Statistics Graduate Level Quiz: Markov And Chebyshev Inequalities

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Markov And Chebyshev InequalitiesQuestion 1 of 10

Let $Y$ have mean $0$ and variance $4$ . For any $c\ge 0$ , Markov's inequality applied to $(Y+c)^2$ yields an upper bound on $P(Y\ge 3)$ . What is the smallest bound obtainable by optimizing over $c$ ?

\frac{4}{13}

\frac{4}{9}

\frac{2}{5}

\frac{4}{49}

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Markov And Chebyshev Inequalities

Practice Markov And Chebyshev Inequalities in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Markov And Chebyshev Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\frac{4}{13}$ (correct answer)
$\frac{4}{9}$
$\frac{2}{5}$
$\frac{4}{49}$

Explanation: When you see a problem asking you to apply Markov's inequality to a transformed variable and then optimize, recognize that this is a variance-bounding technique: by shifting

Y

c

before squaring, you encode both mean and variance information into the bound. Markov's inequality applied to the non-negative random variable

(Y+c)^2

gives:

P\!\left((Y+c)^2 \ge (3+c)^2\right) \ge P(Y \ge 3) \implies P(Y \ge 3) \le \frac{E[(Y+c)^2]}{(3+c)^2}

Now compute the numerator:

E[(Y+c)^2] = E[Y^2] + 2cE[Y] + c^2 = 4 + 0 + c^2 = c^2 + 4

So the bound becomes

f(c) = \frac{c^2 + 4}{(3+c)^2}

. To minimize, take the derivative and set it to zero:

f'(c) = \frac{2c(3+c)^2 - 2(3+c)(c^2+4)}{(3+c)^4} = 0 \implies 2c(3+c) = 2(c^2+4)

6c + 2c^2 = 2c^2 + 8 \implies c = \frac{4}{3}

Substituting back:

f\!\left(\tfrac{4}{3}\right) = \frac{\tfrac{16}{9}+4}{\left(\tfrac{13}{3}\right)^2} = \frac{\tfrac{52}{9}}{\tfrac{169}{9}} = \frac{52}{169} = \frac{4}{13}

That confirms A is correct. Choice B,

\frac{4}{9}

, corresponds to naively using

c=0

(standard Chebyshev-like approach without optimization). Choice C,

\frac{2}{5}

, results from an algebraic error in the derivative. Choice D,

\frac{4}{49}

, is far too small and would violate achievable probability bounds given the variance. Study tip: Whenever a problem says "optimize over a free parameter," always differentiate and solve — don't guess "nice" values of

c

like 0 or 1. The optimal

c

is rarely obvious.

Question 2

Suppose $X_1,\ldots,X_{50}$ are pairwise independent Bernoulli random variables with common success probability $0.2$ . What upper bound does Chebyshev's inequality give for $P(|\overline X-0.2|\ge0.1)$ ?

$0.064$
$0.16$
$0.32$ (correct answer)
$0.40$

Explanation: Chebyshev's inequality applies whenever you know a random variable's mean and variance — and critically, it only requires pairwise independence, making it perfectly valid here. Chebyshev states:

P(|\overline{X} - \mu| \ge \epsilon) \le \frac{\text{Var}(\overline{X})}{\epsilon^2}

Start by finding the ingredients. Each

X_i \sim \text{Bernoulli}(0.2)

, so

\mu = 0.2

and

\text{Var}(X_i) = p(1-p) = 0.2 \times 0.8 = 0.16

. Because the variables are pairwise independent, the variance of the sample mean is:

\text{Var}(\overline{X}) = \frac{\text{Var}(X_i)}{n} = \frac{0.16}{50} = 0.0032

With

\epsilon = 0.1

, plug into Chebyshev:

P(|\overline{X} - 0.2| \ge 0.1) \le \frac{0.0032}{(0.1)^2} = \frac{0.0032}{0.01} = 0.32

So the correct answer is C. As for the distractors: A (0.064) likely comes from forgetting to divide

\text{Var}(X_i)

n

, then misapplying the formula further. B (0.16) results from using

\text{Var}(X_i) = 0.16

directly in the numerator without dividing by

n = 50

— a very common error. D (0.40) may arise from using

\epsilon = 0.1

incorrectly, perhaps forgetting to square it. Study tip: Always remember that Chebyshev uses

\text{Var}(\overline{X}) = \sigma^2/n

, not the population variance

\sigma^2

. Writing out all three quantities —

\text{Var}(\overline{X})

\epsilon

, and

\epsilon^2

— before substituting will prevent the most common arithmetic traps.

