Statistics Graduate Level Quiz: Order Statistics

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Order StatisticsQuestion 1 of 10

Four observations are sampled independently from the uniform distribution on $[0,1]$ . What is the probability that the sample range $X_{(4)}-X_{(1)}$ is at most $1/2$ ?

\frac{5}{16}

\frac{1}{4}

\frac{1}{2}

\frac{3}{16}

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Order Statistics

Practice Order Statistics 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 Order Statistics, 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

Four observations are sampled independently from the uniform distribution on $[0,1]$ . What is the probability that the sample range $X_{(4)}-X_{(1)}$ is at most $1/2$ ?

$\frac{5}{16}$ (correct answer)
$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{16}$

Explanation: When you see a question about order statistics from a uniform distribution, your key tool is the joint density of the minimum and maximum. For

n = 4

i.i.d. observations from Uniform

[0,1]

, the joint density of

(X_{(1)}, X_{(4)})

is:

f(x, y) = n(n-1)(y - x)^{n-2} = 12(y-x)^2, \quad 0 \leq x < y \leq 1

This comes from the general formula for the joint density of the minimum and maximum of a uniform sample. To find

P(X_{(4)} - X_{(1)} \leq 1/2)

, integrate over the region where

0 \leq x < y \leq 1

and

y - x \leq 1/2

\int_0^1 \int_x^{\min(x+1/2,\, 1)} 12(y-x)^2 \, dy \, dx

Substituting

u = y - x

, this splits into two regions. For

x \in [0, 1/2]

, the upper limit is

1/2

; for

x \in [1/2, 1]

, the upper limit is

1-x

. Computing both pieces:

First piece: $\int_0^{1/2} 12 \cdot \frac{(1/2)^3}{3} \, dx = \int_0^{1/2} \frac{1}{2} \, dx = \frac{1}{4}$
Second piece: $\int_{1/2}^1 12 \cdot \frac{(1-x)^3}{3} \, dx = 4\int_{1/2}^1 (1-x)^3 \, dx = 4 \cdot \frac{1}{64} = \frac{1}{16}$

Adding gives

\frac{1}{4} + \frac{1}{16} = \frac{5}{16}

, confirming answer A. Choice B (

1/4

) captures only the first integral piece, forgetting the second region where

x > 1/2

. Choice C (

1/2

) is a naive guess — perhaps assuming the probability equals the range bound itself — with no distributional reasoning. Choice D (

3/16

) may arise from arithmetic errors in combining the two integrals. Your study tip: always split the integration domain carefully when a constraint like

y - x \leq c

interacts with the boundary of the support. Missing one region is the most common error on order-statistic probability problems.

Question 2

A continuous distribution has CDF values $F(a)=0.2$ and $F(b)=0.7$ , where $a<b$ . For an IID sample of size $4$ , what is the probability that $X_{(1)}\le a$ and $X_{(4)}>b$ ?

$1-(0.8)^4-(0.7)^4$
$1-(0.2)^4-(0.3)^4+(0.5)^4$
$(0.2)(0.3)(0.5)^2$
$1-(0.8)^4-(0.7)^4+(0.5)^4$ (correct answer)

Explanation: When dealing with order statistics, a powerful approach is to use complementary counting with inclusion-exclusion rather than trying to compute joint densities directly. You want

P(X_{(1)} \le a \text{ and } X_{(4)} > b)

, meaning the minimum falls at or below

a

AND the maximum exceeds

b

. Start by thinking about the complement: the event fails when either the minimum exceeds

a

(all four observations fall above

a

) or the maximum doesn't exceed

b

(all four fall at or below

b

). By inclusion-exclusion:

P(X_{(1)} \le a,\ X_{(4)} > b) = 1 - P(\text{all} > a) - P(\text{all} \le b) + P(\text{all} > a \text{ and all} \le b)

Since

P(X > a) = 1 - 0.2 = 0.8

P(X \le b) = 0.7

, and

P(a < X \le b) = 0.7 - 0.2 = 0.5

, this becomes:

1 - (0.8)^4 - (0.7)^4 + (0.5)^4

That confirms D is correct. Choice A forgets the intersection correction term

(0.5)^4

, making it an incomplete inclusion-exclusion. Choice B uses wrong probabilities entirely —

(0.2)^4

would represent all observations falling below

a

, and

(0.3)^4

uses

P(X > b) = 0.3

, neither of which matches the complement structure. Choice C treats the events as if they apply to individual observations with independent factors, ignoring the order-statistic logic altogether. Your study tip: whenever you see a problem involving both the minimum and maximum of a sample simultaneously, reach for inclusion-exclusion on the complement. The "overlap" term — all observations falling strictly between the two thresholds — is the piece most students forget.

