Delta Method - Statistics Graduate Level

1. Suppose $\sqrt n\{(\widehat\mu_T,\widehat\mu_C)^\mathsf{T}-(2,1)^\mathsf{T}\}\xrightarrow{d}N(0,\Sigma)$, where $\Sigma=\begin{pmatrix}9&3\\3&4\end{pmatrix}$. For the estimated mean ratio $R_n=\widehat\mu_T/\widehat\mu_C$, what is its limiting distribution?
   Answer: $\sqrt n(R_n-2)\xrightarrow{d}N(0,13)$ 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 $\sqrt{n}(\hat{\theta} - \theta) \xrightarrow{d} N(0, \Sigma)$, then $\sqrt{n}(g(\hat{\theta}) - g(\theta)) \xrightarrow{d} N(0, \nabla g(\theta)^\mathsf{T} \Sigma \nabla g(\theta))$. Here, $g(\mu_T, \mu_C) = \mu_T/\mu_C$, so compute the gradient at the true values $(2, 1)$: $\nabla g = \left(\frac{1}{\mu_C},\ -\frac{\mu_T}{\mu_C^2}\right)^\mathsf{T} = \left(1,\ -2\right)^\mathsf{T}$ Now apply the quadratic form: $\nabla g^\mathsf{T} \Sigma \nabla g = \begin{pmatrix}1 & -2\end{pmatrix}\begin{pmatrix}9&3\\3&4\end{pmatrix}\begin{pmatrix}1\\-2\end{pmatrix}$ First, $\Sigma \nabla g = (9 - 6,\ 3 - 8)^\mathsf{T} = (3,\ -5)^\mathsf{T}$. Then $\nabla g^\mathsf{T}(3, -5)^\mathsf{T} = 3 + 10 = 13$. So $\sqrt{n}(R_n - 2) \xrightarrow{d} N(0, 13)$, confirming B. Choice A gives variance 25, which arises from using only the diagonal entries $9 + 4\cdot4 - 2\cdot3 = 25$ — 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 $n$ instead of $\sqrt{n}$; the Delta Method preserves the $\sqrt{n}$ rate from the original CLT. Study tip: Always write out the gradient explicitly and track signs carefully — the cross-term $2\sigma_{12}\partial_1 g\,\partial_2 g$ is the most common place to make errors on Delta Method questions.
2. At $x=1$, suppose the true coefficients are $\beta_1=3$ and $\beta_0=-3-\log2$, so that $p(1)=1/3$. Also suppose $\sqrt n(\widehat\beta-\beta)\xrightarrow{d}N(0,\Sigma)$, where $\Sigma=\begin{pmatrix}4&1\\1&1\end{pmatrix}$ for $\beta=(\beta_0,\beta_1)^\mathsf{T}$. What is the asymptotic variance of $\sqrt n\{\widehat m(1)-m(1)\}$?
   Answer: $\frac{16}{27}$ When a question asks for the asymptotic variance of a nonlinear function of estimators, your instinct should be the delta method: if $\sqrt{n}(\hat{\beta}-\beta)\xrightarrow{d}N(0,\Sigma)$, then $\sqrt{n}\{\hat{m}(1)-m(1)\}\xrightarrow{d}N(0,\nabla m^\mathsf{T}\Sigma\,\nabla m)$, where $\nabla m$ is the gradient of $m(x)$ evaluated at the true $\beta$. First, compute the true values. With $p(1)=1/3$, the marginal effect is $m(1)=\beta_1 p(1)(1-p(1))=3\cdot\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{3}$. Next, find $\nabla m = \left(\frac{\partial m}{\partial \beta_0},\frac{\partial m}{\partial \beta_1}\right)^\mathsf{T}$. Since $m=\beta_1 p(1-p)$ and $\frac{\partial p}{\partial \beta_j}=x^j p(1-p)$ (with $x=1$): $\frac{\partial m}{\partial \beta_0}=\beta_1 p(1-p)(1-2p)=3\cdot\tfrac{2}{9}\cdot\tfrac{1}{3}=\tfrac{2}{9}$ $\frac{\partial m}{\partial \beta_1}=p(1-p)+\beta_1 p(1-p)(1-2p)=\tfrac{2}{9}+\tfrac{2}{9}=\tfrac{4}{9}$ So $\nabla m = \left(\frac{2}{9},\frac{4}{9}\right)^\mathsf{T}$. The asymptotic variance is: $\nabla m^\mathsf{T}\Sigma\,\nabla m = \tfrac{1}{81}\begin{pmatrix}2&4\end{pmatrix}\begin{pmatrix}4&1\\1&1\end{pmatrix}\begin{pmatrix}2\\4\end{pmatrix}=\tfrac{1}{81}(2,4)\begin{pmatrix}12\\6\end{pmatrix}=\tfrac{48}{81}=\tfrac{16}{27}$ This confirms A is correct. Choice B ($8/27$) likely drops the cross-terms in the quadratic form. Choice C ($4/9$) may use only the $\beta_1$ partial derivative. Choice D ($20/27$) 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 $\Sigma$ — they often flip the answer.
