Chat with us
Statistics 2024 Paper I 50 marks Calculate

Paper I — Q2

(a) Let the joint probability density function of two random variables X and Y be f(x, y) = x/3, for 0 < 2x < 3y < 6; 0…

(a)

Let the joint probability density function of two random variables X and Y be f(x, y) = x/3, for 0 < 2x < 3y < 6; 0, otherwise. Compute the following: (i) E(Y|X = x) (10 marks) (ii) E(var(Y|X = x)) 10 marks

(b)

Find the distribution function of random variable X, for which the characteristic function is φ(t) = e^(-t²), -∞ < t < ∞. Also compute P(X > 2√2) in terms of Φ(z), where Φ(z) = ∫₋∞^z (1/√(2π)) e^(-θ²/2) dθ. 15 marks

(c)

Let X₁, X₂, ..., X₂ₙ be iid N(0, 1) variates. Find the limiting distribution of [(X₁/X₂) + (X₃/X₄) + ... + (X₂ₙ₋₁/X₂ₙ)] / [X₁² + X₂² + ... + Xₙ²]. 15 marks

हिंदी में प्रश्न पढ़ें
(a)

मान लीजिए कि दो यादृच्छिक चरों X तथा Y का संयुक्त प्रायिकता घनत्व फलन f(x, y) = x/3, 0 < 2x < 3y < 6; 0, अन्यथा है। निम्नलिखित की गणना कीजिए : (i) E(Y|X = x) (10) (ii) E(var(Y|X = x)) 10 अंक

(b)

यादृच्छिक चर X का बंटन फलन प्राप्त कीजिए जिसका अभिलक्षण फलन φ(t) = e^(-t²), -∞ < t < ∞ है। इसके अलावा Φ(z) के संदर्भ में P(X > 2√2) की गणना भी कीजिए, जहाँ Φ(z) = ∫₋∞^z (1/√(2π)) e^(-θ²/2) dθ. 15 अंक

(c)

मान लीजिए कि X₁, X₂, ..., X₂ₙ स्वतंत्र और सर्वसम बंति (iid) N(0, 1) विचर हैं। [(X₁/X₂) + (X₃/X₄) + ... + (X₂ₙ₋₁/X₂ₙ)] / [X₁² + X₂² + ... + Xₙ²] का सीमान्त बंटन ज्ञात कीजिए। 15 अंक

Q2 of the 2024 UPSC Mains Statistics Paper I, as printed
The question as printed in the 2024 Statistics paper
Evaluate your answer to this question → See all 2024 Statistics questions

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a)(i) Computing E(Y | X = x)

The joint pdf is f(x,y)=x/3 for 0<2x<3y<6. Rewrite the region: 2x<3y<6 gives 2x/3<y<2, and 2x>0 gives x>0; also 3y<6 gives y<2, while 2x<6 gives x<3. Hence the support is 0<x<3, 2x/3<y<2. First find the marginal pdf of X:

f_X(x) = ∫ from 2x/3 to 2 of (x/3) dy = (x/3)(2-2x/3) = 2x/3 - 2x²/9 = (2/9)x(3-x), 0<x<3.

The conditional pdf of Y given X=x is

f_Y|X(y|x)= f(x,y)/f_X(x) = (x/3) / [(2/9)x(3-x)] = 3/[2(3-x)], for 2x/3<y<2.

This is uniform on the interval [2x/3, 2]. Therefore

E(Y|X=x)= (2x/3 + 2)/2 = x/3 + 1 = (3+x)/3.

Final answer (i): E(Y|X=x) = (3+x)/3.

(a)(ii) Computing E[Var(Y | X = x)]

For a uniform distribution on [a,b], variance = (b-a)²/12. Here a=2x/3, b=2, so

Var(Y|X=x)= [2 - 2x/3]² / 12 = [2(3-x)/3]² / 12 = (4(3-x)²/9)/12 = (3-x)²/27.

Now E[Var(Y|X=x)] = (1/27) E[(3-X)²]. Compute this using f_X(x):

E[(3-X)²] = ∫_0^3 (3-x)² · (2/9)x(3-x) dx = (2/9)∫_0^3 x(3-x)³ dx.

Let u=3-x, so x=3-u and dx=-du; limits change from x=0→3 to u=3→0:

∫_0^3 x(3-x)³ dx = ∫_0^3 (3-u)u³ du = 3∫_0^3 u³ du - ∫_0^3 u⁴ du = 3(81/4) - 243/5 = 243/4 - 243/5 = 243/20.

Thus E[(3-X)²] = (2/9)(243/20) = 27/10. Hence

E[Var(Y|X=x)] = (1/27)(27/10) = 1/10.

Final answer (ii): E[Var(Y|X=x)] = 1/10.

(b) Distribution Function from φ(t) = e^(−t²)

The characteristic function of X is φ(t) = e^(−t²). For a normal distribution N(μ,σ²), the characteristic function is exp(iμt - σ²t²/2). Matching gives μ=0 and σ²/2=1, so σ²=2. Hence X ~ N(0,2).

The distribution function of X is therefore

F_X(x)=P(X≤x)=P(Z≤x/√2)=Φ(x/√2), where Z ~ N(0,1) and Φ is the standard normal CDF.

Now P(X>2√2)=1-P(X≤2√2)=1-Φ(2√2/√2)=1-Φ(2) = 1 − 0.9772 = 0.0228 (from the table, Φ(2) = 0.5 + 0.4772).

