Slutsky's & Continuous Mapping Theorems - Statistics Graduate Level

1. Suppose $nX_n\xrightarrow{p}\lambda$ for a finite constant $\lambda$, and assume $P(X_n>-1)\to1$. What is the probability limit of $(1+X_n)^n$?
   Answer: $e^\lambda,$ obtained by applying Slutsky's theorem to the logarithm of the expression. When you see an expression like $(1+X_n)^n$ where $X_n \to 0$ in probability, your instinct should be to take the logarithm and invoke the classical limit $\lim_{n\to\infty}(1+a_n)^n = e^L$ whenever $na_n \to L$. This is a convergence-in-probability analogue of a fundamental deterministic result. The key move is writing $\log(1+X_n)^n = n\log(1+X_n)$. Since $nX_n \xrightarrow{p} \lambda$, we know $X_n \xrightarrow{p} 0$. Using the Taylor expansion $\log(1+x) \approx x - x^2/2 + \cdots$ near zero, we get $n\log(1+X_n) = nX_n \cdot \frac{\log(1+X_n)}{X_n} \xrightarrow{p} \lambda \cdot 1 = \lambda$, since the ratio $\frac{\log(1+X_n)}{X_n} \xrightarrow{p} 1$ by continuity. Applying the continuous mapping theorem to $e^x$ then gives $(1+X_n)^n \xrightarrow{p} e^\lambda$, confirming B. Choice A is wrong because the first-order binomial approximation $(1+X_n)^n \approx 1 + nX_n$ is only valid when $nX_n$ is small — here $nX_n \to \lambda$, a finite but potentially nonzero constant, and the higher-order terms in the exponential expansion matter crucially. Choice C incorrectly retains only the quadratic term from $\log(1+x) \approx x - x^2/2$, but since $X_n = O(1/n)$, the term $nX_n^2 \to 0$, so the quadratic contribution vanishes and does not alter the exponent. Choice D reflects a common confusion: the fact that the transformation changes with $n$ does not prevent convergence. The continuous mapping theorem and probabilistic limit arguments handle sequences of transformations when the limit is well-defined. Study tip: Whenever you see $(1+\text{small})^n$, immediately take logarithms and use $\log(1+x)/x \to 1$. This logarithm-then-exponentiate strategy is your go-to tool for limits of this form.
2. Let $Z\sim N(0,1)$ and define $X_n=Z/n$ and $J_n=\mathbf{1}\{X_n>0\}$. Although $X_n\xrightarrow{p}0$, which statement correctly describes the behavior of $J_n$?
   Answer: $J_n\xrightarrow{d}\operatorname{Bernoulli}(1/2),$ because its value equals $\mathbf{1}\{Z>0\}$ for every $n$. When studying convergence in probability, a critical subtlety is that the continuous mapping theorem requires the limiting function to be continuous at the limit point — and indicator functions are not. Here, $X_n = Z/n \xrightarrow{p} 0$, but notice something remarkable: for every fixed $n$, $J_n = \mathbf{1}\{X_n > 0\} = \mathbf{1}\{Z/n > 0\} = \mathbf{1}\{Z > 0\}$. The division by $n$ doesn't change the sign of $Z$. So $J_n$ is literally the same random variable for every $n$ — it never changes. Since $\mathbf{1}\{Z > 0\} \sim \operatorname{Bernoulli}(1/2)$, the sequence trivially converges in distribution to $\operatorname{Bernoulli}(1/2)$. Answer C is correct. A is wrong because the continuous mapping theorem does not apply to all bounded measurable functions — only to those continuous at the limit. The indicator $\mathbf{1}\{\cdot > 0\}$ is discontinuous at 0, which is precisely where $X_n$ is converging. So you cannot conclude $J_n \xrightarrow{p} \mathbf{1}\{0 > 0\} = 0\$. **B** is subtly misleading. Yes,$E[$J_n$] = 1/2$for all$n$, but equal expectations do not imply convergence in probability to a constant. Convergence in probability to$1/2$would require$$J_n$$to concentrate around$1/2$, which a Bernoulli random variable never does. **D** is wrong because a limiting distribution clearly exists — as shown,$$J_n$$is constant across$n$$. Strategy tip: Whenever you see convergence applied through a discontinuous function, stop and check whether the continuous mapping theorem actually applies. Evaluate what the transformed sequence looks like directly — sometimes, as here, the sequence is literally constant.
