Statistics Graduate Level Quiz: Method Of Moments And Mle
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Method Of Moments And MleQuestion 1 of 10

Let X1,…,X5X_1,\ldots,X_5 be independent observations from the uniform distribution on [0,θ][0,\theta]. The observed sample mean is 2.12.1, and the sample maximum is 4.04.0. Using the invariance principle separately with the maximum likelihood estimator and the method-of-moments estimator, what are the resulting estimates of Pθ(X>3)P_\theta(X>3)?

MLE: 1/41/4; method of moments: 2/72/7
MLE: 2/72/7; method of moments: 1/41/4
MLE: 00; method of moments: 2/72/7
MLE: 1/41/4; method of moments: 1/21/2
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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Method Of Moments And Mle

Practice Method Of Moments And Mle 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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This quiz focuses on Method Of Moments And Mle, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

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Question 1

Let X1,…,X5X_1,\ldots,X_5 be independent observations from the uniform distribution on [0,θ][0,\theta]. The observed sample mean is 2.12.1, and the sample maximum is 4.04.0. Using the invariance principle separately with the maximum likelihood estimator and the method-of-moments estimator, what are the resulting estimates of Pθ(X>3)P_\theta(X>3)?

  1. MLE: 1/41/4; method of moments: 2/72/7 (correct answer)
  2. MLE: 2/72/7; method of moments: 1/41/4
  3. MLE: 00; method of moments: 2/72/7
  4. MLE: 1/41/4; method of moments: 1/21/2
Explanation: When you see a problem invoking the invariance principle, your job is to first find the estimator for the parameter θ\theta, then plug it into the target function — here, Pθ(X>3)=θ−3θP_\theta(X > 3) = \frac{\theta - 3}{\theta} for θ>3\theta > 3. For the MLE: on a Uniform[0,θ]\text{Uniform}[0,\theta] distribution, the MLE of θ\theta is the sample maximum, θ^MLE=X(5)=4.0\hat{\theta}_{MLE} = X_{(5)} = 4.0. By invariance, the MLE of P(X>3)P(X>3) is 4.0−34.0=14\frac{4.0 - 3}{4.0} = \frac{1}{4}. For the method of moments: since E[X]=θ/2E[X] = \theta/2, setting θ/2=xˉ=2.1\theta/2 = \bar{x} = 2.1 gives θ^MOM=4.2\hat{\theta}_{MOM} = 4.2. By invariance, the MOM estimate is 4.2−34.2=1.24.2=27\frac{4.2 - 3}{4.2} = \frac{1.2}{4.2} = \frac{2}{7}. So the results are MLE: 1/41/4 and MOM: 2/72/7, confirming answer A. Answer B flips the two estimates — a classic trap if you mix up which estimator uses the sample maximum versus the sample mean. Answer C incorrectly computes the MLE estimate as 0, which would only occur if θ^=3\hat{\theta} = 3 exactly — there's no justification for that here. Answer D correctly identifies the MLE estimate but uses θ^MOM=4.2\hat{\theta}_{MOM} = 4.2 incorrectly, arriving at 1/21/2 through a arithmetic error. Study tip: Always identify the estimator for θ\theta first, then apply the invariance principle. For Uniform[0,θ][0,\theta], remember MLE →\to sample max; MOM →\to twice the sample mean.

Question 2

Suppose Xi=a+YiX_i=a+Y_i, where the YiY_i are independent exponential random variables with rate λ\lambda. For an observed sample, the sample mean is 7.47.4, the sample minimum is 2.02.0, and the sample variance computed with divisor nn is 9.09.0. Which pair gives the method-of-moments estimates followed by the maximum likelihood estimates?

