Statistics Graduate Level Quiz: Taylor Expansions For Approximations
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Taylor Expansions For ApproximationsQuestion 1 of 10

Suppose n θ^n→dN(0,σ2)\sqrt{n}\,\hat\theta_n \xrightarrow{d} N(0,\sigma^2), where σ2>0\sigma^2>0. Which limiting distribution follows from an appropriate Taylor expansion of 1−cos⁡(θ^n)1-\cos(\hat\theta_n)?

n{1−cos⁡(θ^n)}→dσ22χ12n\{1-\cos(\hat\theta_n)\}\xrightarrow{d}\frac{\sigma^2}{2}\chi_1^2
n{1−cos⁡(θ^n)}→dσ2χ12n\{1-\cos(\hat\theta_n)\}\xrightarrow{d}\sigma^2\chi_1^2
n{1−cos⁡(θ^n)}→dN(0,σ42)n\{1-\cos(\hat\theta_n)\}\xrightarrow{d}N\left(0,\frac{\sigma^4}{2}\right)
n{1−cos⁡(θ^n)}→d−σ22χ12n\{1-\cos(\hat\theta_n)\}\xrightarrow{d}-\frac{\sigma^2}{2}\chi_1^2
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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Taylor Expansions For Approximations

Practice Taylor Expansions For Approximations in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

Suppose n θ^n→dN(0,σ2)\sqrt{n}\,\hat\theta_n \xrightarrow{d} N(0,\sigma^2), where σ2>0\sigma^2>0. Which limiting distribution follows from an appropriate Taylor expansion of 1−cos⁡(θ^n)1-\cos(\hat\theta_n)?

  1. n{1−cos⁡(θ^n)}→dσ22χ12n\{1-\cos(\hat\theta_n)\}\xrightarrow{d}\frac{\sigma^2}{2}\chi_1^2 (correct answer)
  2. n{1−cos⁡(θ^n)}→dσ2χ12n\{1-\cos(\hat\theta_n)\}\xrightarrow{d}\sigma^2\chi_1^2
  3. n{1−cos⁡(θ^n)}→dN(0,σ42)n\{1-\cos(\hat\theta_n)\}\xrightarrow{d}N\left(0,\frac{\sigma^4}{2}\right)
  4. n{1−cos⁡(θ^n)}→d−σ22χ12n\{1-\cos(\hat\theta_n)\}\xrightarrow{d}-\frac{\sigma^2}{2}\chi_1^2
Explanation: When you see a question like this, think delta method — specifically, what happens when you apply a smooth function to a sequence that's already been shown to converge in distribution. The key tool is a second-order Taylor expansion, used when the first derivative of the function vanishes at zero. Since θ^n→0\hat\theta_n \to 0 in probability (because nθ^n=Op(1)\sqrt{n}\hat\theta_n = O_p(1)), expand 1−cos⁡(θ^n)1 - \cos(\hat\theta_n) around zero: 1−cos⁡(θ)=θ22−θ424+⋯1 - \cos(\theta) = \frac{\theta^2}{2} - \frac{\theta^4}{24} + \cdots So n{1−cos⁡(θ^n)}≈n⋅θ^n22=12(n θ^n)2.n\{1-\cos(\hat\theta_n)\} \approx n \cdot \frac{\hat\theta_n^2}{2} = \frac{1}{2}({\sqrt{n}\,\hat\theta_n})^2. Since n θ^n→dN(0,σ2)\sqrt{n}\,\hat\theta_n \xrightarrow{d} N(0,\sigma^2), we can write n θ^n→dσZ\sqrt{n}\,\hat\theta_n \xrightarrow{d} \sigma Z where Z∼N(0,1)Z \sim N(0,1). Therefore: n{1−cos⁡(θ^n)}→d(σZ)22=σ22Z2=σ22χ12.n\{1-\cos(\hat\theta_n)\} \xrightarrow{d} \frac{(\sigma Z)^2}{2} = \frac{\sigma^2}{2} Z^2 = \frac{\sigma^2}{2}\chi_1^2. This confirms answer A. B is wrong because it drops the factor of 1/21/2 from the Taylor expansion — a classic error from forgetting the denominator in θ2/2\theta^2/2. C is wrong on two levels: the result cannot be normal (it's a squared quantity, hence non-negative), and the distribution family is entirely incorrect. D is wrong because χ12\chi_1^2 is always non-negative, so a negative scaling is impossible; the negative sign has no mathematical justification. Your study tip: whenever the first derivative of your function is zero at the point of expansion, you must use a second-order Taylor expansion — this is the signature of the "second-order delta method," and the 1/21/2 factor is non-negotiable.