3. Let $\widehat\beta_n$ estimate a parameter $\beta_0\in\mathbb R^k$, and suppose $\sqrt n(\widehat\beta_n-\beta_0)\xrightarrow{d}N_k(0,\Sigma)$, where $\Sigma$ is positive definite. If $\widehat\Sigma_n\xrightarrow{p}\Sigma$ and is invertible with probability approaching one, what is the limit of $W_n=n(\widehat\beta_n-\beta_0)^{\mathsf T}\widehat\Sigma_n^{-1}(\widehat\beta_n-\beta_0)?$
   Answer: $W_n\xrightarrow{d}\chi_k^2,$ because consistent matrix inversion and Slutsky's theorem produce a standardized quadratic form. Whenever you see a quadratic form built from an asymptotically normal estimator and a consistently estimated covariance matrix, your instinct should be to apply the continuous mapping theorem paired with Slutsky's theorem. Here's the core reasoning for why D is correct. You know $\sqrt{n}(\widehat{\beta}_n - \beta_0) \xrightarrow{d} Z \sim N_k(0, \Sigma)$. Since $\widehat{\Sigma}_n \xrightarrow{p} \Sigma$ and $\Sigma$ is positive definite, by Slutsky's theorem $\widehat{\Sigma}_n^{-1} \xrightarrow{p} \Sigma^{-1}$. The statistic $W_n$ can be written as $\left[\sqrt{n}(\widehat{\beta}_n - \beta_0)\right]^{\mathsf{T}} \widehat{\Sigma}_n^{-1} \left[\sqrt{n}(\widehat{\beta}_n - \beta_0)\right]$, which converges in distribution to $Z^{\mathsf{T}}\Sigma^{-1}Z$. Because $\Sigma$ is positive definite, write $\Sigma = L L^{\mathsf{T}}$; then $L^{-1}Z \sim N_k(0, I_k)$, so $Z^{\mathsf{T}}\Sigma^{-1}Z = \|L^{-1}Z\|^2 \sim \chi^2_k$. The degrees of freedom equal the dimension $k$ of the parameter vector, confirming D. A contains a grain of truth — $\mathbb{E}[\chi^2_k] = k$ — but convergence in distribution to $\chi^2_k$ is not the same as convergence in probability to the constant $k$. That would require the variance to vanish, which it does not. B is wrong because quadratic forms in normal vectors follow chi-squared distributions, not normal distributions. A $\chi^2_k$ variable has mean $k$ and variance $2k$, but the distribution itself is decidedly non-normal. C confuses dimension reduction with degrees of freedom. Collapsing a $k$-vector into one scalar statistic does not produce $\chi^2_1$ unless $k=1$; the degrees of freedom track the rank of the quadratic form, which is $k$. Study tip: Memorize the chain — asymptotic normality + consistent covariance estimator + Slutsky + CMT = $\chi^2_k$ Wald statistic. This is the backbone of virtually every large-sample hypothesis test you will encounter.
4. Suppose $\theta\neq0$ and $\sqrt n(\widehat\theta_n-\theta)\xrightarrow{d}N(0,\sigma^2)$. Using an algebraic decomposition followed by Slutsky's theorem, what is the limiting distribution of $\sqrt n\left(\widehat\theta_n^{-1}-\theta^{-1}\right)?$
   Answer: $N\left(0,\frac{\sigma^2}{\theta^4}\right),$ because the random denominator converges in probability to $\theta^2$. When you see a question asking for the limiting distribution of a transformed estimator, your first instinct should be the Delta Method — but this problem specifically asks you to use an algebraic decomposition plus Slutsky's theorem, which reveals the same result from first principles. Start by writing the key identity: $\widehat\theta_n^{-1} - \theta^{-1} = \frac{\theta - \widehat\theta_n}{\widehat\theta_n \cdot \theta}$ Multiplying both sides by $\sqrt{n}$: $\sqrt{n}\left(\widehat\theta_n^{-1} - \theta^{-1}\right) = \frac{-\sqrt{n}(\widehat\theta_n - \theta)}{\widehat\theta_n \cdot \theta}$ Now apply Slutsky's theorem. Since $\sqrt{n}(\widehat\theta_n - \theta) \xrightarrow{d} N(0, \sigma^2)$ and $\widehat\theta_n \xrightarrow{p} \theta$ (consistency implied by asymptotic normality), the denominator $\widehat\theta_n \cdot \theta \xrightarrow{p} \theta^2$. Slutsky's theorem lets you replace the random denominator with its probability limit, giving: $\sqrt{n}\left(\widehat\theta_n^{-1} - \theta^{-1}\right) \xrightarrow{d} N\!\left(0,\frac{\sigma^2}{\theta^4}\right)$ This confirms B is correct — the $\theta^4$ in the denominator comes from squaring the $\theta^2$ that appears when you divide. A is wrong because dividing by $\theta^2$ (not $\theta^4$) would correspond to only one power of $\theta$ in the denominator, misapplying the scaling. C inverts the direction of the scaling entirely — a clear algebraic error. D is a red herring; the assumption $\theta \neq 0$ is explicitly given, so the reciprocal map is continuous at the limit point. Study tip: Whenever you apply Slutsky's theorem to a ratio, carefully track every factor in the denominator — each contributes to the final variance through squaring, which is where $\theta^2 \to \theta^4$ trips up many students.