  1. MOM: (a^,λ^)=(4.4,1/3)(\widehat a,\widehat\lambda)=(4.4,1/3); MLE: (2.0,1/5.4)(2.0,1/5.4) (correct answer)
  2. MOM: (a^,λ^)=(2.0,1/3)(\widehat a,\widehat\lambda)=(2.0,1/3); MLE: (4.4,1/5.4)(4.4,1/5.4)
  3. MOM: (a^,λ^)=(4.4,1/9)(\widehat a,\widehat\lambda)=(4.4,1/9); MLE: (2.0,1/7.4)(2.0,1/7.4)
  4. MOM: (a^,λ^)=(7.4,1/3)(\widehat a,\widehat\lambda)=(7.4,1/3); MLE: (2.0,1/3)(2.0,1/3)
Explanation: When a question mixes two estimation methods for a shifted exponential model, your first move should be to recall what each method uses as its raw material: method of moments (MOM) uses population moment equations, while maximum likelihood (MLE) exploits the full likelihood structure. MOM derivation. For Xi=a+YiX_i = a + Y_i with Yi∼Exp(λ)Y_i \sim \text{Exp}(\lambda), the population moments are E[X]=a+1/λE[X] = a + 1/\lambda and Var(X)=1/λ2\text{Var}(X) = 1/\lambda^2. Setting 1/λ2=91/\lambda^2 = 9 gives λ^MOM=1/3\widehat{\lambda}_{MOM} = 1/3, and then a^MOM=Xˉ−1/λ^=7.4−3=4.4\widehat{a}_{MOM} = \bar{X} - 1/\widehat{\lambda} = 7.4 - 3 = 4.4. So MOM yields (4.4, 1/3)(4.4,\ 1/3). MLE derivation. The shifted exponential likelihood is maximized by setting a^MLE=X(1)=2.0\widehat{a}_{MLE} = X_{(1)} = 2.0 (the sample minimum, since aa must not exceed any observation, and the likelihood increases as aa rises). With aa fixed, λ^MLE=1/(Xˉ−a^)=1/(7.4−2.0)=1/5.4\widehat{\lambda}_{MLE} = 1/(\bar{X} - \widehat{a}) = 1/(7.4 - 2.0) = 1/5.4. This gives MLE (2.0, 1/5.4)(2.0,\ 1/5.4), confirming answer A. Answer B swaps the two estimators — assigning the MLE values to MOM and vice versa, a direct reversal of the derivations. Answer C uses the sample variance directly as 1/λ1/\lambda rather than taking its square root, misapplying the variance equation. Answer D incorrectly uses Xˉ\bar{X} as a^\widehat{a} and ignores the MLE's use of the sample minimum entirely. Study tip: Always pair the MLE of aa in a shifted distribution with the sample minimum — it's a recurring pattern on graduate exams. MOM, by contrast, always works through moment equations, never directly using order statistics.

Question 3

Consider a zero-inflated Poisson model in which an observation equals a structural zero with probability π\pi and otherwise follows a Poisson distribution with mean λ\lambda. In a sample, the first empirical factorial moment is n−1∑iXi=1.2n^{-1}\sum_iX_i=1.2 and the second empirical factorial moment is n−1∑iXi(Xi−1)=2.4n^{-1}\sum_iX_i(X_i-1)=2.4. What are the method-of-moments estimates?