Question 2

Suppose n θ^n→dZ\sqrt n\,\hat\theta_n\xrightarrow{d}Z, where Z∼N(0,σ2)Z\sim N(0,\sigma^2), and define g(x)=sin⁡(x)−xg(x)=\sin(x)-x. Which limit follows from the first nonvanishing term in the Taylor expansion of gg at zero?

  1. n3/2g(θ^n)→dZ36n^{3/2}g(\hat\theta_n)\xrightarrow{d}\frac{Z^3}{6}
  2. n3/2g(θ^n)→d−Z6n^{3/2}g(\hat\theta_n)\xrightarrow{d}-\frac{Z}{6}
  3. n3/2g(θ^n)→d−Z36n^{3/2}g(\hat\theta_n)\xrightarrow{d}-\frac{Z^3}{6} (correct answer)
  4. n3/2g(θ^n)→dN(0,σ636)n^{3/2}g(\hat\theta_n)\xrightarrow{d}N\left(0,\frac{\sigma^6}{36}\right)
Explanation: When you see a composition like g(θ^n)g(\hat\theta_n) where θ^n→0\hat\theta_n \to 0 in probability, your instinct should be to Taylor-expand gg around zero and identify the first nonvanishing term — that term governs the limiting distribution. For g(x)=sin⁡(x)−xg(x) = \sin(x) - x, recall the Taylor series: sin⁡(x)=x−x36+⋯\sin(x) = x - \frac{x^3}{6} + \cdots, so g(x)=−x36+O(x5)g(x) = -\frac{x^3}{6} + O(x^5). The leading term is cubic, not linear — this is the key structural fact. Since n θ^n→dZ\sqrt{n}\,\hat\theta_n \xrightarrow{d} Z, we have θ^n≈Z/n\hat\theta_n \approx Z/\sqrt{n}, so: g(θ^n)≈−θ^n36≈−16⋅Z3n3/2g(\hat\theta_n) \approx -\frac{\hat\theta_n^3}{6} \approx -\frac{1}{6}\cdot\frac{Z^3}{n^{3/2}} Multiplying both sides by n3/2n^{3/2} gives n3/2g(θ^n)→d−Z36n^{3/2}g(\hat\theta_n) \xrightarrow{d} -\frac{Z^3}{6}, confirming answer C. A is wrong because it drops the negative sign — sin⁡(x)−x\sin(x) - x has a negative cubic coefficient, so the sign matters critically here. B is wrong because it treats the cubic term as linear in ZZ, which would only be valid if the leading term were g′(0)xg'(0)x. But g′(0)=cos⁡(0)−1=0g'(0) = \cos(0) - 1 = 0, so the linear term vanishes entirely, and you can't apply the standard delta method. D is tempting if you mistakenly apply the delta method (which requires a nonzero first derivative) and then compute the variance — but since g′(0)=0g'(0) = 0, the ordinary delta method breaks down completely here. Study tip: Whenever g′(θ^)=0g'(\hat\theta) = 0 at the limit point, go to higher-order terms and adjust the normalization accordingly — the exponent in nk/2n^{k/2} matches the order of the first nonvanishing derivative.

Question 3

In a regression model, suppose (β^0,β^1)(\hat\beta_0,\hat\beta_1) is approximately unbiased and jointly normal with covariance matrix 1n(4−1−11)\frac{1}{n}\begin{pmatrix}4&-1\\-1&1\end{pmatrix} . A mean response at predictor value x=2x=2 is estimated by m^=exp⁡(β^0+2β^1)\hat m=\exp(\hat\beta_0+2\hat\beta_1).

Writing m=exp⁡(β0+2β1)m=\exp(\beta_0+2\beta_1), which expression is the second-order approximation to E(m^)E(\hat m)?