Question 3

A random variable $X$ has mean $\mu$ , variance $2$ , and fourth central moment $E[(X-\mu)^4]=48$ . Considering both Chebyshev's inequality and Markov's inequality applied to $(X-\mu)^4$ , what is the strongest resulting upper bound for $P(|X-\mu|\ge4)$ ?

$\frac{1}{8}$ (correct answer)
$\frac{3}{16}$
$\frac{1}{4}$
$\frac{3}{8}$

Explanation: When a question asks you to find the "strongest upper bound" using multiple inequalities, your job is to compute each bound and take the minimum — the tightest constraint wins. Chebyshev's inequality states

P(|X-\mu| \ge k) \le \frac{\text{Var}(X)}{k^2}

. With

\text{Var}(X) = 2

and

k = 4

P(|X-\mu| \ge 4) \le \frac{2}{16} = \frac{1}{8}

. Markov's inequality applied to

(X-\mu)^4

is the more creative step. Notice that

|X-\mu| \ge 4

if and only if

(X-\mu)^4 \ge 4^4 = 256

. Treating

(X-\mu)^4

as a non-negative random variable, Markov gives:

P\!\left((X-\mu)^4 \ge 256\right) \le \frac{E[(X-\mu)^4]}{256} = \frac{48}{256} = \frac{3}{16}

. Comparing the two bounds:

\frac{1}{8} = \frac{2}{16} < \frac{3}{16}

, so Chebyshev's inequality yields the tighter bound. The strongest upper bound is

\frac{1}{8}

, confirming answer A. As for the distractors: B (

\frac{3}{16}

) is exactly the Markov bound — correct in isolation, but not the strongest. C (

\frac{1}{4}

) likely comes from misapplying Chebyshev with

k=2

or forgetting to square the denominator. D (

\frac{3}{8}

) suggests an arithmetic error when scaling the fourth-moment calculation. Study tip: When you see higher-order moments given alongside variance, that's a signal to apply Markov's inequality to a power of

(X-\mu)

. Always compute all available bounds and select the minimum — "strongest" always means smallest upper bound.

Question 4

Ten estimators $T_1,\ldots,T_{10}$ are unbiased for the same parameter $\theta$ . Each has variance $1$ , and every pair satisfies $\operatorname{Cov}(T_i,T_j)\le 0.2$ . No independence assumption is made.

Using only the stated moment constraints and Chebyshev's inequality, what is the smallest guaranteed upper bound on $P\left(\left|10^{-1}\sum_{i=1}^{10}T_i-\theta\right|\ge1\right)$ ?

$0.10$
$0.20$
$0.28$ (correct answer)
$0.36$

Explanation: When you see a question combining correlated estimators with Chebyshev's inequality, your first job is to compute the variance of the sample mean exactly — no shortcuts. Let

\bar{T} = \frac{1}{10}\sum_{i=1}^{10} T_i

. Since each

T_i

is unbiased for

\theta

, so is

\bar{T}

, meaning

E[\bar{T}] = \theta

. Now expand the variance:

\operatorname{Var}(\bar{T}) = \frac{1}{100}\left[\sum_{i=1}^{10}\operatorname{Var}(T_i) + 2\sum_{i<j}\operatorname{Cov}(T_i,T_j)\right]

There are 10 variance terms each equal to 1, and

\binom{10}{2} = 45

covariance pairs each bounded above by 0.2. To find the worst-case (largest) variance — which gives the largest probability bound — plug in the maximum covariances:

\operatorname{Var}(\bar{T}) \le \frac{1}{100}[10 + 2(45)(0.2)] = \frac{1}{100}[10 + 18] = \frac{28}{100} = 0.28

Applying Chebyshev's inequality with

\epsilon = 1

P(|\bar{T} - \theta| \ge 1) \le \frac{\operatorname{Var}(\bar{T})}{1^2} \le 0.28

This confirms C) 0.28 as the smallest guaranteed upper bound consistent with all stated constraints. Choice A (0.10) ignores the covariance structure entirely, as if the estimators were independent — giving

10/100

, which underestimates the worst case. Choice B (0.20) accounts for only the covariance terms and drops the variance terms, a partial calculation error. Choice D (0.36) overcounts — perhaps mistakenly using 90 pairs instead of 45, double-counting each covariance. The study tip: always count your pairs carefully. With

n

estimators there are

\binom{n}{2}

covariance pairs, not

n^2 - n

. That factor-of-2 mistake is exactly the trap that produces distractor D.