  1. λ^=2.0\widehat\lambda=2.0 and π^=0.6\widehat\pi=0.6
  2. λ^=1.2\widehat\lambda=1.2 and π^=0.5\widehat\pi=0.5
  3. λ^=2.0\widehat\lambda=2.0 and π^=0.4\widehat\pi=0.4 (correct answer)
  4. λ^=1.5\widehat\lambda=1.5 and π^=0.2\widehat\pi=0.2
Explanation: When you encounter a zero-inflated Poisson (ZIP) model, your first move should be deriving the theoretical moments in terms of the model parameters, then equating them to their empirical counterparts — that's the method-of-moments recipe. For a ZIP model, an observation is zero with probability π\pi (structural zero) and Poisson(λ)(\lambda) with probability (1−π)(1-\pi). The key theoretical factorial moments are: E[X]=(1−π)λE[X] = (1-\pi)\lambda E[X(X−1)]=(1−π)λ2E[X(X-1)] = (1-\pi)\lambda^2 Setting these equal to the sample values gives you: (1−π)λ=1.2and(1−π)λ2=2.4(1-\pi)\lambda = 1.2 \quad \text{and} \quad (1-\pi)\lambda^2 = 2.4 Dividing the second equation by the first eliminates (1−π)(1-\pi) immediately: λ=2.41.2=2.0\lambda = \frac{2.4}{1.2} = 2.0 Then substituting back: (1−π)(2.0)=1.2⇒1−π=0.6⇒π^=0.4(1-\pi)(2.0) = 1.2 \Rightarrow 1-\pi = 0.6 \Rightarrow \widehat{\pi} = 0.4 This confirms answer C: λ^=2.0\widehat{\lambda}=2.0 and π^=0.4\widehat{\pi}=0.4. Answer A gets λ^\widehat{\lambda} right but flips the roles of π\pi and 1−π1-\pi, reporting 0.60.6 instead of 0.40.4 — a classic sign-confusion error. Answer B plugs the first moment in directly as λ^\widehat{\lambda}, ignoring the inflation adjustment entirely. Answer D appears to split the difference between the two moments arithmetically without any principled derivation. A reliable strategy: always divide consecutive factorial moments to isolate λ\lambda first — this sidesteps the need to solve a system simultaneously and reduces the chance of algebraic error. Remember, factorial moments are your friends in any count-data mixture model.

Question 4

Twenty independent subjects have exponentially distributed event times with rate λ\lambda. Eight events are observed, and their observed event times sum to 1919. The other twelve subjects are administratively right-censored at time 33. What is the MLE of λ\lambda?

  1. λ^=8/55\widehat\lambda=8/55, using the number of events divided by total observed exposure (correct answer)
  2. λ^=8/19\widehat\lambda=8/19, using the number of events divided by summed failure times
  3. λ^=20/55\widehat\lambda=20/55, using the sample size divided by total observed exposure
  4. λ^=12/55\widehat\lambda=12/55, using the number censored divided by total observed exposure
Explanation: When survival data includes censoring, the key is to write down the full likelihood correctly. For exponentially distributed event times with rate λ\lambda, each observed event contributes a density λe−λti\lambda e^{-\lambda t_i}, while each censored subject contributes a survival probability e−λcje^{-\lambda c_j}, where cjc_j is their censoring time. The log-likelihood is therefore: ℓ(λ)=dlog⁡λ−λ∑i=1dti−λ∑j=1mcj\ell(\lambda) = d\log\lambda - \lambda\sum_{i=1}^{d} t_i - \lambda\sum_{j=1}^{m} c_j where dd is the number of observed events. Setting dℓ/dλ=0d\ell/d\lambda = 0 gives λ^=d/(∑ti+∑cj)\widehat{\lambda} = d / \left(\sum t_i + \sum c_j\right), which is the number of events divided by total exposure time — the sum of all observed times, whether they ended in an event or censoring. Here, d=8d = 8, the eight observed failure times sum to 1919, and the twelve censored subjects each contribute 33, giving 12×3=3612 \times 3 = 36 additional person-time. Total exposure is 19+36=5519 + 36 = 55, so λ^=8/55\widehat{\lambda} = 8/55. This confirms answer A is correct. Answer B, 8/198/19, ignores the censored subjects entirely, throwing away the information that twelve people survived past time 33 — this severely overestimates λ\lambda. Answer C uses the total sample size 2020 in the numerator instead of the event count 88, which has no basis in the likelihood derivation. Answer D uses the number censored 1212 in the numerator, which is similarly unjustified. The core study tip: in survival analysis, the MLE denominator is always total person-time at risk (events plus censored exposure), not just failure times or sample size. Memorize the formula λ^=events/total exposure\widehat{\lambda} = \text{events}/\text{total exposure}.

Question 5

Suppose X1,…,XnX_1,\ldots,X_n are independent N(θ,1)N(\theta,1) observations. The parameter of interest is ψ=θ2\psi=\theta^2. For one sample, Xˉ=−1.5\bar X=-1.5 and n−1∑iXi2=3.6n^{-1}\sum_iX_i^2=3.6. What are, respectively, the MLE of ψ\psi and the direct method-of-moments estimate obtained by matching the second raw moment?