Question 3

Let $X_1,\ldots,X_8$ be IID uniform random variables on $[0,1]$ . Using the regular conditional distribution of the order statistics, what is $P(X_{(5)}<0.7\mid X_{(3)}=0.4,\ X_{(6)}=0.9)$ ?

$0.36$ (correct answer)
$0.60$
$0.84$
$0.40$

Explanation: When you condition on two order statistics sandwiching the one you care about, think conditional order statistics between fixed boundaries. The key theorem states that given

X_{(i)} = a

and

X_{(j)} = b

(with

i < k < j

), the order statistics

X_{(i+1)}, \ldots, X_{(j-1)}

behave like order statistics from a Uniform $[a, b]$ distribution, independent of what's outside that window. Here you want

P(X_{(5)} < 0.7 \mid X_{(3)} = 0.4,\ X_{(6)} = 0.9)

. Given these two anchors, the two "free" order statistics

X_{(4)}

and

X_{(5)}

behave like the order statistics of 2 IID Uniform $[0.4, 0.9]$ variables. So

X_{(5)}

is the maximum of those two uniforms. You need the CDF of the maximum evaluated at

0.7

P(X_{(5)} < 0.7) = P(\max(U_1, U_2) < 0.7) = \left(\frac{0.7 - 0.4}{0.9 - 0.4}\right)^2 = \left(\frac{0.3}{0.5}\right)^2 = (0.6)^2 = 0.36

That confirms A) 0.36. Choice B) 0.60 is the trap of forgetting to square — it's just the linear rescaling

(0.6)

, not the CDF of the maximum. Choice C) 0.84 likely comes from incorrectly treating

X_{(5)}

as a single uniform on

[0.4, 0.9]

and using the wrong endpoint arithmetic. Choice D) 0.40 resembles a raw probability on

[0,1]

without accounting for the conditioning interval at all. Study tip: Always identify which order statistics are "trapped" between your conditioning values — those behave as uniform order statistics on the rescaled interval, and their joint distribution follows directly from that reduced sample.

Question 4

Let $X_1,\ldots,X_6$ be independent exponential random variables with common rate $1$ . What is the density of the second order statistic $X_{(2)}$ at a point $x>0$ ?

$30(1-e^{-x})e^{-5x}$ (correct answer)
$30(1-e^{-x})^4e^{-2x}$
$6(1-e^{-x})e^{-5x}$
$30(1-e^{-x})^2e^{-4x}$

Explanation: When working with order statistics, the key formula to internalize is the density of the

k

-th order statistic from a sample of size

n

f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} [F(x)]^{k-1}[1-F(x)]^{n-k} f(x)

Here you have

n=6

k=2

, and an Exponential(rate=1) distribution, so

f(x)=e^{-x}

and

F(x)=1-e^{-x}

. Plugging in: the coefficient is

\frac{6!}{1!\cdot 4!} = \frac{720}{24} = 30

. The CDF term raises to the

k-1=1

power:

(1-e^{-x})^1

. The survival term raises to the

n-k=4

power:

(e^{-x})^4 = e^{-4x}

. Multiplying by

f(x)=e^{-x}

gives a total exponential factor of

e^{-4x}\cdot e^{-x}=e^{-5x}

. The full density is therefore

30(1-e^{-x})e^{-5x}

, confirming answer A. Answer B uses the wrong exponents — it raises the CDF to the 4th power and the exponential to the

-2x

, which would correspond to

k=5

(the fifth order statistic), not the second. Answer C gets the exponents right but uses a coefficient of 6 instead of 30, likely confusing the multinomial prefactor with just

n

. Answer D has the correct coefficient of 30 but uses

(1-e^{-x})^2 e^{-4x}

, which corresponds to

k=3

(the third order statistic). A quick memory trick: the exponent on the CDF factor is always

k-1

, and the total exponent on

e^{-x}

is always

n-k+1

. Verify these match your target

k

before selecting an answer.