3. Suppose $\widehat\theta_n\geq0$ and $n\widehat\theta_n\xrightarrow{d}E$, where $E$ has an exponential distribution with rate $1$. If $h(\theta)=\sqrt\theta$, which asymptotic statement is correct?
   Answer: $\sqrt n\,h(\widehat\theta_n)\xrightarrow{d}\sqrt{E}$, because $\sqrt{n}\,h(\widehat\theta_n)=\sqrt{n\widehat\theta_n}$ and the continuous mapping theorem applies When a parameter estimator has a nonstandard convergence rate — here $n\widehat\theta_n \xrightarrow{d} E$ rather than the usual $\sqrt{n}$ rate — you need to track that rate carefully before applying any transformation. The key tool is the continuous mapping theorem (CMT): if $X_n \xrightarrow{d} X$ and $g$ is continuous, then $g(X_n) \xrightarrow{d} g(X)$. Notice the algebraic identity at the heart of option D: $\sqrt{n}\,h(\widehat\theta_n) = \sqrt{n}\,\sqrt{\widehat\theta_n} = \sqrt{n\widehat\theta_n}$. Since $n\widehat\theta_n \xrightarrow{d} E$, you can apply the CMT with the continuous function $g(x) = \sqrt{x}$ to conclude $\sqrt{n\widehat\theta_n} \xrightarrow{d} \sqrt{E}$. This makes D correct. Option A claims $n\,h(\widehat\theta_n) \xrightarrow{d} \sqrt{E}$, but $n\sqrt{\widehat\theta_n} = \sqrt{n}\cdot\sqrt{n\widehat\theta_n}$, which diverges — the rate $n$ is too fast. Option B asserts the limit is still $E$, 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 $h'(\theta_0) \neq 0$ and a regular $\sqrt{n}$-rate estimator. Here $\theta_0 = 0$ lies on the boundary, $h'(0) = \frac{1}{2\sqrt{0}}$ is undefined, and the convergence rate is $n$, not $\sqrt{n}$ — 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.
4. A parameter $\theta$ is defined implicitly as the positive solution of $\theta^2+\lambda\theta-\mu=0$. At the true values, $(\theta,\lambda,\mu)=(1,1,2)$. Suppose $\sqrt n\{(\widehat\lambda,\widehat\mu)^\mathsf{T}-(1,2)^\mathsf{T}\}\xrightarrow{d}N(0,\Sigma)$, where $\Sigma=\begin{pmatrix}1&1/2\\1/2&4\end{pmatrix}$, and let $\widehat\theta$ be the corresponding positive root. What is the asymptotic variance of $\sqrt n(\widehat\theta-1)$?
   Answer: $\frac{4}{9}$, obtained via the implicit function theorem applied to the constraint equation 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 $\nabla_\theta$ in terms of the other parameters, then propagate the asymptotic variance. Your constraint is $g(\theta,\lambda,\mu)=\theta^2+\lambda\theta-\mu=0$. By the IFT, $\frac{\partial\theta}{\partial\lambda}=-\frac{g_\lambda}{g_\theta}$ and $\frac{\partial\theta}{\partial\mu}=-\frac{g_\mu}{g_\theta}$. Compute the partials at $(1,1,2)$: $g_\theta=2\theta+\lambda=3$, $g_\lambda=\theta=1$, $g_\mu=-1$. So the gradient is $\nabla=\left(-\frac{1}{3},\frac{1}{3}\right)^\mathsf{T}$. The asymptotic variance is $\nabla^\mathsf{T}\Sigma\,\nabla$: $\nabla^\mathsf{T}\Sigma\,\nabla = \frac{1}{9}\begin{pmatrix}-1&1\end{pmatrix}\begin{pmatrix}1&1/2\\1/2&4\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}=\frac{1}{9}(1-1/2-1/2+4)=\frac{4}{9}.$ This confirms answer A is correct. B is wrong because it ignores the off-diagonal covariance term $\Sigma_{12}=1/2$. Dropping it gives $\frac{1}{9}(1+4)=\frac{5}{9}$, 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 $\frac{2}{3}$. D ignores $\lambda$ entirely and omits the $1/g_\theta$ scaling, giving an answer of $4$ — essentially computing $\Sigma_{22}$ raw. Study tip: Whenever a parameter is implicitly defined, always compute the full gradient vector and apply $\nabla^\mathsf{T}\Sigma\,\nabla$. Never drop covariance terms without checking the sign structure — the cross-term can reduce or inflate the variance significantly.
