X₁, …, X₁₀ are i.i.d. N(μ, σ²), with sample mean X̄ and sample variance S² (divisor n − 1). Which statements are true?

9S²/σ² has the χ²₉ distribution

√10 (X̄ − μ)/S has the t₁₀ distribution

Which statements about the t, F and chi-square distributions are true?

If F ~ F(m, n), then 1/F ~ F(n, m)

The mean of F(m, n) is n/(n − 2) for n > 2

If F ~ F(m, n), then 1/F ~ F(n, m)

The mean of F(m, n) is n/(n − 2) for n > 2

t_k has variance k/(k − 2) for every k ≥ 1

Joint Distributions II: Functions of Random Vectors, Order Statistics, and the Chi-square, t and F Sampling Distributions

The second chapter for Section 5 of the GATE Statistics paper, on what the section names after the bivariate normal: functions of a random vector and their distributions, by the Jacobian, convolution and MGF methods; the distributions of order statistics, one at a time and jointly; and the sampling distributions — the central chi-square, central t and central F — with the facts about a normal sample that estimation and testing rely on: the independence of X̄ and S², and the chi-square law of (n − 1)S²/σ². The Jacobian, order-statistic and sampling-distribution calculations are where Section 5’s two-mark numericals usually come from.

1. Functions of a random vector

Jacobian method: if (U, V) = g(X, Y) is one-to-one with inverse (x(u, v), y(u, v)), then f_U,V(u, v) = f_X,Y(x(u, v), y(u, v)) |∂(x, y)/∂(u, v)|, on the image of the support. To get the distribution of one function U, introduce a convenient companion V, transform, and integrate V out.

Example: X, Y independent Exp(1); U = X + Y, V = X/(X + Y). Then x = uv, y = u(1 − v), |J| = u, and f_U,V = e^−u · u on u > 0, 0 < v < 1. It factors: U ~ Gamma(2, 1) and V ~ U(0, 1), independent. So P(V ≤ 0.3, U ≤ 1) = 0.3 × (1 − 2e⁻¹) = 0.079. The same calculation for Gamma(a) and Gamma(b) gives a Gamma(a + b) sum independent of a β₁(a, b) proportion.

Convolution: for independent X, Y, f_X+Y(z) = ∫ f_X(x) f_Y(z − x) dx. Two U(0, 1) give the triangle min(z, 2 − z) on (0, 2), with value 0.5 at z = 1.5.
MGF method: M_X+Y = M_X M_Y for independent X, Y, then recognise the product: n independent Exp(λ) give (1 − t/λ)⁻ⁿ, the Gamma(n, λ) MGF.
Ratios of normals: X/Y for independent N(0, 1) is standard Cauchy; Z/√(V/k) with V ~ χ²_k independent is t_k.

2. Order statistics

For an i.i.d. sample from a continuous F with density f, the ordered values X₍₁₎ < … < X₍ₙ₎ have

Distributions of order statistics
Statistic	Density or CDF
Maximum X₍ₙ₎	F₍ₙ₎(x) = F(x)ⁿ
Minimum X₍₁₎	F₍₁₎(x) = 1 − [1 − F(x)]ⁿ
r-th, X₍ᵣ₎	n!/[(r − 1)!(n − r)!] F^r−1(1 − F)^n−r f
Joint (X₍ᵣ₎, X₍ₛ₎), r < s, x < y	n!/[(r−1)!(s−r−1)!(n−s)!] F(x)^r−1[F(y) − F(x)]^s−r−1[1 − F(y)]^n−s f(x)f(y)
All n jointly	n! f(x₁)…f(xₙ) on x₁ < … < xₙ

Uniform samples: X₍ᵣ₎ ~ β₁(r, n − r + 1), so E X₍ᵣ₎ = r/(n + 1); the maximum of 4 has mean 4/5, the middle of 3 is β₁(2, 2) with variance 1/20. The joint density of (X₍₁₎, X₍ₙ₎) is n(n − 1)(y − x)ⁿ⁻² on 0 < x < y < 1, and the range has mean (n − 1)/(n + 1).
Exponential samples: X₍₁₎ ~ Exp(nλ), and the spacings X₍ᵢ₎ − X₍ᵢ₋₁₎ are independent Exp((n − i + 1)λ). So E X₍ₙ₎ = (1/λ)(1 + 1/2 + … + 1/n): for three Exp(1), 11/6.

🧠 Read an order-statistic density as a count

The density of X₍ᵣ₎ at x says: r − 1 observations fall below x (F^r−1), one falls at x (f), and n − r above (1 − F)^n−r, with the multinomial coefficient counting the ways. Written that way, every formula in the table is reconstructed in seconds rather than memorised.