  1. ψ^MLE=−2.25\widehat\psi_{\mathrm{MLE}}=-2.25 and ψ^MOM=4.60\widehat\psi_{\mathrm{MOM}}=4.60
  2. ψ^MLE=1.50\widehat\psi_{\mathrm{MLE}}=1.50 and ψ^MOM=3.60\widehat\psi_{\mathrm{MOM}}=3.60
  3. ψ^MLE=2.60\widehat\psi_{\mathrm{MLE}}=2.60 and ψ^MOM=2.25\widehat\psi_{\mathrm{MOM}}=2.25
  4. ψ^MLE=2.25\widehat\psi_{\mathrm{MLE}}=2.25 and ψ^MOM=2.60\widehat\psi_{\mathrm{MOM}}=2.60 (correct answer)
Explanation: When you encounter a question like this, recognize it's testing two distinct estimation philosophies — the invariance principle of MLEs and the moment-matching logic of MOM — applied to the same parameter. For the MLE, recall that if θ^\hat{\theta} is the MLE of θ\theta, then for any function gg, the MLE of g(θ)g(\theta) is g(θ^)g(\hat{\theta}) by the invariance property. The MLE of θ\theta under N(θ,1)N(\theta,1) is simply θ^=Xˉ=−1.5\hat{\theta} = \bar{X} = -1.5. Therefore, ψ^MLE=θ^2=(−1.5)2=2.25\widehat{\psi}_{\text{MLE}} = \hat{\theta}^2 = (-1.5)^2 = 2.25. For the MOM via the second raw moment, you match the population moment to its sample counterpart. Since X∼N(θ,1)X \sim N(\theta, 1), the second raw moment is E[X2]=Var(X)+(E[X])2=1+θ2=1+ψE[X^2] = \text{Var}(X) + (E[X])^2 = 1 + \theta^2 = 1 + \psi. Setting this equal to the sample version: 1+ψ^=n−1∑iXi2=3.61 + \hat{\psi} = n^{-1}\sum_i X_i^2 = 3.6, so ψ^MOM=3.6−1=2.60\widehat{\psi}_{\text{MOM}} = 3.6 - 1 = 2.60. This confirms answer D. A is wrong because −2.25-2.25 treats the MLE as Xˉ2\bar{X}^2 but keeps the negative sign — you cannot square a number and get a negative result. B confuses the MLE of θ\theta (which is ∣Xˉ∣|\bar{X}|) with the MLE of ψ\psi, and uses the raw second-moment sample value without subtracting 1. C swaps the two estimates, getting the MOM right but assigning it to the MLE label and vice versa. A reliable study tip: always distinguish which parameter a moment equation solves for. Here, E[X2]E[X^2] identifies ψ=θ2\psi = \theta^2, not θ\theta directly — subtracting the variance is essential.

Question 6

Let X1,…,XnX_1,\ldots,X_n be independent random variables from a Laplace distribution with density f(x;μ,b)=(2b)−1exp⁡(−∣x−μ∣/b)f(x;\mu,b)=(2b)^{-1}\exp(-|x-\mu|/b), where b>0b>0. The observed sample is 1,2,2,5,101,2,2,5,10. Which choice gives the MLEs of (μ,b)(\mu,b) followed by the method-of-moments estimates based on matching the population mean and variance?