Question 5

A random variable $X$ has finite mean $\mu$ and satisfies $P(|X-\mu|\ge5)=0.2$ . Based only on this information, what is the strongest universal lower bound on $\operatorname{Var}(X)$ ?

$1$
$5$ (correct answer)
$25$
$125$

Explanation: Whenever you see a question involving a probability bound on deviations from the mean, your instinct should be to reach for Chebyshev's inequality, which states:

P(|X - \mu| \ge k) \le \frac{\operatorname{Var}(X)}{k^2}

This inequality holds for any distribution with finite variance — no additional assumptions needed. The key insight here is that you're given

P(|X-\mu| \ge 5) = 0.2

and asked for the strongest universal lower bound on

\operatorname{Var}(X)

. Plugging into Chebyshev's inequality, you know the bound must hold:

0.2 \le \frac{\operatorname{Var}(X)}{25}

Solving gives

\operatorname{Var}(X) \ge 5

. So B) 5 is the correct answer — it's the tightest lower bound guaranteed by this single probability statement alone. Now let's address the distractors. A) 1 is too weak; while it's technically true that

\operatorname{Var}(X) \ge 1

when

\operatorname{Var}(X) \ge 5

, it's not the strongest bound derivable, making it an inferior answer. C) 25 would be the answer if you mistakenly set

\operatorname{Var}(X) \ge k^2 = 25

, confusing the inequality direction or forgetting to multiply by the probability. D) 125 likely comes from multiplying

25 \times 5

without a valid justification — there's no standard inequality that produces this bound from the given information. A useful study habit: when Chebyshev's inequality appears, always write it out symbolically before plugging in numbers. The direction of the inequality (variance in the numerator) is the most common source of errors, and writing it cleanly prevents costly mistakes.

Question 6

Random variables $X$ and $Y$ have means $\mu_X$ and $\mu_Y$ and variances $1$ and $4$ , respectively. No assumption is made about their dependence. Using Chebyshev's inequality and a union bound, what upper bound is obtained for $P(|X-\mu_X|\ge2\ \text{or}\ |Y-\mu_Y|\ge3)$ ?

$\frac{1}{9}$
$\frac{25}{36}$ (correct answer)
$\frac{5}{12}$
$\frac{4}{5}$

Explanation: When you see a problem asking for a probability bound on a union of events without assuming independence, your two key tools are Chebyshev's inequality and the union bound (Boole's inequality). Chebyshev's inequality states that

P(|X - \mu_X| \ge k) \le \frac{\text{Var}(X)}{k^2}

. Apply it to each event separately. For

X

with variance 1 and threshold 2:

P(|X - \mu_X| \ge 2) \le \frac{1}{4}

. For

Y

with variance 4 and threshold 3:

P(|Y - \mu_Y| \ge 3) \le \frac{4}{9}

. The union bound says

P(A \cup B) \le P(A) + P(B)

for any events, regardless of dependence. Adding the two Chebyshev bounds gives

\frac{1}{4} + \frac{4}{9} = \frac{9}{36} + \frac{16}{36} = \frac{25}{36}

, confirming that B is correct. Now for the distractors. A (

\frac{1}{9}

) might tempt you if you mistakenly apply Chebyshev to

X - Y

as a single variable, or confuse the threshold arithmetic. C (

\frac{5}{12}

) likely arises from averaging the two bounds rather than summing them — averaging is never justified by the union bound. D (

\frac{4}{5}

) could come from incorrectly computing individual Chebyshev bounds, perhaps using variances where standard deviations belong in the denominator. A useful strategy: always separate the union into individual events, bound each with Chebyshev, then sum. Remember,

k

in Chebyshev is the threshold value, not the number of standard deviations — don't forget to square it properly.

Question 7

A nonnegative random variable $X$ has mean $4$ and variance $9$ . Using the sharper of a direct application of Markov's inequality to $X$ and a direct application of the two-sided Chebyshev inequality, what upper bound is obtained for $P(X\ge 10)$ ?