3. The central chi-square, t and F distributions

Sampling distributions
Distribution	Construction	Mean	Variance
χ²_k	Z₁² + … + Z_k², Zᵢ i.i.d. N(0, 1) = Gamma(k/2, rate 1/2)	k	2k
t_k	Z/√(V/k), V ~ χ²_k independent of Z	0 (k > 1)	k/(k − 2) (k > 2)
F(m, n)	(U/m)/(V/n), U ~ χ²_m, V ~ χ²_n independent	n/(n − 2) (n > 2)	2n²(m + n − 2)/[m(n − 2)²(n − 4)] (n > 4)

χ²_m + χ²_n = χ²_m+n for independent summands; χ²₂ is exponential with mean 2, so P(χ²₂ > 4) = e⁻² = 0.135.
t_k → N(0, 1) as k → ∞; t₁ is the standard Cauchy; T ~ t_k ⇒ T² ~ F(1, k).
F ~ F(m, n) ⇒ 1/F ~ F(n, m), so lower F quantiles come from upper ones: F_1−α(m, n) = 1/F_α(n, m).

4. Sampling from a normal population

For X₁, …, Xₙ i.i.d. N(μ, σ²): X̄ ~ N(μ, σ²/n); X̄ and S² are independent; (n − 1)S²/σ² ~ χ²_n−1; √n(X̄ − μ)/S ~ t_n−1. Hence Var(S²) = 2σ⁴/(n − 1). For two independent normal samples, (S₁²/σ₁²)/(S₂²/σ₂²) ~ F(n₁ − 1, n₂ − 1). The independence of X̄ and S² characterises the normal: it holds for no other distribution.

⚠️ n − 1, not n

The degrees of freedom are n − 1 because one linear constraint, Σ(Xᵢ − X̄) = 0, is spent estimating μ. √n(X̄ − μ)/S with n = 10 is t₉, and 9S²/σ² is χ²₉. With μ known, Σ(Xᵢ − μ)²/σ² is χ²_n — the one place n itself appears.

Key takeaways

f_U,V = f_X,Y · |∂(x, y)/∂(u, v)|; for gammas with a common rate the sum is independent of the proportion, which is beta.
X₍ₙ₎ has CDF Fⁿ, X₍₁₎ has 1 − (1 − F)ⁿ; X₍ᵣ₎ density = count below × density × count above.
Uniform: X₍ᵣ₎ ~ β₁(r, n − r + 1), mean r/(n + 1). Exponential: min is Exp(nλ); E max = (1/λ)Σ1/i.
χ²_k: mean k, variance 2k; t_k variance k/(k − 2); F(m, n) mean n/(n − 2); t² ~ F(1, k); t₁ is Cauchy.
Normal sample: X̄ ⊥ S², (n − 1)S²/σ² ~ χ²_n−1, √n(X̄ − μ)/S ~ t_n−1.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