5. Let $X_n\xrightarrow{d}X$, and define $I_n=\mathbf{1}\{X_n\leq 0\}$. Which condition is sufficient for concluding that $I_n\xrightarrow{d}\mathbf{1}\{X\leq 0\}$ by the extended continuous mapping theorem?
   Answer: It is sufficient that $P(X=0)=0$, even if the distributions of the variables $X_n$ have atoms at zero. Whenever you see a question involving convergence in distribution composed with a function, reach for the Continuous Mapping Theorem (CMT). The extended version states that if $X_n \xrightarrow{d} X$ and $g$ is measurable, then $g(X_n) \xrightarrow{d} g(X)$, provided $P(X \in D_g) = 0$, where $D_g$ is the set of discontinuities of $g$. Here, $g(x) = \mathbf{1}\{x \leq 0\}$, which is discontinuous at exactly one point: $x = 0$. The CMT therefore applies as long as $P(X = 0) = 0$ — that is, as long as the limiting variable $X$ places no mass at the single discontinuity of $g$. The distributions of the $X_n$ themselves are irrelevant to this condition; the CMT only cares about the limiting distribution. This confirms that A is correct: $P(X = 0) = 0$ is sufficient, and it doesn't matter whether the $X_n$ have atoms at zero. B is wrong because continuity of each $X_n$'s distribution is neither necessary nor sufficient — the condition must be placed on $X$, not the sequence. C is a fundamental misconception: convergence in distribution absolutely can be preserved by functions via the CMT; convergence in probability is a strictly stronger condition and is not required here. D introduces independence of the $X_n$, which is completely irrelevant — the CMT says nothing about joint structure of the sequence, only about the limit's distribution at discontinuity points. Study tip: For CMT problems, always identify the discontinuity set of $g$ first, then check whether the limit $X$ assigns it zero probability — the sequence's behavior is a red herring.
6. For each $n$, let $X_n$ and $Y_n$ be random variables satisfying $X_n\xrightarrow{d}N(0,1)$ and $Y_n\xrightarrow{d}N(0,1)$. No information about their joint dependence is available. Which conclusion about $X_n+Y_n$ is justified?
   Answer: Its limiting distribution is not determined from the two marginal convergence statements alone. Whenever you see a question involving convergence in distribution of multiple random variables, your first instinct should be: what do we know about their joint distribution? Marginal convergence statements tell you nothing about dependence structure, and sums depend critically on that structure. Here's why D is correct. Consider two extreme cases. If $X_n$ and $Y_n$ are independent with each converging to $N(0,1)$, then $X_n + Y_n \xrightarrow{d} N(0,2)$. But if $Y_n = -X_n$ for every $n$, then both marginals still converge to $N(0,1)$, yet $X_n + Y_n = 0$ for all $n$, which converges to a point mass at zero. You can also construct a coupling where $Y_n = X_n$, giving $X_n + Y_n \xrightarrow{d} N(0,4)$. Since different dependence structures yield completely different limiting distributions, the marginal convergences alone are insufficient to determine the limit of the sum. A is wrong because variances of marginals do not simply add under convergence in distribution when dependence is unknown — that rule applies to independent random variables, not arbitrary ones. B is wrong for the opposite extreme: representing both limits by the same normal variable (i.e., assuming perfect positive dependence) is just one possibility, not a necessity. C is wrong because cancellation to zero requires $Y_n \approx -X_n$, which is a specific dependence assumption that is not given. The key study takeaway: convergence in distribution is a marginal statement. To analyze functions of multiple sequences (like sums), you need joint convergence — typically established via the continuous mapping theorem or Slutsky's theorem, both of which require additional structural assumptions.