  1. m(1+2n)+o(n−1)m\left(1+\frac{2}{n}\right)+o(n^{-1}) (correct answer)
  2. m(1+4n)+o(n−1)m\left(1+\frac{4}{n}\right)+o(n^{-1})
  3. m(1+6n)+o(n−1)m\left(1+\frac{6}{n}\right)+o(n^{-1})
  4. m(1+0n)+o(n−1)m\left(1+\frac{0}{n}\right)+o(n^{-1})
Explanation: When a nonlinear function of an estimator is itself being estimated, you need the delta method — specifically its second-order (bias-correcting) extension — to approximate the expectation. The key formula: if θ^≈θ\hat\theta \approx \theta and gg is smooth, then E[g(θ^)]≈g(θ)+12tr(Hg(θ) Cov(θ^))E[g(\hat\theta)] \approx g(\theta) + \frac{1}{2}\text{tr}\left(H_g(\theta)\,\text{Cov}(\hat\theta)\right), where HgH_g is the Hessian of gg. Here, g(β0,β1)=exp⁡(β0+2β1)g(\beta_0, \beta_1) = \exp(\beta_0 + 2\beta_1). The gradient is ∇g=m (1, 2)T\nabla g = m\,(1,\,2)^T and the Hessian is Hg=m(1224)H_g = m\begin{pmatrix}1&2\\2&4\end{pmatrix} . The covariance matrix of (β^0,β^1)(\hat\beta_0, \hat\beta_1) is $$\Sigma = \frac{1}{n}\begin{pmatrix}4&-1\-1&1\end{pmatrix} The bias correction term is $$\frac{1}{2}\text{tr}(H_g\,\Sigma) = \frac{m}{2}\,\text{tr}\!\left(\begin{pmatrix}1&2\\2&4\end{pmatrix}\frac{1}{n}\begin{pmatrix}4&-1\\-1&1\end{pmatrix}\right)$$. Computing the product's trace: diagonal entries are $$(4-2)/n = 2/n$$ and $$(-2+4)/n = 2/n$$, giving $$\text{tr} = 4/n$$. So the correction is $$\frac{m}{2}\cdot\frac{4}{n} = \frac{2m}{n}$$, and $$E(\hat m) \approx m\!\left(1+\frac{2}{n}\right)$$, confirming **answer A**. Answer B uses $$4/n$$ — forgetting the $$\frac{1}{2}$$ prefactor in the bias formula. Answer C uses $$6/n$$, possibly from summing raw matrix entries incorrectly. Answer D claims zero bias, ignoring the curvature of the exponential entirely. **Study tip:** Whenever you apply the second-order delta method for bias, remember the $$\frac{1}{2}\,\text{tr}(H\Sigma)$$ structure — the factor of $$\frac{1}{2}$$ is the most common place to lose points on this type of calculation.

Question 4

Suppose n{(θ^1,θ^2)T−(2,1)T}→dN(0,Σ)\sqrt{n}\{(\hat\theta_1,\hat\theta_2)^T-(2,1)^T\}\xrightarrow{d}N(0,\Sigma), where Σ=(11/21/22)\Sigma=\begin{pmatrix}1&1/2\\1/2&2\end{pmatrix} . Define Un=θ^1θ^2U_n=\hat\theta_1\hat\theta_2 and Vn=θ^1/θ^2V_n=\hat\theta_1/\hat\theta_2.

What is the asymptotic covariance matrix of n{(Un,Vn)T−(2,2)T}\sqrt{n}\{(U_n,V_n)^T-(2,2)^T\}?

  1. (11777)\begin{pmatrix}11&7\\7&7\end{pmatrix}
  2. (11−7−77)\begin{pmatrix}11&-7\\-7&7\end{pmatrix} (correct answer)
  3. (7−7−711)\begin{pmatrix}7&-7\\-7&11\end{pmatrix}
  4. (9−5−55)\begin{pmatrix}9&-5\\-5&5\end{pmatrix}
Explanation: When you see a transformation of asymptotically normal estimators, reach immediately for the Delta Method: if n(θ^−θ0)→dN(0,Σ)\sqrt{n}(\hat{\boldsymbol{\theta}} - \boldsymbol{\theta}_0) \xrightarrow{d} N(0, \Sigma) and gg is differentiable, then n(g(θ^)−g(θ0))→dN(0,JΣJT)\sqrt{n}(g(\hat{\boldsymbol{\theta}}) - g(\boldsymbol{\theta}_0)) \xrightarrow{d} N(0, J\Sigma J^T), where JJ is the Jacobian of gg evaluated at θ0=(2,1)T\boldsymbol{\theta}_0 = (2,1)^T. Here g(θ1,θ2)=(θ1θ2, θ1/θ2)Tg(\theta_1, \theta_2) = (\theta_1\theta_2,\ \theta_1/\theta_2)^T, so the Jacobian is: J=(θ2θ11/θ2−θ1/θ22)∣(2,1)=(121−2).J = \begin{pmatrix} \theta_2 & \theta_1 \\ 1/\theta_2 & -\theta_1/\theta_2^2 \end{pmatrix}\Bigg|_{(2,1)} = \begin{pmatrix} 1 & 2 \\ 1 & -2 \end{pmatrix}. Now compute JΣJTJ\Sigma J^T. First, $$J\Sigma = \begin{pmatrix}1&2\1&-2\end{pmatrix}\begin{pmatrix}1&1/2\1/2&2\end{pmatrix} = \begin{pmatrix}2&9/2\0&-7/2\end{pmatrix} $$J\Sigma J^T = \begin{pmatrix}2&9/2\\0&-7/2\end{pmatrix}\begin{pmatrix}1&1\\2&-2\end{pmatrix} = \begin{pmatrix}11&-7\\-7&7\end{pmatrix}.$$ This is answer **B**. Choice A has the off-diagonal as $$+7$$ instead of $$-7$$ — a sign error that would arise from mishandling the negative partial derivative $$\partial(\theta_1/\theta_2)/\partial\theta_2 = -\theta_1/\theta_2^2$$. Choice C swaps the diagonal entries, reflecting confusion about which transformation corresponds to which row. Choice D appears to result from using an incorrect Jacobian, perhaps evaluating partial derivatives without applying the quotient rule correctly. Your key study habit: always write the Jacobian carefully and double-check signs on partial derivatives involving quotients. Sign errors in $$J$$ propagate throughout the entire matrix multiplication and are the most common source of mistakes on Delta Method problems.

