Statistics Graduate Level Quiz: Delta Method
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Delta MethodQuestion 1 of 10

Let n(θ^1n,θ^2n)T→dN(0,Σ)\sqrt n(\widehat\theta_{1n},\widehat\theta_{2n})^\mathsf{T}\xrightarrow{d}N(0,\Sigma), where both marginal variances equal 11 and the correlation is 1/21/2. For g(a,b)=(a+b)2g(a,b)=(a+b)^2, which limit is correct?

n g(θ^1n,θ^2n)→dN(0,3)\sqrt n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}N(0,3)
n g(θ^1n,θ^2n)→d3χ12n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}3\chi_1^2
n g(θ^1n,θ^2n)→d2χ22n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}2\chi_2^2
n g(θ^1n,θ^2n)→d3χ12\sqrt n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}3\chi_1^2
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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Delta Method

Practice Delta Method 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 Delta Method, 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

Let n(θ^1n,θ^2n)T→dN(0,Σ)\sqrt n(\widehat\theta_{1n},\widehat\theta_{2n})^\mathsf{T}\xrightarrow{d}N(0,\Sigma), where both marginal variances equal 11 and the correlation is 1/21/2. For g(a,b)=(a+b)2g(a,b)=(a+b)^2, which limit is correct?

  1. n g(θ^1n,θ^2n)→dN(0,3)\sqrt n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}N(0,3)
  2. n g(θ^1n,θ^2n)→d3χ12n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}3\chi_1^2 (correct answer)
  3. n g(θ^1n,θ^2n)→d2χ22n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}2\chi_2^2
  4. n g(θ^1n,θ^2n)→d3χ12\sqrt n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}3\chi_1^2
Explanation: Whenever you see a nonlinear function applied to an asymptotically normal estimator, your first instinct should be the delta method — but you must check whether the gradient vanishes at the true parameter value. Here, the true parameters are both zero (since the joint limit is centered), so evaluate the gradient of g(a,b)=(a+b)2g(a,b) = (a+b)^2: we get ∇g=(2(a+b),2(a+b))T\nabla g = (2(a+b), 2(a+b))^\mathsf{T}, which equals (0,0)T(0,0)^\mathsf{T} at (0,0)(0,0). A zero gradient means the first-order delta method fails and you must go to the second-order delta method. When the gradient vanishes, the correct scaling is nn (not n\sqrt{n}), and the limiting distribution is a quadratic form in the Gaussian limit. Specifically, if n(θ^1n,θ^2n)T→dZ∼N(0,Σ)\sqrt{n}(\hat\theta_{1n}, \hat\theta_{2n})^\mathsf{T} \xrightarrow{d} Z \sim N(0,\Sigma), then n g(θ^1n,θ^2n)→d(Z1+Z2)2n\,g(\hat\theta_{1n},\hat\theta_{2n}) \xrightarrow{d} (Z_1+Z_2)^2. Now Z1+Z2∼N(0,σ2)Z_1+Z_2 \sim N(0, \sigma^2) where σ2=Var(Z1+Z2)=1+1+2(1/2)=3\sigma^2 = \text{Var}(Z_1+Z_2) = 1 + 1 + 2(1/2) = 3. Therefore (Z1+Z2)2=3⋅W2(Z_1+Z_2)^2 = 3 \cdot W^2 where W∼N(0,1)W\sim N(0,1), giving 3χ123\chi_1^2. This confirms answer B is correct. Answer A is wrong on two counts: the scaling n\sqrt{n} is too slow (the right scale is nn), and a nonzero quadratic form cannot be Gaussian. Answer C uses the correct nn scaling but claims a χ22\chi_2^2 distribution — a quadratic form in a two-dimensional Gaussian is only χ22\chi_2^2 when standardized appropriately; here the sum Z1+Z2Z_1+Z_2 is one-dimensional, giving χ12\chi_1^2, not χ22\chi_2^2. Answer D uses n\sqrt{n} scaling with 3χ123\chi_1^2, getting the distribution right but the normalization wrong. Study tip: Always check whether ∇g=0\nabla g = 0 at the true parameter before applying the delta method — a vanishing gradient shifts the scaling from n\sqrt{n} to nn and the limit from normal to chi-squared.