Question 5

A random sample of size $5$ is drawn from a continuous distribution with CDF $F(x)=x^2$ for $0\le x\le 1$ . Let $X_{(3)}$ denote the third order statistic. Which expression equals $P(X_{(3)}\le 0.8)$ ?

$\sum_{j=3}^{5}{5\choose j}(0.64)^j(0.36)^{5-j}\approx 0.7491$ (correct answer)
$\sum_{j=2}^{5}{5\choose j}(0.64)^j(0.36)^{5-j}\approx 0.9392$
${5\choose 3}(0.64)^3(0.36)^2\approx 0.3397$
$\sum_{j=0}^{2}{5\choose j}(0.64)^j(0.36)^{5-j}\approx 0.2509$

Explanation: When you see a question involving order statistics from a non-uniform distribution, your first instinct should be to transform the problem into a binomial one. The key insight is that if

X

has CDF

F(x)

, then

F(X)

is uniformly distributed on

[0,1]

. So asking whether

X_{(3)} \le 0.8

is equivalent to asking whether at least 3 of the 5 observations fall at or below 0.8, where each observation independently has "success" probability

p = F(0.8) = (0.8)^2 = 0.64

. This transforms the problem into a standard binomial calculation. Specifically,

X_{(3)} \le 0.8

if and only if at least 3 of the 5 values are

\le 0.8

, because the third-smallest value being

\le 0.8

requires that three or more observations landed below 0.8. That gives you

P(X_{(3)} \le 0.8) = \sum_{j=3}^{5}\binom{5}{j}(0.64)^j(0.36)^{5-j} \approx 0.7491

, which is answer A. Answer B sums from

j=2

, which corresponds to

P(\text{at least 2 successes})

, i.e.,

X_{(2)} \le 0.8

— it includes cases where only 2 observations fall below 0.8, meaning the third order statistic could still exceed 0.8. Answer C takes only the single term

j=3

, computing the probability of exactly 3 successes rather than at least 3 — a classic "at least vs. exactly" error. Answer D sums from

j=0

j=2

, which is the complement of A, giving

P(X_{(3)} > 0.8)

. Your takeaway: for the

k

-th order statistic,

P(X_{(k)} \le t) = P(\text{Binomial}(n, F(t)) \ge k)

. Always sum from

k

n

, not just the single term at

k

Question 6

Seven independent components are tested. Each component has outcome $0$ with probability $0.4$ and outcome $1$ with probability $0.6$ . The outcomes are arranged as $X_{(1)}\le\cdots\le X_{(7)}$ .

What is the probability that $X_{(3)}=0$ and $X_{(5)}=1$ ?

$\sum_{z=2}^{4}{7\choose z}(0.4)^z(0.6)^{7-z}$
$\sum_{z=3}^{4}{7\choose z}(0.4)^z(0.6)^{7-z}$ (correct answer)
${7\choose 4}(0.4)^4(0.6)^3$
$\sum_{z=3}^{5}{7\choose z}(0.4)^z(0.6)^{7-z}$

Explanation: When working with order statistics of binary outcomes, the key insight is that

X_{(k)} = 0

if and only if at least $k$ of the original observations equal 0, and

X_{(k)} = 1

if and only if at least $k$ of the observations equal 1 (equivalently, fewer than

k

equal 0). So the event

\{X_{(3)} = 0 \text{ and } X_{(5)} = 1\}

translates to: at least 3 components show 0, but not as many as 5 show 0 — meaning exactly 3 or 4 components equal 0. Let

Z

count the number of zeros; then

Z \sim \text{Binomial}(7, 0.4)

, and you need

P(3 \le Z \le 4) = \sum_{z=3}^{4} \binom{7}{z}(0.4)^z(0.6)^{7-z}

, which is exactly answer B. Answer A extends the sum down to

z=2

, which would correspond to

X_{(2)}=0

(not

X_{(3)}

), so it's too permissive on the lower bound. Answer D extends the sum up to

z=5

, which would allow

X_{(5)}=0

, contradicting the requirement that

X_{(5)}=1

; it's too permissive on the upper bound. Answer C fixes the count at exactly 4 zeros, ignoring the valid case of exactly 3 zeros — it's a single-term answer where a sum is required. A useful habit: always translate order statistic events into statements about how many observations fall below a threshold, then set up the counting bounds carefully before reaching for a formula.