Question 5

An estimator θ^n\hat\theta_n is defined implicitly by h(θ^n)=Xˉnh(\hat\theta_n)=\bar X_n, where E(Xˉn)=h(θ)E(\bar X_n)=h(\theta) and Var⁡(Xˉn)=σ2/n\operatorname{Var}(\bar X_n)=\sigma^2/n. Assume hh is three times continuously differentiable near θ\theta, h′(θ)≠0h'(\theta)\ne0, and higher-order moments are sufficiently controlled.

Which expression gives the second-order approximation to the bias E(θ^n)−θE(\hat\theta_n)-\theta?

  1. −h′′(θ)σ22n{h′(θ)}2+o(n−1)-\frac{h''(\theta)\sigma^2}{2n\{h'(\theta)\}^2}+o(n^{-1})
  2. h′′(θ)σ22n{h′(θ)}3+o(n−1)\frac{h''(\theta)\sigma^2}{2n\{h'(\theta)\}^3}+o(n^{-1})
  3. −h′′(θ)σ22n{h′(θ)}3+o(n−1)-\frac{h''(\theta)\sigma^2}{2n\{h'(\theta)\}^3}+o(n^{-1}) (correct answer)
  4. −h′(θ)σ22n{h′′(θ)}3+o(n−1)-\frac{h'(θ)\sigma^2}{2n\{h''(\theta)\}^3}+o(n^{-1})
Explanation: When you see an estimator defined implicitly by h(θ^n)=Xˉnh(\hat\theta_n) = \bar X_n, your instinct should be to apply the delta method in reverse — specifically, a second-order Taylor expansion of h−1h^{-1} around h(θ)h(\theta). This is the classic bias calculation for a nonlinear transformation of an unbiased (or nearly unbiased) estimator. Since h(θ^n)=Xˉnh(\hat\theta_n) = \bar X_n, write θ^n=h−1(Xˉn)\hat\theta_n = h^{-1}(\bar X_n). Expand h−1h^{-1} around μ=h(θ)\mu = h(\theta) to second order. Setting g=h−1g = h^{-1}, we have g′(μ)=1/h′(θ)g'(\mu) = 1/h'(\theta) and, by differentiating the identity h(g(y))=yh(g(y))=y twice, g′′(μ)=−h′′(θ)/[h′(θ)]3g''(\mu) = -h''(\theta)/[h'(\theta)]^3. The second-order bias formula for a smooth function of a sample mean is: E[θ^n]−θ≈12g′′(μ)⋅Var⁡(Xˉn)=12⋅(−h′′(θ)[h′(θ)]3)⋅σ2nE[\hat\theta_n] - \theta \approx \frac{1}{2}g''(\mu)\cdot\operatorname{Var}(\bar X_n) = \frac{1}{2}\cdot\left(-\frac{h''(\theta)}{[h'(\theta)]^3}\right)\cdot\frac{\sigma^2}{n} which simplifies directly to −h′′(θ)σ22n[h′(θ)]3-\dfrac{h''(\theta)\sigma^2}{2n[h'(\theta)]^3}, confirming C. A is wrong because it carries [h′(θ)]2[h'(\theta)]^2 in the denominator instead of [h′(θ)]3[h'(\theta)]^3 — this would arise from a first-derivative-only calculation, forgetting that differentiating g′g' introduces an extra factor of h′(θ)h'(\theta). B has the correct denominator [h′(θ)]3[h'(\theta)]^3 but the wrong sign; the negative sign is essential and comes directly from the chain-rule calculation of g′′g''. D inverts the roles of h′h' and h′′h'' entirely — a clear indicator of confusing the numerator and denominator structure. Study tip: Always derive g′′=−h′′/[h′]3g'' = -h''/[h']^3 explicitly by differentiating the chain-rule identity twice. The sign and the cube in the denominator are the two most common errors on these bias approximation problems.