Final answer (b): F_X(x)=Φ(x/√2); P(X>2√2)=1-Φ(2).

(c) Limiting Distribution: Standard Cauchy

Let Yᵢ = X₂ᵢ₋₁/X₂ᵢ, i = 1, 2, …, n. Since the X's are iid N(0,1), each Y_i is the ratio of two independent standard normal variables, so Y_i has the standard Cauchy distribution C(0,1) with characteristic function φ_Y(t) = e^(−|t|). The pairs (X₂ᵢ₋₁, X₂ᵢ) are disjoint, so the Y_i are independent.

Let Sₙ = Σᵢ₌₁ⁿ Yᵢ. Then Sₙ has characteristic function e^(−n|t|), which is the characteristic function of C(0,n). Therefore Sₙ/n has characteristic function e^(−|t|), so Sₙ/n ~ C(0,1) exactly for every n. Hence Sₙ/n converges in distribution to the standard Cauchy distribution.

Let Dₙ = Σⱼ₌₁ⁿ Xⱼ². Since the X_j² are iid with mean 1, the Weak Law of Large Numbers gives Dₙ/n → 1 in probability. By Slutsky's theorem,

Sₙ/Dₙ = (Sₙ/n)/(Dₙ/n) →d C(0,1)/1 = C(0,1).

Thus the limiting distribution is the standard Cauchy distribution with density

f(z)=1/[π(1+z²)], -∞<z<∞.

Final answer (c): standard Cauchy distribution with pdf f(z)=1/[π(1+z²)], -∞<z<∞.

How this answer will be evaluated

Framework: Setup > Assumptions > Derivation or test > Computation and interpretation

What earns full marks: Rigorous derivation with correct identification of distributions and theorems.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a)(i) (a)(i) For f(x,y)=x/3 on 0<2x<3y<6, compute E(Y | X = x) (10 marks) 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Rewrites the support as 0<x<3, 2x/3<y<2
    • Marginal f_X(x)=(2/9)x(3−x) on 0<x<3
    • Conditional density 3/[2(3−x)] on (2x/3, 2), i.e. uniform
    • E(Y | X = x)=(3+x)/3

    Loses marks

    • Wrong limits for y (e.g. 0 to 2)
    • Leaving the answer as an unevaluated integral

    Earns more

    • Checks that the marginal integrates to 1
    • Sketches or describes the triangular support

    Extra mark

    • Verifies E[E(Y|X)] equals E(Y) computed directly
  2. (a)(ii) (a)(ii) Compute E[Var(Y | X = x)] (10 marks) 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Var(Y | X = x)=(b−a)²/12=(3−x)²/27
    • E[Var(Y|X)]=(1/27)E[(3−X)²] using the marginal of X
    • Integral (2/9)∫₀³ x(3−x)³ dx=(2/9)(243/20)=27/10
    • Final answer 1/10

    Loses marks

    • Returning (3−x)²/27 as the final answer without taking the expectation
    • Using the uniform variance formula (b−a)²/2 or similar slip

    Earns more

    • Clean substitution u=3−x
    • Recognises the expectation is over the distribution of X

    Extra mark

    • Links to the decomposition Var(Y)=E[Var(Y|X)]+Var[E(Y|X)]
  3. (b) (b) Distribution function of X with φ(t)=e^(−t²); P(X>2√2) in terms of Φ (15 marks) 15 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Recalls the N(μ,σ²) characteristic function exp(iμt−σ²t²/2)
    • Matches to get μ=0, σ²=2, so X ~ N(0,2)
    • Invokes the uniqueness (inversion) theorem for characteristic functions
    • F(x)=Φ(x/√2) and P(X>2√2)=1−Φ(2)≈0.0228

    Loses marks

    • Taking σ²=1 (confusing e^(−t²) with e^(−t²/2))
    • Standardising with σ=2 instead of √2

    Earns more

    • Writes the density (1/(2√π))e^(−x²/4)
    • Uses the supplied table value 0.4772

    Extra mark

    • Derives the density directly by the inversion formula
  4. (c) (c) Limiting distribution of (X₁/X₂+…+X₂ₙ₋₁/X₂ₙ)/(X₁²+…+Xₙ²) (15 marks) 15 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Each Yᵢ=X₂ᵢ₋₁/X₂ᵢ is standard Cauchy and the Yᵢ are independent
    • Characteristic function of ΣYᵢ is e^(−n|t|), so (ΣYᵢ)/n ~ C(0,1) exactly
    • Denominator/n → 1 in probability by WLLN (E Xⱼ²=1)
    • Slutsky's theorem gives the standard Cauchy limit with density 1/[π(1+z²)]

    Loses marks

    • Applying CLT to the Cauchy sum
    • Concluding the limit is degenerate at 0

    Earns more

    • Explains why Slutsky applies although numerator and denominator share variables
    • States that Cauchy has no mean, so the numerator does not obey WLLN

    Extra mark

    • Derives the Cauchy law of X₁/X₂ by transformation
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

All UPSC directive words, compared →

Practice this exact question

Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.

Evaluate my answer →

More from Statistics 2024 Paper I

Write this answer and get it marked in about 20 seconds

Free account: 5 AI evaluations a month, marked against the rubric above.

Start free

Already have an account? Log in