5. For each $n$, let the true parameter be $\theta_n=c/\sqrt n$, where $c$ is fixed. Suppose $\sqrt n(\widehat\theta_n-\theta_n)\xrightarrow{d}Z$ with $Z\sim N(0,\sigma^2)$. For $g(\theta)=\theta^2$, what is the limit of $n\{g(\widehat\theta_n)-g(\theta_n)\}$?
   Answer: $(c+Z)^2-c^2$, because both Taylor terms contribute at this rate When a parameter sequence drifts with the sample size — here $\theta_n = c/\sqrt{n}$ — you're in a local/contiguous alternatives setting. The key insight is that both the estimator and the true parameter are shrinking at rate $1/\sqrt{n}$, so you must track what happens to $g(\hat\theta_n) - g(\theta_n)$ when scaled by $n$, not $\sqrt{n}$. Apply a second-order Taylor expansion of $g(\hat\theta_n)$ around $\theta_n$: $g(\hat\theta_n) - g(\theta_n) \approx g'(\theta_n)(\hat\theta_n - \theta_n) + \tfrac{1}{2}g''(\theta_n)(\hat\theta_n - \theta_n)^2$ For $g(\theta) = \theta^2$, we have $g'(\theta) = 2\theta$ and $g''(\theta) = 2$. Multiplying through by $n$: $n\{g(\hat\theta_n)-g(\theta_n)\} \approx 2\theta_n \cdot \sqrt{n}(\hat\theta_n-\theta_n) + [\sqrt{n}(\hat\theta_n-\theta_n)]^2$ Since $\sqrt{n}\,\theta_n = c$ and $\sqrt{n}(\hat\theta_n - \theta_n) \xrightarrow{d} Z$, the limit is $2cZ + Z^2 = (c+Z)^2 - c^2$, confirming answer B. Choice A fails because it drops the second-order term — but here that term is $O(1)$, not negligible; both terms survive at the same scale. Choice C would be correct if $\theta_n \to 0$ fast enough to kill the first-order term, but $\sqrt{n}\,\theta_n = c \neq 0$, so it persists. Choice D mistakes the distribution of the limit for the limit itself; $(c+Z)^2 - c^2$ is not Gaussian. Study tip: Whenever $\theta_n \sim 1/\sqrt{n}$, always keep the second-order Taylor term — the local drift $c$ prevents cancellation that would otherwise simplify the limit.
6. Let $\widehat\theta_n$ be the sample mean of independent $N(0,1)$ observations, and consider inference for $g(\theta)=\theta^2$ at $\theta=0$. A researcher uses the plug-in delta-method standard error $\widehat{\operatorname{se}}\{g(\widehat\theta_n)\}=2|\widehat\theta_n|/\sqrt n$ and forms $W_n=\{g(\widehat\theta_n)-g(0)\}/\widehat{\operatorname{se}}\{g(\widehat\theta_n)\}$ whenever $\widehat\theta_n\neq0$. What is the limiting distribution of $W_n$?
   Answer: $\frac12|Z|$, where $Z\sim N(0,1)$ 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 $g(\theta) = \theta^2$, we have $g'(\theta) = 2\theta$, which equals zero at $\theta = 0$. This kills the usual delta-method approximation. Let's work through what actually happens. Write $\widehat\theta_n = Z/\sqrt{n}$ where $Z \sim N(0,1)$. Then: $W_n = \frac{\widehat\theta_n^2 - 0}{2|\widehat\theta_n|/\sqrt{n}} = \frac{Z^2/n}{2|Z|/n} = \frac{Z^2}{2|Z|} = \frac{|Z|}{2}$ This is an exact finite-sample identity (not just a limit), so the limiting distribution is $\frac{1}{2}|Z|$, 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 $1/\sqrt{n}$), but here the denominator also depends on $\widehat\theta_n$ in a way that cancels one power, leaving a half-|Z| rather than |Z|. C confuses the distribution of the numerator — yes, $n\widehat\theta_n^2 \to \chi^2_1$ — 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 $g'(\theta_0) = 0$, always rework the asymptotics from scratch using the second-order expansion, and track how both numerator and denominator scale with $n$.