Question 6

Let p^=X/n\hat p=X/n, where X∼Binomial⁡(n,p)X\sim\operatorname{Binomial}(n,p) and 0<p<10<p<1. Ignore boundary events whose probabilities vanish asymptotically.

A second-order Taylor expansion is used to bias-correct the log-odds estimator log⁡{p^/(1−p^)}\log\{\hat p/(1-\hat p)\} through order n−1n^{-1}. Which corrected estimator is obtained by replacing unknown quantities with p^\hat p?

  1. log⁡p^1−p^−2p^−12np^(1−p^)\log\frac{\hat p}{1-\hat p}-\frac{2\hat p-1}{2n\hat p(1-\hat p)} (correct answer)
  2. log⁡p^1−p^+2p^−12np^(1−p^)\log\frac{\hat p}{1-\hat p}+\frac{2\hat p-1}{2n\hat p(1-\hat p)}
  3. log⁡p^1−p^−12np^(1−p^)\log\frac{\hat p}{1-\hat p}-\frac{1}{2n\hat p(1-\hat p)}
  4. log⁡p^1−p^−2p^−12n p^(1−p^)\log\frac{\hat p}{1-\hat p}-\frac{2\hat p-1}{2\sqrt n\,\hat p(1-\hat p)}
Explanation: When you need to bias-correct an estimator using a Taylor expansion, the key tool is the delta method extended to second order. For a function g(p^)g(\hat p) of p^=X/n\hat p = X/n, the expected value expands as: E[g(p^)]≈g(p)+12g′′(p)⋅p(1−p)nE[g(\hat p)] \approx g(p) + \frac{1}{2}g''(p)\cdot\frac{p(1-p)}{n} So the bias is approximately 12g′′(p)⋅p(1−p)n\frac{1}{2}g''(p)\cdot\frac{p(1-p)}{n}, and a bias-corrected estimator subtracts this bias from g(p^)g(\hat p). Here g(p)=log⁡p1−pg(p) = \log\frac{p}{1-p}. Computing derivatives: g′(p)=1p(1−p)g'(p) = \frac{1}{p(1-p)}, and differentiating again gives g′′(p)=2p−1p2(1−p)2g''(p) = \frac{2p-1}{p^2(1-p)^2}. The bias is therefore: Bias≈12⋅2p−1p2(1−p)2⋅p(1−p)n=2p−12n p(1−p)\text{Bias} \approx \frac{1}{2}\cdot\frac{2p-1}{p^2(1-p)^2}\cdot\frac{p(1-p)}{n} = \frac{2p-1}{2n\,p(1-p)} The bias-corrected estimator subtracts this bias and replaces pp with p^\hat p: log⁡p^1−p^−2p^−12np^(1−p^)\log\frac{\hat p}{1-\hat p} - \frac{2\hat p - 1}{2n\hat p(1-\hat p)} This is exactly answer A. Answer B adds the bias instead of subtracting it — a sign error that would inflate rather than remove the bias. Answer C uses 11 in the numerator instead of 2p^−12\hat p - 1, which would only be correct if g′′g'' were 1/[p2(1−p)2]1/[p^2(1-p)^2] — a miscalculation of the second derivative. Answer D divides by n\sqrt{n} rather than nn, corresponding to a first-order (standard error) correction rather than a second-order bias correction. Study tip: Always track the sign when bias-correcting — you subtract the bias. And remember that second-order corrections scale as n−1n^{-1}, not n−1/2n^{-1/2}; if you see n\sqrt{n} in a bias-correction context, that's a red flag.

Question 7

For each nn, let the true parameter be θn=c/n\theta_n=c/\sqrt n, where cc is fixed. Suppose n(θ^n−θn)→dZ\sqrt n(\hat\theta_n-\theta_n)\xrightarrow{d}Z with Z∼N(0,σ2)Z\sim N(0,\sigma^2).

What is the limiting distribution of n(θ^n2−θn2)n(\hat\theta_n^2-\theta_n^2)?

