $80$ and standard deviation $2$. Give the interval within $4$ standard deviations of the mean, with both endpoints included.
This task has no paper form; do it on a device.
You can apply Markov and Chebyshev, convert a threshold into standard deviations, and report a bound as a ceiling rather than an estimate. Say in your own words why a bound that holds for every distribution has to be loose for almost all of them.
9. Your turn: a variable has mean $50$ and standard deviation $5$; bound the probability of being outside $35$ to $65$, step 3