X and Y are independent U(0, 1). The density of X + Y at the point 1.5, correct to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
f_X+Y(z) = ∫ 1{0 < x < 1} 1{0 < z − x < 1} dx = length of (z − 1, 1) = 2 − z for 1 < z < 2, so 0.5 at z = 1.5. Treating the sum as U(0, 2) gives a flat 0.5 everywhere — right here by coincidence, wrong at z = 1, where the density is 1.
X₁, …, X₄ are i.i.d. U(0, 1). The expected value of max(X₁, …, X₄), correct to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.8
F₍₄₎(x) = x⁴, density 4x³, mean ∫4x⁴ dx = 4/5 = 0.8 — the general r/(n + 1) with r = n = 4. Answering 1, the largest possible value, confuses the expected maximum with the supremum.
X₁, X₂, X₃ are i.i.d. U(0, 1). The variance of the sample median X₍₂₎, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.05
X₍₂₎ has density 3!/(1!1!) x(1 − x) = 6x(1 − x), i.e. β₁(2, 2), with variance ab/[(a + b)²(a + b + 1)] = 4/(16 × 5) = 0.05. Quoting the variance of a single uniform, 1/12 = 0.083, ignores that the median of three is concentrated near 1/2.
X₁, X₂, X₃ are i.i.d. exponential with rate 2. The expected value of min(X₁, X₂, X₃), correct to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.167
P(min > x) = (e^−2x)³ = e^−6x, so the minimum is Exp(6) with mean 1/6 = 0.1667 → 0.167. Dividing the mean 1/2 by 3 gives the same number here, but the reason is that the rates add — for a non-exponential distribution that shortcut fails.
X₁, …, X₄ are i.i.d. U(0, 1). The expected value of the sample range X₍₄₎ − X₍₁₎, correct to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.6
E X₍₄₎ = 4/5 and E X₍₁₎ = 1/5, so E(range) = 3/5 = 0.6 = (n − 1)/(n + 1). Linearity of expectation does the work; the dependence between the maximum and the minimum does not matter for the mean.
X₁, X₂, X₃ are i.i.d. exponential with mean 1. The expected value of max(X₁, X₂, X₃), correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.83
Spacings: the first failure among 3 takes Exp(3) (mean 1/3), then among 2, Exp(2) (1/2), then Exp(1) (1). E max = 1/3 + 1/2 + 1 = 11/6 = 1.833 → 1.83. Answering 3 adds three means of 1 as if the variables were waited for one after another.
X₁, …, X₁₀ are i.i.d. N(μ, σ²), with sample mean X̄ and sample variance S² (divisor n − 1). Which statements are true?
1. X̄ and S² are independent
2. 9S²/σ² has the χ²₉ distribution
3. √10 (X̄ − μ)/S has the t₁₀ distribution
4. Var(S²) = 2σ⁴/9
Show answer
Answer: A — X̄ and S² are independent; B — 9S²/σ² has the χ²₉ distribution; D — Var(S²) = 2σ⁴/9
(A) and (B) are the normal-sample theorem, with n − 1 = 9. (C) False: the statistic is t with n − 1 = 9 degrees of freedom, not 10. (D) S² = σ²χ²₉/9, so Var S² = σ⁴ × 2 × 9/81 = 2σ⁴/9.
Which statements about the t, F and chi-square distributions are true?
1. If T ~ t_k, then T² ~ F(1, k)
2. If F ~ F(m, n), then 1/F ~ F(n, m)
3. The mean of F(m, n) is n/(n − 2) for n > 2
4. t_k has variance k/(k − 2) for every k ≥ 1
Show answer
Answer: A — If T ~ t_k, then T² ~ F(1, k); B — If F ~ F(m, n), then 1/F ~ F(n, m); C — The mean of F(m, n) is n/(n − 2) for n > 2
(A) T² = Z²/(V/k) = (χ²₁/1)/(χ²_k/k). (B) Invert the ratio of the two scaled chi-squares. (C) E[U/m] = 1 and E[n/V] = n/(n − 2) by independence. (D) False: the variance is finite only for k > 2 — t₁ (Cauchy) has no mean, and t₂ has infinite variance.
T has the t distribution with 6 degrees of freedom. Var(T), correct to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.5
Var(t_k) = k/(k − 2) = 6/4 = 1.5 — larger than the normal’s 1 because S is itself random. Answering 1 treats t₆ as if it were already standard normal.
The mean of the F(4, 10) distribution, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.25
E F(m, n) = n/(n − 2) = 10/8 = 1.25, depending only on the denominator degrees of freedom. Using m/(m − 2) = 2 takes the wrong one.
X and Y are independent Exp(1). Let U = X + Y and V = X/(X + Y). P(V ≤ 0.3, U ≤ 1), correct to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.079
With x = uv, y = u(1 − v), |J| = u and f_U,V = u e^−u on u > 0, 0 < v < 1: U ~ Gamma(2, 1) and V ~ U(0, 1) are independent. P(U ≤ 1) = 1 − e⁻¹(1 + 1) = 0.2642, so the answer is 0.3 × 0.2642 = 0.0793 → 0.079. Using P(U ≤ 1) = 1 − e⁻¹, as if U were Exp(1), gives 0.190.
X₁, …, Xₙ are i.i.d. exponential with rate λ. The distribution of X₁ + … + Xₙ is:
1. Gamma with shape n and rate λ
2. Exponential with rate nλ
3. Exponential with rate λ/n
4. Gamma with shape λ and rate n
Show answer
Answer: A — Gamma with shape n and rate λ
The MGF of the sum is [λ/(λ − t)]ⁿ = (1 − t/λ)⁻ⁿ, the Gamma(n, λ) MGF. Exponential with rate nλ is the distribution of the minimum, not the sum; the sum has mean n/λ, which no exponential with the listed rates matches.
X has the chi-square distribution with 2 degrees of freedom. P(X > 4), correct to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.135
χ²₂ = Gamma(1, rate 1/2) is exponential with mean 2, so P(X > 4) = e^−4/2 = e⁻² = 0.1353 → 0.135. Using mean 1 (rate 1) gives e⁻⁴ = 0.018.
For an i.i.d. sample of size n ≥ 2 from U(0, 1), the joint density of (X₍₁₎, X₍ₙ₎) at 0 < x < y < 1 is:
1. n(n − 1)(y − x)ⁿ⁻²
2. n²xⁿ⁻¹(1 − y)ⁿ⁻¹
3. n(y − x)ⁿ⁻¹
4. n! (y − x)ⁿ⁻²
Show answer
Answer: A — n(n − 1)(y − x)^{n−2}
One observation at x, one at y and the other n − 2 in between: n!/(0! (n − 2)! 0!) (y − x)ⁿ⁻² = n(n − 1)(y − x)ⁿ⁻². It integrates to 1 over the triangle. The product of the two marginal densities is wrong because the minimum and maximum are dependent; n! over-counts the arrangements of the middle n − 2.