  1. MLE: (μ^,b^)=(4,5.4)(\widehat\mu,\widehat b)=(4,\sqrt{5.4}); MOM: (μ^,b^)=(2,2.4)(\widehat\mu,\widehat b)=(2,2.4)
  2. MLE: (μ^,b^)=(2,2.4)(\widehat\mu,\widehat b)=(2,2.4); MOM: (μ^,b^)=(4,5.4)(\widehat\mu,\widehat b)=(4,\sqrt{5.4}) (correct answer)
  3. MLE: (μ^,b^)=(2,3.0)(\widehat\mu,\widehat b)=(2,3.0); MOM: (μ^,b^)=(4,6.75)(\widehat\mu,\widehat b)=(4,\sqrt{6.75})
  4. MLE: (μ^,b^)=(2,2.4)(\widehat\mu,\widehat b)=(2,\sqrt{2.4}); MOM: (μ^,b^)=(4,5.4)(\widehat\mu,\widehat b)=(4,5.4)
Explanation: When you encounter a question combining MLE and method-of-moments for the Laplace distribution, tackle each estimation method separately and systematically. For MLE, the log-likelihood is ℓ(μ,b)=−nln⁡(2b)−1b∑∣xi−μ∣\ell(\mu,b) = -n\ln(2b) - \frac{1}{b}\sum|x_i - \mu|. Maximizing over μ\mu means minimizing ∑∣xi−μ∣\sum|x_i - \mu|, which is achieved by the sample median. For the data 1,2,2,5,101,2,2,5,10, the median is 22. With μ^=2\hat{\mu}=2, you then maximize over bb: setting the derivative to zero gives b^=1n∑∣xi−μ^∣=∣1−2∣+∣2−2∣+∣2−2∣+∣5−2∣+∣10−2∣5=1+0+0+3+85=125=2.4\hat{b} = \frac{1}{n}\sum|x_i - \hat{\mu}| = \frac{|1-2|+|2-2|+|2-2|+|5-2|+|10-2|}{5} = \frac{1+0+0+3+8}{5} = \frac{12}{5} = 2.4. So the MLE is (μ^,b^)=(2,2.4)(\hat{\mu}, \hat{b}) = (2, 2.4). For MOM, the Laplace distribution has mean μ\mu and variance 2b22b^2. Matching to sample moments: the sample mean is xˉ=1+2+2+5+105=4\bar{x} = \frac{1+2+2+5+10}{5} = 4, giving μ^=4\hat{\mu}=4. The sample variance is s2=∑(xi−xˉ)2n=9+4+4+1+365=545=10.8s^2 = \frac{\sum(x_i-\bar{x})^2}{n} = \frac{9+4+4+1+36}{5} = \frac{54}{5} = 10.8, so 2b^2=10.8⇒b^=5.42\hat{b}^2 = 10.8 \Rightarrow \hat{b} = \sqrt{5.4}. MOM gives (4,5.4)(4, \sqrt{5.4}). This matches choice B. Choice A reverses the two methods — a classic trap. Choice C uses incorrect formulas, producing b^=3.0\hat{b}=3.0 for MLE (wrong divisor) and 6.75\sqrt{6.75} for MOM (likely using n−1n-1 variance). Choice D confuses the variance formula by not taking the square root properly for MLE. Key tip: For Laplace MLE, always remember median → mean absolute deviation. For MOM, remember that the Laplace variance is 2b22b^2, not b2b^2 — mixing these up is the most common error on this type of question.

Question 7

A sample is drawn from a gamma distribution with shape α\alpha and rate β\beta. The observed summaries are Xˉ=2\bar X=2, n−1∑i=1n(Xi−Xˉ)2=2n^{-1}\sum_{i=1}^n(X_i-\bar X)^2=2, and n−1∑i=1nlog⁡Xi=0.3n^{-1}\sum_{i=1}^n\log X_i=0.3. Which statement correctly gives the method-of-moments estimates and the equations characterizing the MLE, where ψ\psi is the digamma function?