$\frac{1}{4}$ (correct answer)
$\frac{2}{5}$
$\frac{9}{100}$
$\frac{3}{20}$

Explanation: When bounding tail probabilities, you have two classical tools: Markov's inequality and Chebyshev's inequality. The key skill here is applying both correctly and selecting the tighter bound. Markov's inequality applies to any nonnegative random variable:

P(X \ge a) \le \frac{E[X]}{a}

. With

E[X] = 4

and

a = 10

, this gives

P(X \ge 10) \le \frac{4}{10} = \frac{2}{5}

. Chebyshev's inequality (two-sided) states:

P(|X - \mu| \ge k) \le \frac{\sigma^2}{k^2}

. Here

\mu = 4

\sigma^2 = 9

, and we want

X \ge 10

, which means

X - 4 \ge 6

, so

|X - 4| \ge 6

(since the deviation is in one direction). Setting

k = 6

P(|X-4| \ge 6) \le \frac{9}{36} = \frac{1}{4}

. Since

\{X \ge 10\} \subseteq \{|X-4| \ge 6\}

, this bound also applies to

P(X \ge 10)

. Comparing:

\frac{1}{4} = 0.25 < \frac{2}{5} = 0.40

, so Chebyshev gives the sharper (smaller) upper bound of $\frac{1}{4}$ , confirming answer A. Choice B (

\frac{2}{5}

) is the Markov bound alone — valid but not the sharper one. Choice C (

\frac{9}{100}

) comes from mistakenly using

k = 10

in Chebyshev instead of

k = 6

(the distance from the mean to 10). Choice D (

\frac{3}{20}

) has no clean derivation from either inequality applied correctly. Study tip: When applying Chebyshev to a one-sided event like

X \ge c

, always compute

k = c - \mu

, not

k = c

itself — that's the most common arithmetic trap on these problems.

Question 8

Let $\mathcal G$ be a sigma-field. Suppose $E[X\mid\mathcal G]=M$ and $\operatorname{Var}(X\mid\mathcal G)\le V$ almost surely, where $V$ is nonnegative and $E[V]=3$ .

Which unconditional upper bound follows by applying Chebyshev's inequality conditionally and then averaging for $P(|X-M|\ge2)$ ?

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$ (correct answer)
$1$ only

Explanation: Whenever you see conditional probability bounds combined with averaging, think of the smoothing/tower property of conditional expectation paired with a classical inequality applied at the conditional level first. Here's the core reasoning. Chebyshev's inequality applied conditionally given

\mathcal{G}

states:

P(|X - M| \ge 2 \mid \mathcal{G}) \le \frac{\operatorname{Var}(X \mid \mathcal{G})}{2^2} = \frac{\operatorname{Var}(X \mid \mathcal{G})}{4}

almost surely. Since

\operatorname{Var}(X \mid \mathcal{G}) \le V

a.s., we get:

P(|X - M| \ge 2 \mid \mathcal{G}) \le \frac{V}{4} \quad \text{a.s.}

Now apply the tower property by taking unconditional expectations of both sides:

P(|X - M| \ge 2) = E\!\left[P(|X - M| \ge 2 \mid \mathcal{G})\right] \le E\!\left[\frac{V}{4}\right] = \frac{E[V]}{4} = \frac{3}{4}

This gives C,

\frac{3}{4}

, as the correct upper bound. Choice A (

\frac{1}{4}

) would result from forgetting to divide by

2^2 = 4

and instead treating

E[V]

as if you divide by

4^2

, or equivalently misreading the threshold as

t = 4

. Choice B (

\frac{1}{2}

) arises from halving

E[V]

without proper application of the denominator

t^2

, perhaps confusing Chebyshev with a one-sided bound. Choice D (

1

only) reflects abandoning the inequality entirely and concluding no useful bound exists — but Chebyshev and the tower property give a tighter result. Your key strategy: conditional → unconditional is a two-step move. Apply the inequality inside the conditioning, then take expectations. The denominator is always

t^2

, never

t

Question 9

Among the following distributions, which one has mean $0$ and variance $4$ and attains equality in Chebyshev's bound for $P(|X|\ge3)$ ?