  1. Z2−c2Z^2-c^2
  2. 2cZ2cZ
  3. (c+Z)2(c+Z)^2
  4. 2cZ+Z22cZ+Z^2 (correct answer)
Explanation: When you see a question involving a sequence of parameters that shrink with nn, your first instinct should be the Delta Method combined with careful algebraic expansion — don't try to apply the Delta Method blindly to the transformed quantity. Instead, work directly from what you're given. You know n(θ^n−θn)→dZ\sqrt{n}(\hat\theta_n - \theta_n) \xrightarrow{d} Z, so θ^n=θn+Z/n+op(1/n)\hat\theta_n = \theta_n + Z/\sqrt{n} + o_p(1/\sqrt{n}). Now expand n(θ^n2−θn2)n(\hat\theta_n^2 - \theta_n^2): n(θ^n2−θn2)=n(θ^n−θn)(θ^n+θn)n(\hat\theta_n^2 - \theta_n^2) = n(\hat\theta_n - \theta_n)(\hat\theta_n + \theta_n) Write θ^n+θn=2θn+(θ^n−θn)=2cn+Zn+op(1/n)\hat\theta_n + \theta_n = 2\theta_n + (\hat\theta_n - \theta_n) = \frac{2c}{\sqrt{n}} + \frac{Z}{\sqrt{n}} + o_p(1/\sqrt{n}). Multiplying: n⋅Zn⋅(2c+Zn)=Z(2c+Z)=2cZ+Z2n \cdot \frac{Z}{\sqrt{n}} \cdot \left(\frac{2c + Z}{\sqrt{n}}\right) = Z(2c + Z) = 2cZ + Z^2 This confirms D is correct. A (Z2−c2Z^2 - c^2) has no algebraic justification — there's no mechanism to produce a −c2-c^2 term from this expansion. B (2cZ2cZ) is a common partial-answer trap: it captures only the first-order term from the product, ignoring the Z2Z^2 contribution from (θ^n−θn)2(\hat\theta_n - \theta_n)^2. Since θn→0\theta_n \to 0, the quadratic term survives at rate nn and cannot be dropped. C ((c+Z)2=c2+2cZ+Z2(c+Z)^2 = c^2 + 2cZ + Z^2) over-counts by adding an extra c2c^2, which would require nθn2→c2n\theta_n^2 \to c^2 to appear — but that term cancels out because you're looking at the difference θ^n2−θn2\hat\theta_n^2 - \theta_n^2, not θ^n2\hat\theta_n^2 alone. Study tip: When parameters shrink at rate 1/n1/\sqrt{n}, don't discard quadratic fluctuation terms — they survive scaling and contribute to the limit. Always expand fully before taking limits.

Question 8

Suppose n{(θ^n−θ,s^n−s)T}→dN(0,(τ2κκω2))\sqrt n\{(\hat\theta_n-\theta,\hat s_n-s)^T\}\xrightarrow{d}N\left(0,\begin{pmatrix}\tau^2&\kappa\\\kappa&\omega^2\end{pmatrix}\right), where s>0s>0. Consider the studentized statistic Tn=n(θ^n−θ)/s^nT_n=\sqrt n(\hat\theta_n-\theta)/\hat s_n.

Which is the limiting distribution of TnT_n?