  1. MOM: (2,1)(2,1); MLE satisfies β=2α\beta=2\alpha and log⁡α−ψ(α)=0.3−log⁡2\log\alpha-\psi(\alpha)=0.3-\log2
  2. MOM: (1,2)(1,2); MLE satisfies β=α/2\beta=\alpha/2 and ψ(α)−log⁡α=log⁡2−0.3\psi(\alpha)-\log\alpha=\log2-0.3
  3. MOM: (2,1)(2,1); MLE satisfies β=α/2\beta=\alpha/2 and log⁡α−ψ(α)=log⁡2−0.3\log\alpha-\psi(\alpha)=\log2-0.3 (correct answer)
  4. MOM: (2,1)(2,1); MLE satisfies β=2/α\beta=2/\alpha and log⁡α+ψ(α)=log⁡2+0.3\log\alpha+\psi(\alpha)=\log2+0.3
Explanation: When working with the gamma distribution, you need two separate derivations: method of moments (MOM) uses the mean and variance, while MLE requires differentiating the log-likelihood with respect to both parameters. MOM: For a Gamma(α,β)(\alpha, \beta) with rate parameterization, the mean is μ=α/β\mu = \alpha/\beta and variance is σ2=α/β2\sigma^2 = \alpha/\beta^2. Setting Xˉ=α/β=2\bar{X} = \alpha/\beta = 2 and s2=α/β2=2s^2 = \alpha/\beta^2 = 2, dividing gives 1/β=11/\beta = 1, so β^=1\hat{\beta} = 1, and then α^=2\hat{\alpha} = 2. The MOM estimates are (2,1)(2, 1), reported as (α^,β^)(\hat{\alpha}, \hat{\beta}). MLE: The log-likelihood yields two score equations. Differentiating with respect to β\beta gives β^=α/Xˉ=α/2\hat{\beta} = \alpha/\bar{X} = \alpha/2. Differentiating with respect to α\alpha gives log⁡α−ψ(α)=Xˉ−1log⁡X‾⋅Xˉ−log⁡X‾\log\alpha - \psi(\alpha) = \bar{X}^{-1}\overline{\log X} \cdot \bar{X} - \overline{\log X}... more carefully: the equation becomes log⁡α−ψ(α)=log⁡Xˉ−log⁡X‾=log⁡2−0.3\log\alpha - \psi(\alpha) = \log\bar{X} - \overline{\log X} = \log 2 - 0.3. This matches answer C exactly. Answer A reports the correct MOM pair (2,1)(2,1) but flips the MLE relationship to β=2α\beta = 2\alpha and gets the sign wrong in the digamma equation. Answer B swaps the MOM estimates to (1,2)(1, 2), confusing α\alpha and β\beta, and also inverts the rate equation. Answer D keeps the right MOM estimates but writes β=2/α\beta = 2/\alpha (inconsistent with Xˉ=α/β\bar{X} = \alpha/\beta) and incorrectly adds rather than subtracts in the digamma equation. A reliable strategy: always derive the MLE β\beta equation first (it's algebraically clean), then substitute into the α\alpha equation to isolate the digamma condition. Keeping log⁡α−ψ(α)=log⁡Xˉ−log⁡X‾\log\alpha - \psi(\alpha) = \log\bar{X} - \overline{\log X} as a template will prevent sign errors under exam pressure.

Question 8

Let X1,…,XnX_1,\ldots,X_n be independent uniform random variables on [θ,θ+1][\theta,\theta+1]. An observed sample has minimum 2.22.2, maximum 2.92.9, and mean 2.552.55. Which statement correctly describes the MLE and the method-of-moments estimate of θ\theta?

  1. The unique MLE is 1.91.9, and the moment estimate is 2.552.55
  2. The unique MLE is 2.22.2, and the moment estimate is 2.052.05
  3. Every θ∈[2.2,2.9]\theta\in[2.2,2.9] is an MLE, and the moment estimate is 2.402.40
  4. Every θ∈[1.9,2.2]\theta\in[1.9,2.2] is an MLE, and the moment estimate is 2.052.05 (correct answer)
Explanation: When you see a uniform distribution on [θ,θ+1][\theta, \theta+1], your first instinct should be to identify the constraints that θ\theta must satisfy given the observed data, then maximize the likelihood over the feasible region. MLE: The likelihood function is L(θ)=1L(\theta) = 1 if all observations fall in [θ,θ+1][\theta, \theta+1], and 00 otherwise. For every XiX_i to lie in [θ,θ+1][\theta, \theta+1], you need θ≤X(1)\theta \leq X_{(1)} (the minimum) and θ≥X(n)−1\theta \geq X_{(n)} - 1 (the maximum minus 1). With X(1)=2.2X_{(1)} = 2.2 and X(n)=2.9X_{(n)} = 2.9, the feasible set is θ≥2.9−1=1.9\theta \geq 2.9 - 1 = 1.9 and θ≤2.2\theta \leq 2.2. So every θ∈[1.9,2.2]\theta \in [1.9, 2.2] yields L(θ)=1L(\theta) = 1, and the MLE is non-unique — a whole interval. Method of Moments: The population mean of Uniform[θ,θ+1]\text{Uniform}[\theta, \theta+1] is θ+12\theta + \frac{1}{2}. Setting this equal to the sample mean: θ+0.5=2.55\theta + 0.5 = 2.55, giving θ^MOM=2.05\hat{\theta}_{MOM} = 2.05. This is answer D. Why the others fail: A uses 1.91.9 as a unique MLE — the MLE isn't unique here. B claims 2.22.2 is the unique MLE — also wrong, and it only identifies the upper boundary of the feasible region. C misidentifies the feasible interval entirely: [2.2,2.9][2.2, 2.9] violates the constraint derivation (those values of θ\theta would place the maximum outside the support). Study tip: For uniform distribution MLEs, always write out the support constraints explicitly — the likelihood is flat over its feasible region, so the MLE is often an interval, not a unique point.