7. Suppose $X_n\xrightarrow{d}N(1,4)$ and $Y_n\xrightarrow{p}2$. The two sequences may be dependent. What is the limiting distribution of $W_n=Y_nX_n+Y_n^{-1}?$
   Answer: $W_n\xrightarrow{d}N\left(\frac{5}{2},16\right),$ obtained by applying the limiting affine transformation to $X_n$. When you see a sequence converging in distribution combined with another sequence converging in probability, your immediate instinct should be to reach for Slutsky's Theorem: if $X_n \xrightarrow{d} X$ and $Y_n \xrightarrow{p} c$ (a constant), then $g(X_n, Y_n) \xrightarrow{d} g(X, c)$ for continuous functions $g$. Critically, this holds regardless of dependence between the sequences — that's what makes Slutsky so powerful. Here, apply Slutsky directly: since $Y_n \xrightarrow{p} 2$, replace every $Y_n$ with its probability limit of 2. The function $W_n = Y_n X_n + Y_n^{-1}$ converges in distribution to $2X + 2^{-1} = 2X + \frac{1}{2}$, where $X \sim N(1,4)$. For a normal random variable, an affine transformation $aX + b$ gives $N(a\mu + b,\ a^2\sigma^2)$. So the limit is $N\!\left(2(1)+\tfrac{1}{2},\ 4(4)\right) = N\!\left(\tfrac{5}{2}, 16\right)$. This confirms answer A is correct. Answer B incorrectly ignores that multiplying $X_n$ by 2 scales the variance — it only accounts for the mean shift from the reciprocal term. Answer C gets the mean right but miscalculates the variance: multiplying by 2 multiplies variance by $2^2 = 4$, giving 16, not 8. Answer D is the most tempting trap — many students believe Slutsky requires independence. It does not; independence is sufficient but not necessary when one sequence converges to a constant in probability. Your study tip: memorize that Slutsky's Theorem demands only $Y_n \xrightarrow{p} c$, not independence. Exam questions are specifically designed to make you doubt this by mentioning potential dependence — don't take the bait.
8. Suppose $X_n\xrightarrow{d}N(0,1)$ and $Y_n\xrightarrow{p}0$, with arbitrary dependence between the sequences. Define $M_n=\max(X_n,Y_n)$. Which function is the cumulative distribution function of the limiting random variable?
   Answer: $F(t)=0$ for $t<0$ and $F(t)=\Phi(t)$ for $t\geq0$, so the limit is a mixture with a point mass of $\tfrac{1}{2}$ at zero. When you see a maximum of two sequences where one converges in distribution and the other converges in probability to a constant, your instinct should be to carefully track how the maximum behaves on each side of that constant — Slutsky's theorem alone won't carry you through. Here, $Y_n \xrightarrow{p} 0$, so for large $n$, $Y_n$ is negligible except when $X_n$ is also near zero. The key insight is that $M_n = \max(X_n, Y_n)$. For $t < 0$: since $Y_n \xrightarrow{p} 0$, the event $\{M_n \leq t\}$ requires both $X_n \leq t$ and $Y_n \leq t$, but $P(Y_n \leq t) \to 0$ for $t < 0$, so $F_{M_n}(t) \to 0$. For $t \geq 0$: $P(M_n \leq t) = P(X_n \leq t, Y_n \leq t)$. Since $Y_n \xrightarrow{p} 0 \leq t$, the event $\{Y_n \leq t\}$ has probability approaching 1, leaving $P(M_n \leq t) \to \Phi(t)$. This gives a CDF that jumps from 0 to $\Phi(0) = \tfrac{1}{2}$ at $t = 0$, confirming answer B — a mixture with a point mass of $\tfrac{1}{2}$ at zero. A misapplies Slutsky: that theorem handles sums and products of sequences, not maxima, and it ignores the truncation effect below zero. C incorrectly assumes independence and identical marginals, producing $\Phi(t)^2$, which would only be valid if both variables were i.i.d. standard normal. D describes a half-normal, which has no mass at zero but gets the behavior for $t > 0$ wrong — it confuses $|Z|$ with $\max(Z, 0)$. The study tip: for maximum operations in limit theorems, always evaluate the CDF separately on each region, checking where the "degenerate" sequence controls the bound.