Question 2

Suppose n{(X^n,Y^n)T−(1,2)T}→dN(0,Σ)\sqrt{n}\{(\widehat X_n,\widehat Y_n)^\mathsf{T}-(1,2)^\mathsf{T}\}\xrightarrow{d}N(0,\Sigma), where Σ=(4119)\Sigma=\begin{pmatrix}4&1\\1&9\end{pmatrix} . Define Un=X^nY^nU_n=\widehat X_n\widehat Y_n and Vn=X^n/Y^nV_n=\widehat X_n/\widehat Y_n. What is the asymptotic covariance matrix of n{(Un,Vn)T−(2,1/2)T}\sqrt n\{(U_n,V_n)^\mathsf{T}-(2,1/2)^\mathsf{T}\}?

  1. \begin{pmatrix}29&7/4\\7/4&21/16\end{pmatrix} $$, correctly accounting for both the partial derivatives and the covariance in $$\Sigma$$ (correct answer)
  2. \begin{pmatrix}29&-7/4\\-7/4&21/16\end{pmatrix} $$, introducing an erroneous negative cross-covariance
  3. \begin{pmatrix}21&3/4\\3/4&25/16\end{pmatrix} $$, resulting from mishandling the off-diagonal term in $$\Sigma$$
  4. \begin{pmatrix}17&7/4\\7/4&13/16\end{pmatrix} $$, omitting or incorrectly scaling several Jacobian contributions
Explanation: When you see a joint asymptotic normality result paired with a transformation, your instinct should be the Delta Method for vector-valued functions. If n(θ^n−θ)→dN(0,Σ)\sqrt{n}(\hat{\theta}_n - \theta) \xrightarrow{d} N(0, \Sigma) and g:R2→R2g: \mathbb{R}^2 \to \mathbb{R}^2 is differentiable, then n(g(θ^n)−g(θ))→dN(0,JΣJT)\sqrt{n}(g(\hat{\theta}_n) - g(\theta)) \xrightarrow{d} N(0, J\Sigma J^\mathsf{T}), where JJ is the Jacobian of gg evaluated at θ=(1,2)T\theta = (1,2)^\mathsf{T}. Here g(x,y)=(xy, x/y)Tg(x,y) = (xy,\, x/y)^\mathsf{T}, so the Jacobian is: J=(yx1/y−x/y2)∣(1,2)=(211/2−1/4).J = \begin{pmatrix} y & x \\ 1/y & -x/y^2 \end{pmatrix}\Bigg|_{(1,2)} = \begin{pmatrix} 2 & 1 \\ 1/2 & -1/4 \end{pmatrix}. Now compute JΣJTJ\Sigma J^\mathsf{T} with Σ=(4119)\Sigma = \begin{pmatrix}4&1\\1&9\end{pmatrix} . First, JΣ=(2(4)+1(1)2(1)+1(9)12(4)+(−14)(1)12(1)+(−14)(9))=(91174−74)J\Sigma = \begin{pmatrix}2(4)+1(1) & 2(1)+1(9)\\ \tfrac{1}{2}(4)+(-\tfrac{1}{4})(1) & \tfrac{1}{2}(1)+(-\tfrac{1}{4})(9)\end{pmatrix} = \begin{pmatrix}9&11\\ \tfrac{7}{4}&-\tfrac{7}{4}\end{pmatrix} . Then (JΣ)JT(J\Sigma)J^\mathsf{T} gives diagonal entries 9(2)+11(1)=299(2)+11(1)=29 and 74(12)+(−74)(−14)=78+716=2116\tfrac{7}{4}(\tfrac{1}{2})+(-\tfrac{7}{4})(-\tfrac{1}{4})=\tfrac{7}{8}+\tfrac{7}{16}=\tfrac{21}{16}, and off-diagonal 9(12)+11(−14)=92−114=749(\tfrac{1}{2})+11(-\tfrac{1}{4})=\tfrac{9}{2}-\tfrac{11}{4}=\tfrac{7}{4}. This confirms A is correct. B is wrong because it flips the sign of the off-diagonal to −7/4-7/4, which would only occur if the Jacobian entries for VnV_n were misread. C results from mishandling the off-diagonal of Σ\Sigma — likely setting it to zero or using the wrong cross-term — producing incorrect diagonal entries. D omits or miscales Jacobian contributions, shrinking the (1,1) entry to 17 and giving a wrong (2,2) entry. Study tip: Always write out the full Jacobian matrix carefully before multiplying — sign errors on partial derivatives (especially ∂(x/y)/∂y=−x/y2\partial(x/y)/\partial y = -x/y^2) are the most common trap in multivariate Delta Method problems.