Question 9

Let X1,…,XnX_1,\ldots,X_n have density f(x;α)=αxα−1f(x;\alpha)=\alpha x^{\alpha-1} for 0<x<10<x<1 and α>0\alpha>0. In an observed sample, Xˉ=0.6\bar X=0.6 and n−1∑ilog⁡Xi=−0.8n^{-1}\sum_i\log X_i=-0.8. Using the MLE and the method-of-moments estimator separately, what are the corresponding plug-in estimates of Pα(X>1/2)P_\alpha(X>1/2)?

  1. MLE: 1−2−1.51-2^{-1.5}; method of moments: 1−2−1.251-2^{-1.25}
  2. MLE: 1−2−1.251-2^{-1.25}; method of moments: 1−2−1.51-2^{-1.5} (correct answer)
  3. MLE: 2−1.252^{-1.25}; method of moments: 2−1.52^{-1.5}
  4. MLE: 1−2−0.81-2^{-0.8}; method of moments: 1−2−0.61-2^{-0.6}
Explanation: When you see a question combining MLE and method of moments with a parametric family, your first job is to derive each estimator separately, then plug into the survival function. For this power distribution, Pα(X>1/2)=1−(1/2)α=1−2−αP_\alpha(X > 1/2) = 1 - (1/2)^\alpha = 1 - 2^{-\alpha}, so everything hinges on estimating α\alpha. MLE: The log-likelihood gives score equation n/α+∑ilog⁡Xi=0n/\alpha + \sum_i \log X_i = 0, so α^MLE=−1/log⁡Xˉ=−1/(−0.8)=1.25\hat\alpha_{\text{MLE}} = -1/\bar{\log X} = -1/(-0.8) = 1.25. Plugging in: 1−2−1.251 - 2^{-1.25}. Method of Moments: The population mean is E[X]=α/(α+1)E[X] = \alpha/(\alpha+1). Setting this equal to Xˉ=0.6\bar{X} = 0.6: α/(α+1)=0.6⇒α=0.6α+0.6⇒0.4α=0.6⇒α^MOM=1.5\alpha/(\alpha+1) = 0.6 \Rightarrow \alpha = 0.6\alpha + 0.6 \Rightarrow 0.4\alpha = 0.6 \Rightarrow \hat\alpha_{\text{MOM}} = 1.5. Plugging in: 1−2−1.51 - 2^{-1.5}. So the MLE gives 1−2−1.251-2^{-1.25} and MOM gives 1−2−1.51-2^{-1.5}, confirming answer B. Answer A simply swaps the two estimators — a classic trap when you solve both correctly but mislabel which is which. Answer C forgets to subtract from 1; it computes P(X<1/2)=(1/2)αP(X < 1/2) = (1/2)^\alpha rather than the survival probability P(X>1/2)P(X > 1/2). Answer D skips the estimation entirely and plugs the raw sample statistics Xˉ=0.6\bar{X}=0.6 and log⁡X‾=−0.8\overline{\log X}=-0.8 directly as if they were α\alpha estimates without solving for α\alpha. Always derive the estimator formula explicitly before substituting numbers — confusing Xˉ\bar{X} with α^\hat\alpha is a frequent careless error under exam pressure.

