Statistics Graduate Level Quiz: Likelihood Manipulation
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Likelihood ManipulationQuestion 1 of 10

A random sample is drawn from a uniform distribution on [θ,θ+1][\theta,\theta+1]. The observed sample minimum is 0.420.42 and the observed sample maximum is 1.111.11. Which value is the largest maximum likelihood estimator of θ\theta?

0.110.11
0.420.42
0.7650.765
1.111.11
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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Likelihood Manipulation

Practice Likelihood Manipulation 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 Likelihood Manipulation, 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

A random sample is drawn from a uniform distribution on [θ,θ+1][\theta,\theta+1]. The observed sample minimum is 0.420.42 and the observed sample maximum is 1.111.11. Which value is the largest maximum likelihood estimator of θ\theta?

  1. 0.110.11
  2. 0.420.42 (correct answer)
  3. 0.7650.765
  4. 1.111.11
Explanation: When working with MLE for a uniform distribution on [θ,θ+1][\theta, \theta+1], your first instinct should be to write down the likelihood function explicitly. Since the density is f(x∣θ)=1f(x|\theta) = 1 for x∈[θ,θ+1]x \in [\theta, \theta+1] and 0 otherwise, the likelihood for a sample of size nn is simply L(θ)=1L(\theta) = 1 whenever all observations fall within [θ,θ+1][\theta, \theta+1], and 0 otherwise. The MLE is any θ\theta that keeps all data inside the interval — so you need to find the range of valid θ\theta values, then identify the largest one. For every observation xix_i to satisfy θ≤xi≤θ+1\theta \leq x_i \leq \theta + 1, you need two constraints simultaneously: θ≤x(1)\theta \leq x_{(1)} (the minimum) and θ≥x(n)−1\theta \geq x_{(n)} - 1 (the maximum). With x(1)=0.42x_{(1)} = 0.42 and x(n)=1.11x_{(n)} = 1.11, the valid range is 1.11−1≤θ≤0.421.11 - 1 \leq \theta \leq 0.42, i.e., 0.11≤θ≤0.420.11 \leq \theta \leq 0.42. The likelihood equals 1 throughout this entire interval, so every value in [0.11,0.42][0.11, 0.42] is an MLE. The question asks for the largest MLE, which is 0.42\mathbf{0.42} — answer B. Answer A (0.11) is the smallest MLE, not the largest — it's the lower bound of the valid interval. Answer C (0.765) is simply the midpoint of the two observations, which corresponds to the method-of-moments estimator, not the MLE. Answer D (1.11) violates the upper constraint: if θ=1.11\theta = 1.11, then x(1)=0.42<1.11=θx_{(1)} = 0.42 < 1.11 = \theta, placing data outside the support. The key study tip: for uniform distribution MLEs, always write the constraint region explicitly. The likelihood is flat (not peaked), so the MLE is a set, and exam questions will test whether you can identify specific boundary values within it.

Question 2

Independent subjects have exponential event times with hazard λ\lambda. For subject ii, one observes ti=min⁡(Ti,Ci)t_i=\min(T_i,C_i) and δi=I(Ti≤Ci)\delta_i=I(T_i\le C_i), where censoring is independent and its distribution does not involve λ\lambda. Let D=∑i=1nδiD=\sum_{i=1}^n\delta_i and R=∑i=1ntiR=\sum_{i=1}^n t_i. Assuming D>0D>0, which estimator maximizes the likelihood based on the observed data?

