Independent X (2k−1)~Bin(1,θ),X (2k)~Bin(1,1−θ),0 0. Put S 2n= k=1n(X 2k-1-X 2k+1-2), W n= k=1nY k2 and σ (2n)²=2nθ(1−θ). Which limits hold?

[nS_(2n)+σ_(2n)W_n]/[nσ_(2n)] ⇒ N(0,1/[λ²(1+λ)²])

GATE 2026 ST – Question 54

Convergence of Random Variables · Laws of Large Numbers and Central Limit Theorem · 2 marks · Multiple select

Independent X_(2k−1)~Bin(1,θ),X_(2k)~Bin(1,1−θ),0<θ<1. The Y_k are iid Poisson(λ),λ>0. Put $S_{2n}=\sum_{k=1}^n(X_{2k-1}-X_{2k}+1-2\theta)$, $W_n=\sum_{k=1}^nY_k^2$ and σ_(2n)²=2nθ(1−θ). Which limits hold?

nS_(2n)/(σ_(2n)W_n) ⇒ N(0,1)
nS_(2n)/(σ_(2n)W_n) ⇒ N(0,1/[λ²(1+λ)²])
[nS_(2n)+σ_(2n)W_n]/[nσ_(2n)] ⇒ N(λ(1+λ),1)
[nS_(2n)+σ_(2n)W_n]/[nσ_(2n)] ⇒ N(0,1/[λ²(1+λ)²])

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Show answer and explanation

Correct answer: (B) nS_(2n)/(σ_(2n)W_n) ⇒ N(0,1/[λ²(1+λ)²]); (C) [nS_(2n)+σ_(2n)W_n]/[nσ_(2n)] ⇒ N(λ(1+λ),1)

Explanation

Each paired summand has mean zero and variance 2θ(1−θ), so the iid CLT gives S_(2n)/σ_(2n)⇒N(0,1). The law of large numbers gives W_n/n→E(Y²)=λ+λ² in probability. Slutsky’s theorem gives the first ratio’s variance 1/[λ²(1+λ)²]. The second expression is S_(2n)/σ_(2n)+W_n/n and tends to N(λ(1+λ),1). **B,C**.

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