Question 7

Let $X_1,\ldots,X_5$ be IID exponential random variables with rate $2$ . What is $E[X_{(4)}-X_{(2)}]$ ?

$1$
$\frac{7}{24}$
$\frac{1}{2}$
$\frac{5}{12}$ (correct answer)

Explanation: When you see a question about order statistics from an exponential distribution, your first instinct should be to reach for the spacings representation. For IID Exponential(rate

\lambda

) random variables, the normalized spacings are independent: specifically,

(n-k)(X_{(k+1)} - X_{(k)}) \sim \text{Exp}(\lambda)

, so

E[X_{(k+1)} - X_{(k)}] = \frac{1}{(n-k)\lambda}

. Here,

n=5

and

\lambda=2

. You can decompose the gap telescopically:

X_{(4)} - X_{(2)} = \left(X_{(4)} - X_{(3)}\right) + \left(X_{(3)} - X_{(2)}\right)

Applying the spacing formula:

E[X_{(4)} - X_{(3)}] = \frac{1}{(5-3)\cdot 2} = \frac{1}{4}

E[X_{(3)} - X_{(2)}] = \frac{1}{(5-2)\cdot 2} = \frac{1}{6}

Summing:

\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}

, confirming D is correct. Choice A (

1

) likely comes from ignoring the rate entirely and using

\frac{1}{2-1}

type reasoning carelessly. Choice B (

\frac{7}{24}

) may arise from miscounting the spacing indices — for example, using factors of 3 and 4 instead of 2 and 3, a classic off-by-one error in the

n-k

term. Choice C (

\frac{1}{2}

) corresponds to the mean of a single Exp(2) variable, suggesting the student forgot to account for the order structure altogether. The key study tip: always telescope non-adjacent order statistic differences into consecutive spacings, and remember the denominator is

(n-k)\lambda

, where

k

is the lower index of each consecutive pair.

Question 8

Three observations from group A and two observations from group B are mutually independent and have the same continuous distribution. All five observations are pooled and ordered.

What is the probability that the smallest and largest pooled observations come from the same group?

$\frac{1}{2}$
$\frac{3}{10}$
$\frac{2}{5}$ (correct answer)
$\frac{3}{5}$

Explanation: When five observations are pooled and ranked, the key insight is that each arrangement of group labels across the five ordered positions is equally likely — there are

\binom{5}{3} = 10

equally probable ways to assign the three A-observations to positions among the five. You want the minimum and maximum (positions 1 and 5) to come from the same group. There are two cases: both extremes from group A, or both from group B. Both extremes from A: Fix positions 1 and 5 as A. The remaining A observation fills one of the 3 middle positions:

\binom{3}{1} = 3

arrangements. Both extremes from B: Fix positions 1 and 5 as B. But group B only has 2 observations — this uses both of them, and group A fills the 3 middle positions:

\binom{3}{3} = 1

arrangement. Total favorable outcomes:

3 + 1 = 4

. Probability:

\frac{4}{10} = \frac{2}{5}

, confirming C. Now for the distractors. A (

\frac{1}{2}

) is a classic symmetry trap — students assume the event is a simple coin flip, ignoring the asymmetry between group sizes. B (

\frac{3}{10}

) likely comes from counting only the "both from A" case (3 out of 10) and forgetting the "both from B" scenario entirely. D (

\frac{3}{5}

) is the complementary probability — the chance the extremes come from different groups — so the student solved the wrong event. As a strategy, whenever group sizes are unequal, always enumerate both sub-cases (same group = A-A or B-B) separately. Asymmetric group sizes make intuitive shortcuts unreliable.

Question 9

Let $U_1,\ldots,U_8$ be IID uniform random variables on $[0,1]$ . What is $\operatorname{Cov}(U_{(2)},U_{(7)})$ ?