7. Suppose $\sqrt n\,\widehat\theta_n\xrightarrow{d}N(0,4)$, and define $T_n=|\widehat\theta_n|$. Which statement correctly describes the asymptotic behavior of $T_n$?
   Answer: $\sqrt nT_n\xrightarrow{d}2|Z|$, where $Z\sim N(0,1)$ 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 $g(x) = |x|$ is not differentiable at zero, so if $\widehat{\theta}_n \xrightarrow{p} 0$, the classical delta method fails and you must work directly from the limiting distribution. Here, $\sqrt{n}\,\widehat{\theta}_n \xrightarrow{d} N(0,4)$, which we can write as $\sqrt{n}\,\widehat{\theta}_n \xrightarrow{d} 2Z$ where $Z \sim N(0,1)$. Since the absolute value function is continuous everywhere, the continuous mapping theorem applies directly: $\sqrt{n}\,T_n = |\sqrt{n}\,\widehat{\theta}_n| \xrightarrow{d} |2Z| = 2|Z|$. That confirms C is correct. Choice A fails because $g(x) = |x|$ 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 $2|Z|$, not $|Z|$. Forgetting to carry the standard deviation through is a classic algebra slip. Choice D is tempting because $(\sqrt{n}\,\widehat{\theta}_n)^2/4 \xrightarrow{d} \chi_1^2$, so $nT_n^2 \xrightarrow{d} 4\chi_1^2$, not $2\chi_1^2$ — 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.
8. Suppose $\sqrt{n}\{(\widehat X_n,\widehat Y_n)^\mathsf{T}-(1,2)^\mathsf{T}\}\xrightarrow{d}N(0,\Sigma)$, where $\Sigma=\begin{pmatrix}4&1\\1&9\end{pmatrix}$. Define $U_n=\widehat X_n\widehat Y_n$ and $V_n=\widehat X_n/\widehat Y_n$. What is the asymptotic covariance matrix of $\sqrt n\{(U_n,V_n)^\mathsf{T}-(2,1/2)^\mathsf{T}\}$?
   Answer: $\begin{pmatrix}29&7/4\\7/4&21/16\end{pmatrix}$, correctly accounting for both the partial derivatives and the covariance in $\Sigma$ When you see a joint asymptotic normality result paired with a transformation, your instinct should be the Delta Method for vector-valued functions. If $\sqrt{n}(\hat{\theta}_n - \theta) \xrightarrow{d} N(0, \Sigma)$ and $g: \mathbb{R}^2 \to \mathbb{R}^2$ is differentiable, then $\sqrt{n}(g(\hat{\theta}_n) - g(\theta)) \xrightarrow{d} N(0, J\Sigma J^\mathsf{T})$, where $J$ is the Jacobian of $g$ evaluated at $\theta = (1,2)^\mathsf{T}$. Here $g(x,y) = (xy,\, x/y)^\mathsf{T}$, so the Jacobian is: $J = \begin{pmatrix} y & x \\ 1/y & -x/y^2 \end{pmatrix}\Bigg|_{(1,2)} = \begin{pmatrix} 2 & 1 \\ 1/2 & -1/4 \end{pmatrix}.$ Now compute $J\Sigma J^\mathsf{T}$ with $\Sigma = \begin{pmatrix}4&1\\1&9\end{pmatrix}$. First, $J\Sigma = \begin{pmatrix}2(4)+1(1) & 2(1)+1(9)\\ \tfrac{1}{2}(4)+(-\tfrac{1}{4})(1) & \tfrac{1}{2}(1)+(-\tfrac{1}{4})(9)\end{pmatrix} = \begin{pmatrix}9&11\\ \tfrac{7}{4}&-\tfrac{7}{4}\end{pmatrix}$. Then $(J\Sigma)J^\mathsf{T}$ gives diagonal entries $9(2)+11(1)=29$ and $\tfrac{7}{4}(\tfrac{1}{2})+(-\tfrac{7}{4})(-\tfrac{1}{4})=\tfrac{7}{8}+\tfrac{7}{16}=\tfrac{21}{16}$, and off-diagonal $9(\tfrac{1}{2})+11(-\tfrac{1}{4})=\tfrac{9}{2}-\tfrac{11}{4}=\tfrac{7}{4}$. This confirms A is correct. B is wrong because it flips the sign of the off-diagonal to $-7/4$, which would only occur if the Jacobian entries for $V_n$ were misread. C results from mishandling the off-diagonal of $\Sigma$ — 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 $\partial(x/y)/\partial y = -x/y^2$) are the most common trap in multivariate Delta Method problems.