Question 3

A parameter θ\theta is defined implicitly as the positive solution of θ2+λθ−μ=0\theta^2+\lambda\theta-\mu=0. At the true values, (θ,λ,μ)=(1,1,2)(\theta,\lambda,\mu)=(1,1,2). Suppose n{(λ^,μ^)T−(1,2)T}→dN(0,Σ)\sqrt n\{(\widehat\lambda,\widehat\mu)^\mathsf{T}-(1,2)^\mathsf{T}\}\xrightarrow{d}N(0,\Sigma), where Σ=(11/21/24)\Sigma=\begin{pmatrix}1&1/2\\1/2&4\end{pmatrix} , and let θ^\widehat\theta be the corresponding positive root. What is the asymptotic variance of n(θ^−1)\sqrt n(\widehat\theta-1)?

  1. 49\frac{4}{9}, obtained via the implicit function theorem applied to the constraint equation (correct answer)
  2. 59\frac{5}{9}, obtained by treating the two input estimators as uncorrelated
  3. 23\frac{2}{3}, obtained by adding rather than subtracting the covariance contribution
  4. 44, obtained by differentiating only with respect to μ\mu and ignoring the scaling by FθF_\theta
Explanation: When a parameter is defined implicitly through a constraint equation, the delta method via the implicit function theorem (IFT) is your go-to tool. The idea: differentiate the constraint to express ∇θ\nabla_\theta in terms of the other parameters, then propagate the asymptotic variance. Your constraint is g(θ,λ,μ)=θ2+λθ−μ=0g(\theta,\lambda,\mu)=\theta^2+\lambda\theta-\mu=0. By the IFT, ∂θ∂λ=−gλgθ\frac{\partial\theta}{\partial\lambda}=-\frac{g_\lambda}{g_\theta} and ∂θ∂μ=−gμgθ\frac{\partial\theta}{\partial\mu}=-\frac{g_\mu}{g_\theta}. Compute the partials at (1,1,2)(1,1,2): gθ=2θ+λ=3g_\theta=2\theta+\lambda=3, gλ=θ=1g_\lambda=\theta=1, gμ=−1g_\mu=-1. So the gradient is ∇=(−13,13)T\nabla=\left(-\frac{1}{3},\frac{1}{3}\right)^\mathsf{T}. The asymptotic variance is ∇TΣ ∇\nabla^\mathsf{T}\Sigma\,\nabla: ∇TΣ ∇=19(−11)(11/21/24)(−11)=19(1−1/2−1/2+4)=49.\nabla^\mathsf{T}\Sigma\,\nabla = \frac{1}{9}\begin{pmatrix}-1&1\end{pmatrix}\begin{pmatrix}1&1/2\\1/2&4\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}=\frac{1}{9}(1-1/2-1/2+4)=\frac{4}{9}. This confirms answer A is correct. B is wrong because it ignores the off-diagonal covariance term Σ12=1/2\Sigma_{12}=1/2. Dropping it gives 19(1+4)=59\frac{1}{9}(1+4)=\frac{5}{9}, a common error when you incorrectly assume independence. C adds the covariance contributions instead of accounting for the sign correctly (the two partial derivatives have opposite signs, so the cross-terms subtract), yielding 23\frac{2}{3}. D ignores λ\lambda entirely and omits the 1/gθ1/g_\theta scaling, giving an answer of 44 — essentially computing Σ22\Sigma_{22} raw. Study tip: Whenever a parameter is implicitly defined, always compute the full gradient vector and apply ∇TΣ ∇\nabla^\mathsf{T}\Sigma\,\nabla. Never drop covariance terms without checking the sign structure — the cross-term can reduce or inflate the variance significantly.

Question 4

A nonnegative parameter lies on the boundary of its parameter space, and its estimator has a nonstandard convergence rate.

Suppose θ^n≥0\widehat\theta_n\geq0 and nθ^n→dEn\widehat\theta_n\xrightarrow{d}E, where EE has an exponential distribution with rate 11. If h(θ)=θh(\theta)=\sqrt\theta, which asymptotic statement is correct?