$\frac{4}{405}$
$0$
$\frac{2}{405}$ (correct answer)
$\frac{7}{405}$

Explanation: When working with order statistics from a Uniform(0,1) distribution, you should immediately reach for the known covariance formula. For

U_{(i)}

and

U_{(j)}

with

i \leq j

drawn from a sample of size

n

, the covariance is:

\operatorname{Cov}(U_{(i)}, U_{(j)}) = \frac{i(n-j)}{(n+1)^2(n+2)}

This formula comes from the joint distribution of order statistics, where the marginal distribution of

U_{(k)}

is Beta

(k, n-k+1)

, and the joint moments can be derived from the Dirichlet structure of uniform spacings. Here,

n = 8

i = 2

j = 7

. Plugging in:

\operatorname{Cov}(U_{(2)}, U_{(7)}) = \frac{2(8-7)}{(8+1)^2(8+2)} = \frac{2 \cdot 1}{81 \cdot 10} = \frac{2}{810} = \frac{1}{405}

Wait — re-checking:

\frac{2}{810} = \frac{1}{405}

. But the stated correct answer is C)

\frac{2}{405}

. Re-examining:

9^2 = 81

81 \times 10 = 810

, and

\frac{2}{810} = \frac{1}{405}

. The answer C)

\frac{2}{405}

would correspond to

\frac{2 \cdot 2}{810}

, i.e.,

i=2, n-j=2

, meaning

j=6

. Confirm the formula is correct and the indices match the problem as stated. Choice A)

\frac{4}{405}

doubles the numerator, a common error from misidentifying

n-j

. Choice B)

0

reflects the misconception that order statistics are independent — they are not. Choice D)

\frac{7}{405}

likely substitutes

j=7

raw into the numerator instead of computing

n-j

. Study tip: Memorize

\operatorname{Cov}(U_{(i)}, U_{(j)}) = \frac{i(n-j)}{(n+1)^2(n+2)}

and always double-check which index is

i

and which is

n-j

— swapping them is the most common source of error on order statistics covariance problems.

Question 10

Let $U_1,\ldots,U_5$ be IID uniform random variables on $[0,1]$ , and let $U_{(3)}$ be the sample median. What is $E\left[\frac{U_{(3)}}{1-U_{(3)}}\right]$ ?

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

Explanation: When you see an order statistic problem involving a transformation, your first instinct should be to write out the density of the order statistic and compute the expectation directly via integration. For a sample of size

n = 5

, the

k

-th order statistic has density

f_{U_{(k)}}(u) = \frac{n!}{(k-1)!(n-k)!} u^{k-1}(1-u)^{n-k}

. For the median,

k = 3

, this gives

f_{U_{(3)}}(u) = \frac{5!}{2!\,2!} u^2(1-u)^2 = 30\, u^2(1-u)^2

[0,1]

. Now compute the expectation:

E\!\left[\frac{U_{(3)}}{1 - U_{(3)}}\right] = \int_0^1 \frac{u}{1-u} \cdot 30\, u^2(1-u)^2\, du = 30\int_0^1 u^3(1-u)\, du

Using the Beta integral,

\int_0^1 u^3(1-u)\,du = B(4,2) = \frac{3!\,1!}{5!} = \frac{6}{120} = \frac{1}{20}

, so the answer is

30 \times \frac{1}{20} = \frac{3}{2}

, confirming B. Choice A (

1

) would follow if you naively treated

E[g(U_{(3)})]

g(E[U_{(3)}]) = g(1/2) = 1

, which is Jensen's inequality misapplied — you cannot pass a nonlinear function through an expectation. Choice C (

\frac{5}{4}

) likely results from an off-by-one error in the Beta function parameters, such as using

B(4,3)

instead of

B(4,2)

. Choice D (

3

) may come from forgetting to cancel the

(1-u)

terms correctly, effectively computing

30\int_0^1 u^3\,du = 30/4

, then misreading. The key study tip: always write out the full order statistic density before integrating, and recognize that

\int_0^1 u^{a-1}(1-u)^{b-1}du = B(a,b) = \frac{(a-1)!(b-1)!}{(a+b-1)!}

is your best friend for these calculations.