  1. N(0,τ2+ω2s2−2κs)N\left(0,\tau^2+\frac{\omega^2}{s^2}-\frac{2\kappa}{s}\right)
  2. N(0,τ2s2−2θκs3+θ2ω2s4)N\left(0,\frac{\tau^2}{s^2}-\frac{2\theta\kappa}{s^3}+\frac{\theta^2\omega^2}{s^4}\right)
  3. N(0,τ2s2−2κs3+ω2s4)N\left(0,\frac{\tau^2}{s^2}-\frac{2\kappa}{s^3}+\frac{\omega^2}{s^4}\right)
  4. N(0,τ2s2)N\left(0,\frac{\tau^2}{s^2}\right) (correct answer)
Explanation: Whenever you see a studentized statistic, your instinct should be to apply the Delta Method — specifically, treat Tn=n(θ^n−θ)/s^nT_n = \sqrt{n}(\hat\theta_n - \theta)/\hat s_n as a function of the joint estimator (θ^n,s^n)(\hat\theta_n, \hat s_n). Define g(a,b)=a/bg(a, b) = a/b, so Tn=g(n(θ^n−θ), s^n)T_n = g(\sqrt{n}(\hat\theta_n - \theta),\, \hat s_n)... but more carefully, write Tn=n(θ^n−θ)⋅1s^nT_n = \sqrt{n}(\hat\theta_n - \theta) \cdot \frac{1}{\hat s_n}. Since s^n→ps>0\hat s_n \xrightarrow{p} s > 0, Slutsky's theorem applies directly: 1s^n→p1s\frac{1}{\hat s_n} \xrightarrow{p} \frac{1}{s}. Therefore, Tn=n(θ^n−θ)s^n→d1s⋅Z,Z∼N(0,τ2),T_n = \frac{\sqrt{n}(\hat\theta_n - \theta)}{\hat s_n} \xrightarrow{d} \frac{1}{s} \cdot Z, \quad Z \sim N(0, \tau^2), which gives Tn→dN ⁣(0,τ2s2)T_n \xrightarrow{d} N\!\left(0, \frac{\tau^2}{s^2}\right), confirming D. The key insight is that only the numerator n(θ^n−θ)\sqrt{n}(\hat\theta_n - \theta) is Op(1)O_p(1) (i.e., stochastically bounded at rate n\sqrt{n}); the denominator s^n\hat s_n is already Op(1)O_p(1) and converges to a constant. Slutsky lets you "pull the limit through" the division without any covariance correction. A is wrong because it applies a Delta Method gradient for g(a,b)=a−bg(a,b) = a - b, mixing the two components incorrectly. B mistakenly applies g(θ,s)=θ/sg(\theta, s) = \theta/s as if you were estimating a ratio of parameters, dragging θ\theta into the variance. C correctly uses g(a,b)=a/bg(a,b) = a/b with gradient (1/s, −θ/s2)(1/s,\, -\theta/s^2)... wait — actually C applies the Delta Method to the joint scaled quantity n(θ^n,s^n)\sqrt{n}(\hat\theta_n, \hat s_n), treating both components as fluctuating at rate n\sqrt{n}, which is the critical error. Only θ^n−θ\hat\theta_n - \theta is scaled by n\sqrt{n}; s^n\hat s_n is not centered, so its fluctuation is asymptotically negligible relative to the numerator. Study tip: When a statistic is a ratio where the denominator converges to a nonzero constant (not zero), Slutsky's theorem is your most efficient tool — no covariance terms needed, and the variance simplifies to just the numerator's variance divided by the squared limiting constant.

Question 9

Let (Xˉ,Yˉ)(\bar X,\bar Y) be sample means satisfying E(Xˉ)=4E(\bar X)=4, E(Yˉ)=2E(\bar Y)=2, Var⁡(Yˉ)=9/n\operatorname{Var}(\bar Y)=9/n, and Cov⁡(Xˉ,Yˉ)=3/n\operatorname{Cov}(\bar X,\bar Y)=3/n. Assume the remaining regularity conditions needed for a second-order Taylor approximation and that Yˉ\bar Y stays away from zero with probability tending to one.

What is the second-order approximation to E(Xˉ/Yˉ)E(\bar X/\bar Y) through terms of order n−1n^{-1}?

  1. 2+154n+o(n−1)2+\frac{15}{4n}+o(n^{-1}) (correct answer)
  2. 2+92n+o(n−1)2+\frac{9}{2n}+o(n^{-1})
  3. 2−34n+o(n−1)2-\frac{3}{4n}+o(n^{-1})
  4. 2+214n+o(n−1)2+\frac{21}{4n}+o(n^{-1})
Explanation: When you see a ratio of sample means like Xˉ/Yˉ\bar X/\bar Y, the tool to reach for is the delta method — specifically its second-order extension, which captures bias terms of order n−1n^{-1}. Let g(x,y)=x/yg(x,y) = x/y. You need the partial derivatives at (μx,μy)=(4,2)(\mu_x, \mu_y) = (4, 2): gx=1/yg_x = 1/y, gy=−x/y2g_y = -x/y^2, gxx=0g_{xx} = 0, gyy=2x/y3g_{yy} = 2x/y^3, gxy=−1/y2g_{xy} = -1/y^2. Evaluating at the means: gx=1/2g_x = 1/2, gy=−1g_y = -1, gyy=1g_{yy} = 1, gxy=−1/4g_{xy} = -1/4. The second-order approximation to E[g(Xˉ,Yˉ)]E[g(\bar X, \bar Y)] is: g(μx,μy)+12gxxVar⁡(Xˉ)+12gyyVar⁡(Yˉ)+gxyCov⁡(Xˉ,Yˉ)g(\mu_x,\mu_y) + \tfrac{1}{2}g_{xx}\operatorname{Var}(\bar X) + \tfrac{1}{2}g_{yy}\operatorname{Var}(\bar Y) + g_{xy}\operatorname{Cov}(\bar X,\bar Y) Plugging in: the base value is 4/2=24/2 = 2. The gxxg_{xx} term vanishes. The gyyg_{yy} term gives 12(1)(9/n)=9/(2n)\tfrac{1}{2}(1)(9/n) = 9/(2n). The cross term gives (−1/4)(3/n)=−3/(4n)(-1/4)(3/n) = -3/(4n). Summing: 2+9/(2n)−3/(4n)=2+18/(4n)−3/(4n)=2+15/(4n)2 + 9/(2n) - 3/(4n) = 2 + 18/(4n) - 3/(4n) = 2 + 15/(4n), confirming answer A. Answer B omits the cross-derivative term entirely, keeping only 9/(2n)9/(2n). Answer C keeps only the cross term −3/(4n)-3/(4n), dropping the gyyg_{yy} contribution. Answer D adds rather than subtracts the cross term, computing 18/(4n)+3/(4n)=21/(4n)18/(4n) + 3/(4n) = 21/(4n) — a sign error on gxyg_{xy}. Study tip: Always compute all three second-order correction terms — variance in numerator, variance in denominator, and the covariance cross term. Sign errors on the cross term (gxy=−1/y2<0g_{xy} = -1/y^2 < 0) are the most common trap here.