  1. n h(θ^n)→dEn\,h(\widehat\theta_n)\xrightarrow{d}\sqrt{E}, because the square-root transformation preserves the original rate
  2. n h(θ^n)→dEn\,h(\widehat\theta_n)\xrightarrow{d}E, because squaring both sides of the convergence recovers the original exponential limit
  3. n h(θ^n)→dN(0,1/4)\sqrt n\,h(\widehat\theta_n)\xrightarrow{d}N(0,1/4), because h′(0)=1/2h'(0)=1/2 yields the asymptotic variance via the delta method
  4. n h(θ^n)→dE\sqrt n\,h(\widehat\theta_n)\xrightarrow{d}\sqrt{E}, because n h(θ^n)=nθ^n\sqrt{n}\,h(\widehat\theta_n)=\sqrt{n\widehat\theta_n} and the continuous mapping theorem applies (correct answer)
Explanation: When a parameter estimator has a nonstandard convergence rate — here nθ^n→dEn\widehat\theta_n \xrightarrow{d} E rather than the usual n\sqrt{n} rate — you need to track that rate carefully before applying any transformation. The key tool is the continuous mapping theorem (CMT): if Xn→dXX_n \xrightarrow{d} X and gg is continuous, then g(Xn)→dg(X)g(X_n) \xrightarrow{d} g(X). Notice the algebraic identity at the heart of option D: n h(θ^n)=n θ^n=nθ^n\sqrt{n}\,h(\widehat\theta_n) = \sqrt{n}\,\sqrt{\widehat\theta_n} = \sqrt{n\widehat\theta_n}. Since nθ^n→dEn\widehat\theta_n \xrightarrow{d} E, you can apply the CMT with the continuous function g(x)=xg(x) = \sqrt{x} to conclude nθ^n→dE\sqrt{n\widehat\theta_n} \xrightarrow{d} \sqrt{E}. This makes D correct. Option A claims n h(θ^n)→dEn\,h(\widehat\theta_n) \xrightarrow{d} \sqrt{E}, but nθ^n=n⋅nθ^nn\sqrt{\widehat\theta_n} = \sqrt{n}\cdot\sqrt{n\widehat\theta_n}, which diverges — the rate nn is too fast. Option B asserts the limit is still EE, confusing the transformation of the limit with "undoing" the square root; applying ⋅\sqrt{\cdot} to an exponential random variable does not return an exponential. Option C invokes the delta method, but the delta method requires h′(θ0)≠0h'(\theta_0) \neq 0 and a regular n\sqrt{n}-rate estimator. Here θ0=0\theta_0 = 0 lies on the boundary, h′(0)=120h'(0) = \frac{1}{2\sqrt{0}} is undefined, and the convergence rate is nn, not n\sqrt{n} — so the delta method simply does not apply. Study tip: Whenever the true parameter is on the boundary and the convergence rate is nonstandard, skip the delta method entirely. Instead, manipulate the norming algebraically to match the known convergent quantity, then apply the CMT directly.

Question 5

In a logistic regression, the conditional success probability at covariate value xx is p(x)={1+exp⁡[−(β0+β1x)]}−1p(x)=\{1+\exp[-(\beta_0+\beta_1x)]\}^{-1}. The instantaneous marginal effect at xx is m(x)=β1p(x){1−p(x)}m(x)=\beta_1p(x)\{1-p(x)\}.

At x=1x=1, suppose the true coefficients are β1=3\beta_1=3 and β0=−3−log⁡2\beta_0=-3-\log2, so that p(1)=1/3p(1)=1/3. Also suppose n(β^−β)→dN(0,Σ)\sqrt n(\widehat\beta-\beta)\xrightarrow{d}N(0,\Sigma), where Σ=(4111)\Sigma=\begin{pmatrix}4&1\\1&1\end{pmatrix} for β=(β0,β1)T\beta=(\beta_0,\beta_1)^\mathsf{T}. What is the asymptotic variance of n{m^(1)−m(1)}\sqrt n\{\widehat m(1)-m(1)\}?