Question 10

Let X1,…,XnX_1,\ldots,X_n be independent uniform random variables on [0,θ][0,\theta]. Define θ^M=X(n)\widehat\theta_M=X_{(n)} and θ^Q=2Xˉ\widehat\theta_Q=2\bar X. Which joint asymptotic statement is correct?

  1. n(θ−θ^M)/θ⇒Exp⁡(1)n(\theta-\widehat\theta_M)/\theta\Rightarrow\operatorname{Exp}(1) and n(θ^Q−θ)⇒N(0,θ2/3)\sqrt n(\widehat\theta_Q-\theta)\Rightarrow N(0,\theta^2/3) (correct answer)
  2. n(θ^M−θ)⇒N(0,θ2/12)\sqrt n(\widehat\theta_M-\theta)\Rightarrow N(0,\theta^2/12) and n(θ^Q−θ)/θ⇒Exp⁡(1)n(\widehat\theta_Q-\theta)/\theta\Rightarrow\operatorname{Exp}(1)
  3. n(θ^M−θ)/θ⇒Exp⁡(1)n(\widehat\theta_M-\theta)/\theta\Rightarrow\operatorname{Exp}(1) and n(θ^Q−θ)⇒N(0,θ2/12)\sqrt n(\widehat\theta_Q-\theta)\Rightarrow N(0,\theta^2/12)
  4. n(θ−θ^M)⇒N(0,θ2/3)\sqrt n(\theta-\widehat\theta_M)\Rightarrow N(0,\theta^2/3) and n(θ^Q−θ)⇒N(0,θ2/3)\sqrt n(\widehat\theta_Q-\theta)\Rightarrow N(0,\theta^2/3)
Explanation: When comparing two estimators with different convergence rates, your first instinct should be to ask: how fast does each estimator concentrate around the truth? Order statistics and sample means behave very differently asymptotically, and this question tests whether you know both. For θ^M=X(n)\widehat{\theta}_M = X_{(n)}, the maximum of a uniform sample, the key fact is that the bias shrinks at rate 1/n1/n, not 1/n1/\sqrt{n}. Specifically, E[X(n)]=nn+1θ\mathbb{E}[X_{(n)}] = \frac{n}{n+1}\theta, so X(n)X_{(n)} underestimates θ\theta by order 1/n1/n. The exact CDF of X(n)X_{(n)} is (x/θ)n(x/\theta)^n, which gives P(n(θ−X(n))/θ>t)=(1−t/n)n→e−tP(n(\theta - X_{(n)})/\theta > t) = (1 - t/n)^n \to e^{-t}. So n(θ−θ^M)/θ⇒Exp(1)n(\theta - \widehat{\theta}_M)/\theta \Rightarrow \text{Exp}(1). Note the centering is θ−X(n)\theta - X_{(n)} (positive), since the maximum always undershoots. For θ^Q=2Xˉ\widehat{\theta}_Q = 2\bar{X}, since Var(Xi)=θ2/12\text{Var}(X_i) = \theta^2/12, the variance of 2Xˉ2\bar{X} is 4⋅θ2/(12n)=θ2/(3n)4 \cdot \theta^2/(12n) = \theta^2/(3n). By the CLT, n(θ^Q−θ)⇒N(0,θ2/3)\sqrt{n}(\widehat{\theta}_Q - \theta) \Rightarrow N(0, \theta^2/3). This confirms answer A. B is wrong because it swaps the two convergence rates — the maximum converges at rate nn, not n\sqrt{n}. C uses the correct rate nn for the maximum and the correct variance θ2/3\theta^2/3 for θ^Q\widehat{\theta}_Q, but writes θ2/12\theta^2/12 instead — that's the variance of a single observation, not of 2Xˉ2\bar{X}. D incorrectly applies CLT-rate convergence to the maximum, which is the most common trap here. Remember: whenever you see an order statistic estimating a boundary parameter, the convergence rate is nn, not n\sqrt{n} — this is a signature of non-regular estimation problems.