Question 10

Let X∼Poisson⁡(λ)X\sim\operatorname{Poisson}(\lambda), where λ\lambda is large, and consider Tc=2X+cT_c=2\sqrt{X+c}. Using a second-order Taylor expansion in XX about λ\lambda, followed by an expansion in powers of λ−1/2\lambda^{-1/2}, which value of cc eliminates the leading bias term of order λ−1/2\lambda^{-1/2} in estimating 2λ2\sqrt\lambda?

  1. c=0c=0
  2. c=14c=\frac{1}{4} (correct answer)
  3. c=38c=\frac{3}{8}
  4. c=12c=\frac{1}{2}
Explanation: This question tests your ability to apply variance-stabilizing transformations via Taylor expansion — a key technique for making estimators approximately unbiased and normal. When you see Tc=2X+cT_c = 2\sqrt{X+c}, think: "expand around the mean, collect bias terms, then choose cc to cancel them." Start with a second-order Taylor expansion of f(X)=2X+cf(X) = 2\sqrt{X+c} around X=λX = \lambda: E[Tc]≈2λ+c+12f′′(λ+c)⋅Var⁡(X)E[T_c] \approx 2\sqrt{\lambda+c} + \frac{1}{2}f''(\lambda+c)\cdot\operatorname{Var}(X) Since f′′(x)=−12(x+c)−3/2f''(x) = -\tfrac{1}{2}(x+c)^{-3/2} and Var⁡(X)=λ\operatorname{Var}(X) = \lambda, the expectation becomes: E[Tc]≈2λ+c−λ4(λ+c)3/2E[T_c] \approx 2\sqrt{\lambda+c} - \frac{\lambda}{4(\lambda+c)^{3/2}} Now expand 2λ+c=2λ(1+cλ)1/2≈2λ+cλ−c24λ3/2+⋯2\sqrt{\lambda+c} = 2\sqrt{\lambda}\left(1 + \tfrac{c}{\lambda}\right)^{1/2} \approx 2\sqrt{\lambda} + \frac{c}{\sqrt{\lambda}} - \frac{c^2}{4\lambda^{3/2}} + \cdots and the second term ≈14λ\approx \frac{1}{4\sqrt{\lambda}}. The bias relative to 2λ2\sqrt{\lambda} at order λ−1/2\lambda^{-1/2} is: Bias≈cλ−14λ=c−14λ\text{Bias} \approx \frac{c}{\sqrt{\lambda}} - \frac{1}{4\sqrt{\lambda}} = \frac{c - \frac{1}{4}}{\sqrt{\lambda}} Setting this to zero gives c=14c = \tfrac{1}{4}, confirming answer B is correct. Choice A (c=0c=0) leaves a bias of −14λ-\tfrac{1}{4\sqrt{\lambda}} — this is the naive square-root transformation with no bias correction. Choice C (c=38c=\tfrac{3}{8}) is the famous Anscombe transformation that stabilizes variance and corrects skewness to second order, but it does not minimize the leading bias term alone. Choice D (c=12c=\tfrac{1}{2}) overcorrects, leaving a positive residual bias. Study tip: On problems involving stabilizing transformations, always track which order of bias you're eliminating — different values of cc optimize different criteria (bias, skewness, variance stability), and exams love to mix these as distractors.