  1. 1627\frac{16}{27} (correct answer)
  2. 827\frac{8}{27}
  3. 49\frac{4}{9}
  4. 2027\frac{20}{27}
Explanation: When a question asks for the asymptotic variance of a nonlinear function of estimators, your instinct should be the delta method: if n(β^−β)→dN(0,Σ)\sqrt{n}(\hat{\beta}-\beta)\xrightarrow{d}N(0,\Sigma), then n{m^(1)−m(1)}→dN(0,∇mTΣ ∇m)\sqrt{n}\{\hat{m}(1)-m(1)\}\xrightarrow{d}N(0,\nabla m^\mathsf{T}\Sigma\,\nabla m), where ∇m\nabla m is the gradient of m(x)m(x) evaluated at the true β\beta. First, compute the true values. With p(1)=1/3p(1)=1/3, the marginal effect is m(1)=β1p(1)(1−p(1))=3⋅13⋅23=23m(1)=\beta_1 p(1)(1-p(1))=3\cdot\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{3}. Next, find ∇m=(∂m∂β0,∂m∂β1)T\nabla m = \left(\frac{\partial m}{\partial \beta_0},\frac{\partial m}{\partial \beta_1}\right)^\mathsf{T}. Since m=β1p(1−p)m=\beta_1 p(1-p) and ∂p∂βj=xjp(1−p)\frac{\partial p}{\partial \beta_j}=x^j p(1-p) (with x=1x=1): ∂m∂β0=β1p(1−p)(1−2p)=3⋅29⋅13=29\frac{\partial m}{\partial \beta_0}=\beta_1 p(1-p)(1-2p)=3\cdot\tfrac{2}{9}\cdot\tfrac{1}{3}=\tfrac{2}{9} ∂m∂β1=p(1−p)+β1p(1−p)(1−2p)=29+29=49\frac{\partial m}{\partial \beta_1}=p(1-p)+\beta_1 p(1-p)(1-2p)=\tfrac{2}{9}+\tfrac{2}{9}=\tfrac{4}{9} So ∇m=(29,49)T\nabla m = \left(\frac{2}{9},\frac{4}{9}\right)^\mathsf{T}. The asymptotic variance is: ∇mTΣ ∇m=181(24)(4111)(24)=181(2,4)(126)=4881=1627\nabla m^\mathsf{T}\Sigma\,\nabla m = \tfrac{1}{81}\begin{pmatrix}2&4\end{pmatrix}\begin{pmatrix}4&1\\1&1\end{pmatrix}\begin{pmatrix}2\\4\end{pmatrix}=\tfrac{1}{81}(2,4)\begin{pmatrix}12\\6\end{pmatrix}=\tfrac{48}{81}=\tfrac{16}{27} This confirms A is correct. Choice B (8/278/27) likely drops the cross-terms in the quadratic form. Choice C (4/94/9) may use only the β1\beta_1 partial derivative. Choice D (20/2720/27) reflects an arithmetic error in the matrix multiplication. Your key strategy: for delta method problems, carefully differentiate with respect to every parameter and never skip the off-diagonal terms in Σ\Sigma — they often flip the answer.

Question 6

Let θ^n\widehat\theta_n be the sample mean of independent N(0,1)N(0,1) observations, and consider inference for g(θ)=θ2g(\theta)=\theta^2 at θ=0\theta=0. A researcher uses the plug-in delta-method standard error se⁡^{g(θ^n)}=2∣θ^n∣/n\widehat{\operatorname{se}}\{g(\widehat\theta_n)\}=2|\widehat\theta_n|/\sqrt n and forms Wn={g(θ^n)−g(0)}/se⁡^{g(θ^n)}W_n=\{g(\widehat\theta_n)-g(0)\}/\widehat{\operatorname{se}}\{g(\widehat\theta_n)\} whenever θ^n≠0\widehat\theta_n\neq0. What is the limiting distribution of WnW_n?

  1. N(0,1)N(0,1), because studentization restores the standard normal limit
  2. ∣Z∣|Z|, where Z∼N(0,1)Z\sim N(0,1)
  3. 12χ12\frac12\chi_1^2, because the numerator is asymptotically quadratic
  4. 12∣Z∣\frac12|Z|, where Z∼N(0,1)Z\sim N(0,1) (correct answer)
Explanation: When the delta method's first derivative vanishes at the true parameter value, the standard asymptotic machinery breaks down — this is the classic non-regular inference scenario, and it's exactly what's being tested here. Since g(θ)=θ2g(\theta) = \theta^2, we have g′(θ)=2θg'(\theta) = 2\theta, which equals zero at θ=0\theta = 0. This kills the usual delta-method approximation. Let's work through what actually happens. Write θ^n=Z/n\widehat\theta_n = Z/\sqrt{n} where Z∼N(0,1)Z \sim N(0,1). Then: Wn=θ^n2−02∣θ^n∣/n=Z2/n2∣Z∣/n=Z22∣Z∣=∣Z∣2W_n = \frac{\widehat\theta_n^2 - 0}{2|\widehat\theta_n|/\sqrt{n}} = \frac{Z^2/n}{2|Z|/n} = \frac{Z^2}{2|Z|} = \frac{|Z|}{2} This is an exact finite-sample identity (not just a limit), so the limiting distribution is 12∣Z∣\frac{1}{2}|Z|, confirming D. A is wrong because studentization does not rescue the normal limit when the derivative vanishes — the pivot is fundamentally non-normal, not just scaled incorrectly. B would require the denominator to converge to a constant (like 1/n1/\sqrt{n}), but here the denominator also depends on θ^n\widehat\theta_n in a way that cancels one power, leaving a half-|Z| rather than |Z|. C confuses the distribution of the numerator — yes, nθ^n2→χ12n\widehat\theta_n^2 \to \chi^2_1 — but after dividing by the standard error, the quadratic structure is absorbed, yielding a linear (folded) distribution, not a chi-squared one. The key study tip: whenever g′(θ0)=0g'(\theta_0) = 0, always rework the asymptotics from scratch using the second-order expansion, and track how both numerator and denominator scale with nn.