$P(X=-1)=\frac{4}{5}$ and $P(X=4)=\frac{1}{5}$
$P(X=-2)=P(X=2)=\frac{1}{2}$ and no mass elsewhere
$P(X=-4)=P(X=4)=\frac{1}{8}$ and $P(X=0)=\frac{3}{4}$
$P(X=-3)=P(X=3)=\frac{2}{9}$ and $P(X=0)=\frac{5}{9}$ (correct answer)

Explanation: Chebyshev's inequality states that for any distribution with mean

\mu

and variance

\sigma^2

P(|X - \mu| \ge k) \le \frac{\sigma^2}{k^2}

. A question like this asks you to simultaneously verify three conditions: mean zero, variance 4, and equality in Chebyshev's bound. That last condition is the key constraint most students overlook. For a distribution with

\mu = 0

and

\sigma^2 = 4

, Chebyshev gives

P(|X| \ge 3) \le \frac{4}{9}

. Equality is achieved only when all probability mass outside

[-3, 3)

sits exactly at the boundary points

\pm 3

, with the remaining mass at

\mu = 0

. This is a known characterization: the extremal distribution places mass at 0 and at

\pm k

. Option D has

E[X] = 0

(symmetric),

E[X^2] = 16 \cdot \frac{2}{9} + 16 \cdot \frac{2}{9} = \frac{64}{9} \cdot \frac{9}{8}

... more carefully:

E[X^2] = 9 \cdot \frac{2}{9} + 9 \cdot \frac{2}{9} = 4

. So

\sigma^2 = 4

✓. And

P(|X| \ge 3) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}

, exactly matching the Chebyshev bound ✓. Option A has mean

-1 \cdot \frac{4}{5} + 4 \cdot \frac{1}{5} = 0

and variance

= \frac{16}{5} \ne 4

. It fails the variance condition. Option B has correct mean and variance (

E[X^2] = 4

), but

P(|X| \ge 3) = 0

, far below

\frac{4}{9}

— no equality. Option C has

E[X^2] = 16 \cdot \frac{1}{4} = 4

✓, but

P(|X| \ge 3) = \frac{1}{4} \ne \frac{4}{9}

— again no equality. The strategy: when a question asks for equality in Chebyshev's bound at threshold

k

, look for all off-center mass concentrated exactly at

\pm k

with the remainder at the mean.

Question 10

Over all nonnegative random variables $X$ satisfying $E[X]=6$ , what is the largest possible value of $P(X\ge10)$ , and which two-point distribution attains it?

$\frac{3}{4}$ , attained by $P(X=0)=\frac{1}{4}$ and $P(X=8)=\frac{3}{4}$
$\frac{1}{2}$ , attained by $P(X=0)=\frac{1}{2}$ and $P(X=12)=\frac{1}{2}$
$\frac{2}{5}$ , attained by $P(X=0)=\frac{3}{5}$ and $P(X=15)=\frac{2}{5}$
$\frac{3}{5}$ , attained by $P(X=0)=\frac{2}{5}$ and $P(X=10)=\frac{3}{5}$ (correct answer)

Explanation: Whenever you see a problem asking for the maximum probability of a tail event given a fixed mean, think of Markov's inequality — but more importantly, think about which distribution saturates that bound. The key insight is that to maximize

P(X \ge 10)

, you should concentrate all mass on exactly two points: zero (below the threshold) and the threshold itself (at 10), then use the mean constraint to solve for the weights. Let

P(X = 0) = 1 - p

and

P(X = 10) = p

. The mean constraint gives:

E[X] = 0 \cdot (1-p) + 10 \cdot p = 10p = 6 \implies p = \frac{3}{5}

So the maximum is

P(X \ge 10) = \frac{3}{5}

, attained by

P(X=0) = \frac{2}{5}

and

P(X=10) = \frac{3}{5}

. This is answer D, and it matches exactly what Markov's inequality predicts:

P(X \ge 10) \le \frac{E[X]}{10} = \frac{6}{10} = \frac{3}{5}

. Choice A places mass at

X = 8

, which is below the threshold of 10, so it contributes nothing to

P(X \ge 10)

— a wasted opportunity. Choice B uses

X = 12 > 10

, which is valid, but gives only

p = 1/2 < 3/5

— moving mass above the threshold forces

p

down to satisfy the mean constraint. Choice C similarly uses

X = 15

, yielding

p = 2/5

, even lower. The strategy tip: Markov's inequality is tight precisely when mass sits at zero and the threshold. Any mass above the threshold wastes efficiency, and any mass below but nonzero also reduces

p

. Always anchor one point at the threshold to maximize the tail probability.