9. Suppose an estimator satisfies $\sqrt{n}(\widehat{\theta}_n-\theta)\xrightarrow{d}N(0,V),$ where $0<V<\infty$, and suppose $\widehat{V}_n\xrightarrow{p}V$. No independence between $\widehat{\theta}_n$ and $\widehat{V}_n$ is assumed. What is the limiting distribution of $T_n=\frac{\sqrt{n}(\widehat{\theta}_n-\theta)}{\sqrt{\widehat{V}_n}}?$
   Answer: $T_n\xrightarrow{d}N(0,1),$ because the denominator converges in probability to the positive constant $\sqrt V$. When you see a question involving a standardized statistic where a variance parameter is replaced by a consistent estimator, your instinct should be to reach for Slutsky's theorem: if $X_n \xrightarrow{d} X$ and $Y_n \xrightarrow{p} c$ (a constant), then $X_n / Y_n \xrightarrow{d} X/c$, regardless of any dependence structure between $X_n$ and $Y_n$. Here, the numerator $\sqrt{n}(\widehat{\theta}_n - \theta) \xrightarrow{d} N(0,V)$, and since $\widehat{V}_n \xrightarrow{p} V > 0$, the continuous mapping theorem gives $\sqrt{\widehat{V}_n} \xrightarrow{p} \sqrt{V}$. Applying Slutsky's theorem, $T_n = \frac{\sqrt{n}(\widehat{\theta}_n-\theta)}{\sqrt{\widehat{V}_n}} \xrightarrow{d} \frac{N(0,V)}{\sqrt{V}} = N\!\left(0, \frac{V}{V}\right) = N(0,1).$ This confirms A is correct. B is wrong because dividing by $\sqrt{\widehat{V}_n}$ does alter the limiting variance — it normalizes it to 1, not leaves it at $V$. C is wrong because the numerator does not converge in probability to zero; it diverges in distribution to a non-degenerate normal after the $\sqrt{n}$ scaling. The ratio stabilizes to a non-trivial limit. D is the most tempting distractor — students often believe Slutsky's theorem requires independence. It does not; independence is sufficient but not necessary. Slutsky's theorem only requires convergence in distribution for one sequence and convergence in probability for the other. Your takeaway: Slutsky's theorem does not require independence. Anytime you see a consistent estimator plugged into a denominator (or any continuous function), you can apply it freely — this is the backbone of virtually every asymptotic $t$-statistic argument in statistics.
10. Suppose $T_n\xrightarrow{d}\chi_k^2$ and a data-dependent critical value satisfies $c_n\xrightarrow{p}q_{1-\alpha}$, where $q_{1-\alpha}$ is the upper $\alpha$ critical value of the $\chi_k^2$ distribution. No independence between $T_n$ and $c_n$ is assumed. What is the limit of $P(T_n>c_n)$?
    Answer: $\alpha,$ because Slutsky's theorem gives $T_n-c_n\xrightarrow{d}\chi_k^2-q_{1-\alpha}$ and the boundary has probability zero. When a test statistic converges in distribution and a critical value converges in probability, your instinct should be to reach for Slutsky's theorem: if $X_n \xrightarrow{d} X$ and $Y_n \xrightarrow{p} c$, then $X_n - Y_n \xrightarrow{d} X - c$, regardless of the dependence structure between $X_n$ and $Y_n$. This is the key insight the question is testing. Applying Slutsky directly, $T_n - c_n \xrightarrow{d} \chi_k^2 - q_{1-\alpha}$. Then: $P(T_n > c_n) = P(T_n - c_n > 0) \to P(\chi_k^2 - q_{1-\alpha} > 0) = P(\chi_k^2 > q_{1-\alpha}) = \alpha$ The last step uses the fact that $q_{1-\alpha}$ is defined as the upper-$\alpha$ quantile of $\chi_k^2$, and since $\chi_k^2$ is continuous, the boundary point $\{X = q_{1-\alpha}\}$ has probability zero. This confirms C is correct. A is wrong because it confuses almost-sure behavior with distributional convergence — $c_n \xrightarrow{p} q_{1-\alpha}$ does not mean $c_n$ eventually exceeds $T_n$ with certainty; the probability approaches $\alpha$, not 0. B is wrong because $1 - \alpha$ is the CDF evaluated at $q_{1-\alpha}$, i.e., $P(\chi_k^2 \leq q_{1-\alpha})$. The question asks for the tail probability $P(T_n > c_n)$, which is the complement. D is the most tempting distractor. Slutsky's theorem is precisely powerful because it requires no independence assumption — only that one sequence converges in probability to a constant. Study tip: Whenever you see $\xrightarrow{d}$ paired with $\xrightarrow{p}$ to a constant, Slutsky's theorem applies unconditionally — independence is never required when the second limit is a constant.

Start this quiz