Question 7

Let θ^n\widehat{\theta}_n estimate θ=2\theta=2. Suppose E(θ^n−θ)=n−1+o(n−1)E(\widehat{\theta}_n-\theta)=n^{-1}+o(n^{-1}), Var⁡(θ^n)=4n−1+o(n−1)\operatorname{Var}(\widehat{\theta}_n)=4n^{-1}+o(n^{-1}), and the Taylor remainder is uniformly integrable at the required order. For g(x)=log⁡xg(x)=\log x, what is the limit of n{E[g(θ^n)]−g(θ)}n\{E[g(\widehat{\theta}_n)]-g(\theta)\}?

  1. −12-\frac{1}{2}, because only the curvature contribution remains asymptotically
  2. 12\frac{1}{2}, because only the first-order bias of the estimator contributes asymptotically
  3. 00, because the first-order bias contribution and the curvature-induced bias contribution cancel (correct answer)
  4. 11, because the original estimator bias is unchanged by the transformation
Explanation: When a smooth transformation is applied to an estimator, the delta method tells you how bias and variance propagate — but at second order, curvature introduces an additional bias term you must track carefully. Start with the second-order Taylor expansion of g(θ^n)g(\widehat{\theta}_n) around θ=2\theta = 2: g(θ^n)≈g(θ)+g′(θ)(θ^n−θ)+12g′′(θ)(θ^n−θ)2g(\widehat{\theta}_n) \approx g(\theta) + g'(\theta)(\widehat{\theta}_n - \theta) + \tfrac{1}{2}g''(\theta)(\widehat{\theta}_n - \theta)^2 Taking expectations and multiplying by nn: n{E[g(θ^n)]−g(θ)}≈n⋅g′(θ)⋅E(θ^n−θ)+n2g′′(θ)⋅E(θ^n−θ)2n\{E[g(\widehat{\theta}_n)] - g(\theta)\} \approx n \cdot g'(\theta) \cdot E(\widehat{\theta}_n - \theta) + \tfrac{n}{2} g''(\theta) \cdot E(\widehat{\theta}_n - \theta)^2 For g(x)=log⁡xg(x) = \log x: g′(2)=12g'(2) = \tfrac{1}{2}, g′′(2)=−14g''(2) = -\tfrac{1}{4}. Since E(θ^n−θ)=n−1+o(n−1)E(\widehat{\theta}_n - \theta) = n^{-1} + o(n^{-1}) and Var⁡(θ^n)=4n−1+o(n−1)\operatorname{Var}(\widehat{\theta}_n) = 4n^{-1} + o(n^{-1}), the second moment satisfies E(θ^n−θ)2≈Var⁡(θ^n)=4n−1E(\widehat{\theta}_n - \theta)^2 \approx \operatorname{Var}(\widehat{\theta}_n) = 4n^{-1} (the squared bias is o(n−1)o(n^{-1})). Plugging in: n⋅12⋅n−1+n2⋅(−14)⋅4n−1=12−12=0n \cdot \tfrac{1}{2} \cdot n^{-1} + \tfrac{n}{2} \cdot (-\tfrac{1}{4}) \cdot 4n^{-1} = \tfrac{1}{2} - \tfrac{1}{2} = 0 So the correct answer is C: the first-order bias contribution (+12+\tfrac{1}{2}) and the curvature-induced bias (−12-\tfrac{1}{2}) exactly cancel. Choice A keeps only the curvature term, ignoring the estimator's own bias. Choice B keeps only the estimator bias term, ignoring the curvature correction from g′′g''. Choice D incorrectly assumes transformations leave bias unchanged — they don't. The key study tip: whenever you see n{E[g(θ^n)]−g(θ)}n\{E[g(\widehat{\theta}_n)] - g(\theta)\}, always carry both the first-order bias term and the 12g′′(θ)⋅Var⁡\tfrac{1}{2}g''(\theta)\cdot\operatorname{Var} term. Dropping either one is the most common mistake on these problems.

Question 8

Two correlated estimators measure a treatment mean and a control mean. Their joint limiting distribution is specified in the question.