9. Let $\widehat{\theta}_n$ estimate $\theta=2$. Suppose $E(\widehat{\theta}_n-\theta)=n^{-1}+o(n^{-1})$, $\operatorname{Var}(\widehat{\theta}_n)=4n^{-1}+o(n^{-1})$, and the Taylor remainder is uniformly integrable at the required order. For $g(x)=\log x$, what is the limit of $n\{E[g(\widehat{\theta}_n)]-g(\theta)\}$?
   Answer: $0$, because the first-order bias contribution and the curvature-induced bias contribution cancel 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 $g(\widehat{\theta}_n)$ around $\theta = 2$: $g(\widehat{\theta}_n) \approx g(\theta) + g'(\theta)(\widehat{\theta}_n - \theta) + \tfrac{1}{2}g''(\theta)(\widehat{\theta}_n - \theta)^2$ Taking expectations and multiplying by $n$: $n\{E[g(\widehat{\theta}_n)] - g(\theta)\} \approx n \cdot g'(\theta) \cdot E(\widehat{\theta}_n - \theta) + \tfrac{n}{2} g''(\theta) \cdot E(\widehat{\theta}_n - \theta)^2$ For $g(x) = \log x$: $g'(2) = \tfrac{1}{2}$, $g''(2) = -\tfrac{1}{4}$. Since $E(\widehat{\theta}_n - \theta) = n^{-1} + o(n^{-1})$ and $\operatorname{Var}(\widehat{\theta}_n) = 4n^{-1} + o(n^{-1})$, the second moment satisfies $E(\widehat{\theta}_n - \theta)^2 \approx \operatorname{Var}(\widehat{\theta}_n) = 4n^{-1}$ (the squared bias is $o(n^{-1})$). Plugging in: $n \cdot \tfrac{1}{2} \cdot n^{-1} + \tfrac{n}{2} \cdot (-\tfrac{1}{4}) \cdot 4n^{-1} = \tfrac{1}{2} - \tfrac{1}{2} = 0$ So the correct answer is C: the first-order bias contribution ($+\tfrac{1}{2}$) and the curvature-induced bias ($-\tfrac{1}{2}$) 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 $g''$. Choice D incorrectly assumes transformations leave bias unchanged — they don't. The key study tip: whenever you see $n\{E[g(\widehat{\theta}_n)] - g(\theta)\}$, always carry both the first-order bias term and the $\tfrac{1}{2}g''(\theta)\cdot\operatorname{Var}$ term. Dropping either one is the most common mistake on these problems.
10. Let $\sqrt n(\widehat\theta_{1n},\widehat\theta_{2n})^\mathsf{T}\xrightarrow{d}N(0,\Sigma)$, where both marginal variances equal $1$ and the correlation is $1/2$. For $g(a,b)=(a+b)^2$, which limit is correct?
    Answer: $n\,g(\widehat\theta_{1n},\widehat\theta_{2n})\xrightarrow{d}3\chi_1^2$ 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 $g(a,b) = (a+b)^2$: we get $\nabla g = (2(a+b), 2(a+b))^\mathsf{T}$, which equals $(0,0)^\mathsf{T}$ at $(0,0)$. 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 $n$ (not $\sqrt{n}$), and the limiting distribution is a quadratic form in the Gaussian limit. Specifically, if $\sqrt{n}(\hat\theta_{1n}, \hat\theta_{2n})^\mathsf{T} \xrightarrow{d} Z \sim N(0,\Sigma)$, then $n\,g(\hat\theta_{1n},\hat\theta_{2n}) \xrightarrow{d} (Z_1+Z_2)^2$. Now $Z_1+Z_2 \sim N(0, \sigma^2)$ where $\sigma^2 = \text{Var}(Z_1+Z_2) = 1 + 1 + 2(1/2) = 3$. Therefore $(Z_1+Z_2)^2 = 3 \cdot W^2$ where $W\sim N(0,1)$, giving $3\chi_1^2$. This confirms answer B is correct. Answer A is wrong on two counts: the scaling $\sqrt{n}$ is too slow (the right scale is $n$), and a nonzero quadratic form cannot be Gaussian. Answer C uses the correct $n$ scaling but claims a $\chi_2^2$ distribution — a quadratic form in a two-dimensional Gaussian is only $\chi_2^2$ when standardized appropriately; here the sum $Z_1+Z_2$ is one-dimensional, giving $\chi_1^2$, not $\chi_2^2$. Answer D uses $\sqrt{n}$ scaling with $3\chi_1^2$, getting the distribution right but the normalization wrong. Study tip: Always check whether $\nabla g = 0$ at the true parameter before applying the delta method — a vanishing gradient shifts the scaling from $\sqrt{n}$ to $n$ and the limit from normal to chi-squared.

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