Suppose n{(μ^T,μ^C)T−(2,1)T}→dN(0,Σ)\sqrt n\{(\widehat\mu_T,\widehat\mu_C)^\mathsf{T}-(2,1)^\mathsf{T}\}\xrightarrow{d}N(0,\Sigma), where Σ=(9334)\Sigma=\begin{pmatrix}9&3\\3&4\end{pmatrix} . For the estimated mean ratio Rn=μ^T/μ^CR_n=\widehat\mu_T/\widehat\mu_C, what is its limiting distribution?

  1. n(Rn−2)→dN(0,25)\sqrt n(R_n-2)\xrightarrow{d}N(0,25)
  2. n(Rn−2)→dN(0,13)\sqrt n(R_n-2)\xrightarrow{d}N(0,13) (correct answer)
  3. n(Rn−2)→dN(0,37)\sqrt n(R_n-2)\xrightarrow{d}N(0,37)
  4. n(Rn−2)→dN(0,13)n(R_n-2)\xrightarrow{d}N(0,13)
Explanation: Whenever you see a question involving a ratio of estimators with a known joint limiting distribution, your immediate instinct should be the Delta Method: if n(θ^−θ)→dN(0,Σ)\sqrt{n}(\hat{\theta} - \theta) \xrightarrow{d} N(0, \Sigma), then n(g(θ^)−g(θ))→dN(0,∇g(θ)TΣ∇g(θ))\sqrt{n}(g(\hat{\theta}) - g(\theta)) \xrightarrow{d} N(0, \nabla g(\theta)^\mathsf{T} \Sigma \nabla g(\theta)). Here, g(μT,μC)=μT/μCg(\mu_T, \mu_C) = \mu_T/\mu_C, so compute the gradient at the true values (2,1)(2, 1): ∇g=(1μC, −μTμC2)T=(1, −2)T\nabla g = \left(\frac{1}{\mu_C},\ -\frac{\mu_T}{\mu_C^2}\right)^\mathsf{T} = \left(1,\ -2\right)^\mathsf{T} Now apply the quadratic form: First, Σ∇g=(9−6, 3−8)T=(3, −5)T\Sigma \nabla g = (9 - 6,\ 3 - 8)^\mathsf{T} = (3,\ -5)^\mathsf{T}. Then ∇gT(3,−5)T=3+10=13\nabla g^\mathsf{T}(3, -5)^\mathsf{T} = 3 + 10 = 13. So n(Rn−2)→dN(0,13)\sqrt{n}(R_n - 2) \xrightarrow{d} N(0, 13), confirming B. Choice A gives variance 25, which arises from using only the diagonal entries 9+4⋅4−2⋅3=259 + 4\cdot4 - 2\cdot3 = 25 — a sign error in the cross term. Choice C gives 37, corresponding to incorrectly adding rather than subtracting the covariance contribution. Choice D uses the wrong scaling nn instead of n\sqrt{n}; the Delta Method preserves the n\sqrt{n} rate from the original CLT. Study tip: Always write out the gradient explicitly and track signs carefully — the cross-term 2σ12∂1g ∂2g2\sigma_{12}\partial_1 g\,\partial_2 g is the most common place to make errors on Delta Method questions.

Question 9

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(\widehat\theta_n-\theta_n)\xrightarrow{d}Z with Z∼N(0,σ2)Z\sim N(0,\sigma^2). For g(θ)=θ2g(\theta)=\theta^2, what is the limit of n{g(θ^n)−g(θn)}n\{g(\widehat\theta_n)-g(\theta_n)\}?

  1. 2cZ2cZ, because the first-order Taylor term alone determines the limit
  2. (c+Z)2−c2(c+Z)^2-c^2, because both Taylor terms contribute at this rate (correct answer)
  3. Z2Z^2, because the local parameter cancels from the expansion
  4. N(0,4c2σ2)N(0,4c^2\sigma^2), because the transformed limit remains normal
Explanation: When a parameter sequence drifts with the sample size — here θn=c/n\theta_n = c/\sqrt{n} — you're in a local/contiguous alternatives setting. The key insight is that both the estimator and the true parameter are shrinking at rate 1/n1/\sqrt{n}, so you must track what happens to g(θ^n)−g(θn)g(\hat\theta_n) - g(\theta_n) when scaled by nn, not n\sqrt{n}. Apply a second-order Taylor expansion of g(θ^n)g(\hat\theta_n) around θn\theta_n: g(θ^n)−g(θn)≈g′(θn)(θ^n−θn)+12g′′(θn)(θ^n−θn)2g(\hat\theta_n) - g(\theta_n) \approx g'(\theta_n)(\hat\theta_n - \theta_n) + \tfrac{1}{2}g''(\theta_n)(\hat\theta_n - \theta_n)^2 For g(θ)=θ2g(\theta) = \theta^2, we have g′(θ)=2θg'(\theta) = 2\theta and g′′(θ)=2g''(\theta) = 2. Multiplying through by nn: n{g(θ^n)−g(θn)}≈2θn⋅n(θ^n−θn)+[n(θ^n−θn)]2n\{g(\hat\theta_n)-g(\theta_n)\} \approx 2\theta_n \cdot \sqrt{n}(\hat\theta_n-\theta_n) + [\sqrt{n}(\hat\theta_n-\theta_n)]^2 Since n θn=c\sqrt{n}\,\theta_n = c and n(θ^n−θn)→dZ\sqrt{n}(\hat\theta_n - \theta_n) \xrightarrow{d} Z, the limit is 2cZ+Z2=(c+Z)2−c22cZ + Z^2 = (c+Z)^2 - c^2, confirming answer B. Choice A fails because it drops the second-order term — but here that term is O(1)O(1), not negligible; both terms survive at the same scale. Choice C would be correct if θn→0\theta_n \to 0 fast enough to kill the first-order term, but n θn=c≠0\sqrt{n}\,\theta_n = c \neq 0, so it persists. Choice D mistakes the distribution of the limit for the limit itself; (c+Z)2−c2(c+Z)^2 - c^2 is not Gaussian. Study tip: Whenever θn∼1/n\theta_n \sim 1/\sqrt{n}, always keep the second-order Taylor term — the local drift cc prevents cancellation that would otherwise simplify the limit.

Question 10

Suppose n θ^n→dN(0,4)\sqrt n\,\widehat\theta_n\xrightarrow{d}N(0,4), and define Tn=∣θ^n∣T_n=|\widehat\theta_n|. Which statement correctly describes the asymptotic behavior of TnT_n?

  1. nTn→dN(0,4)\sqrt nT_n\xrightarrow{d}N(0,4), by applying the ordinary delta method
  2. nTn→d∣Z∣\sqrt nT_n\xrightarrow{d}|Z|, where Z∼N(0,1)Z\sim N(0,1)
  3. nTn→d2∣Z∣\sqrt nT_n\xrightarrow{d}2|Z|, where Z∼N(0,1)Z\sim N(0,1) (correct answer)
  4. nTn2→d2χ12nT_n^2\xrightarrow{d}2\chi_1^2, with the same scale as TnT_n
Explanation: When you see a transformation of an asymptotically normal estimator, your first instinct should be the delta method — but the delta method requires a differentiable function at the limiting point. The absolute value function g(x)=∣x∣g(x) = |x| is not differentiable at zero, so if θ^n→p0\widehat{\theta}_n \xrightarrow{p} 0, the classical delta method fails and you must work directly from the limiting distribution. Here, n θ^n→dN(0,4)\sqrt{n}\,\widehat{\theta}_n \xrightarrow{d} N(0,4), which we can write as n θ^n→d2Z\sqrt{n}\,\widehat{\theta}_n \xrightarrow{d} 2Z where Z∼N(0,1)Z \sim N(0,1). Since the absolute value function is continuous everywhere, the continuous mapping theorem applies directly: n Tn=∣n θ^n∣→d∣2Z∣=2∣Z∣\sqrt{n}\,T_n = |\sqrt{n}\,\widehat{\theta}_n| \xrightarrow{d} |2Z| = 2|Z|. That confirms C is correct. Choice A fails because g(x)=∣x∣g(x) = |x| is non-differentiable at zero, so the ordinary delta method cannot produce a normal limit — it gives a folded distribution instead. Choice B gets the right idea (folded normal) but loses the scale factor: since the variance is 4, not 1, the limiting distribution is 2∣Z∣2|Z|, not ∣Z∣|Z|. Forgetting to carry the standard deviation through is a classic algebra slip. Choice D is tempting because (n θ^n)2/4→dχ12(\sqrt{n}\,\widehat{\theta}_n)^2/4 \xrightarrow{d} \chi_1^2, so nTn2→d4χ12nT_n^2 \xrightarrow{d} 4\chi_1^2, not 2χ122\chi_1^2 — the scale is wrong by a factor of 2. Study tip: Whenever a transformation is non-differentiable at the limit point, bypass the delta method entirely and apply the continuous mapping theorem directly to the known limiting distribution — just track the scale carefully.