  1. λ^=DR\widehat{\lambda}=\frac{D}{R} (correct answer)
  2. λ^=nR\widehat{\lambda}=\frac{n}{R}
  3. λ^=D∑i=1nδiti\widehat{\lambda}=\frac{D}{\sum_{i=1}^n\delta_i t_i}
  4. λ^=RD\widehat{\lambda}=\frac{R}{D}
Explanation: When you encounter survival analysis questions involving exponential event times and censoring, your first instinct should be to write down the likelihood function carefully — the MLE falls out naturally from there. For exponential survival times with hazard λ\lambda, each subject contributes λe−λti\lambda e^{-\lambda t_i} if their event is observed (δi=1\delta_i = 1) and e−λtie^{-\lambda t_i} if censored (δi=0\delta_i = 0). The full likelihood is therefore: L(λ)=∏i=1nλδie−λti=λDe−λRL(\lambda) = \prod_{i=1}^n \lambda^{\delta_i} e^{-\lambda t_i} = \lambda^D e^{-\lambda R} Taking the log: ℓ(λ)=Dlog⁡λ−λR\ell(\lambda) = D\log\lambda - \lambda R. Differentiating and setting equal to zero gives dℓdλ=Dλ−R=0\frac{d\ell}{d\lambda} = \frac{D}{\lambda} - R = 0, which immediately yields λ^=D/R\widehat{\lambda} = D/R. This is answer A — the number of observed events divided by the total accumulated time (observed plus censored), which is the empirical event rate per unit of exposure. Answer B uses nn instead of DD, treating all subjects as having experienced the event regardless of censoring — this ignores the censoring indicator entirely and will overestimate λ\lambda. Answer C uses only the event times of uncensored subjects in the denominator (∑δiti\sum \delta_i t_i), discarding the valuable exposure time contributed by censored individuals, which is a fundamental misuse of the survival likelihood. Answer D inverts the correct formula, giving an estimate of mean survival time rather than the hazard rate. A useful memory anchor: in survival analysis, censored observations still contribute information through their risk exposure. The MLE always reflects total person-time at risk (RR) in the denominator, not just event times.

Question 3

Suppose X1,…,X6X_1,\ldots,X_6 are independent normal observations with unknown variance σ2\sigma^2 and mean constrained by μ≥0\mu\ge 0. The data satisfy Xˉ=−1\bar X=-1 and ∑i=16(Xi−Xˉ)2=18\sum_{i=1}^6(X_i-\bar X)^2=18. What is the maximum likelihood estimate of σ2\sigma^2 under the constraint?

  1. σ^2=3\widehat{\sigma}^2=3
  2. σ^2=185\widehat{\sigma}^2=\frac{18}{5}
  3. σ^2=4\widehat{\sigma}^2=4 (correct answer)
  4. σ^2=245\widehat{\sigma}^2=\frac{24}{5}
Explanation: When you encounter constrained MLE problems, your first instinct should be to find the unconstrained MLE, then check whether it satisfies the constraint. If it doesn't, the constrained MLE lives on the boundary of the parameter space. Here, the unconstrained MLEs are μ^=Xˉ=−1\hat{\mu} = \bar{X} = -1 and σ^2=1n∑(Xi−Xˉ)2=186=3\hat{\sigma}^2 = \frac{1}{n}\sum(X_i - \bar{X})^2 = \frac{18}{6} = 3. But μ^=−1\hat{\mu} = -1 violates μ≥0\mu \geq 0, so we must impose the boundary condition μ=0\mu = 0. With μ\mu fixed at 0, the log-likelihood is maximized over σ2\sigma^2 alone, giving the constrained MLE: σ^2=1n∑i=16Xi2=16[∑(Xi−Xˉ)2+nXˉ2]=18+6(1)6=246=4\widehat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^6 X_i^2 = \frac{1}{6}\left[\sum(X_i - \bar{X})^2 + n\bar{X}^2\right] = \frac{18 + 6(1)}{6} = \frac{24}{6} = 4 This confirms the correct answer is C. Choice A (σ^2=3\widehat{\sigma}^2 = 3) is the unconstrained MLE of σ2\sigma^2, which pairs with an infeasible μ^=−1\hat{\mu} = -1. You cannot use this estimate because the joint MLE violates the constraint. Choice B (18/518/5) suggests dividing by n−1=5n-1 = 5, which is the unbiased estimator of σ2\sigma^2, not the MLE — a classic confusion. Choice D (24/524/5) similarly divides the constrained sum of squares by 5 instead of 6, mixing unbiased estimation logic with the constrained setup. The key study tip: when a constraint forces you to the boundary, recompute the MLE of all remaining free parameters from scratch using that boundary value — don't just patch the unconstrained answer.

Question 4

Event times have a gamma distribution with known shape 22 and unknown rate θ\theta, with density f(x;θ)=θ2xe−θxf(x;\theta)=\theta^2xe^{-\theta x} for x>0x>0. Only event times exceeding a fixed threshold c>0c>0 are observed, and the sample consists of nn independent times xi>cx_i>c. Which likelihood equation must an interior maximum likelihood estimate satisfy?

  1. 2nθ−∑i=1n(xi−c)−nc1+θc=0\frac{2n}{\theta}-\sum_{i=1}^n(x_i-c)-\frac{nc}{1+\theta c}=0 (correct answer)
  2. 2nθ−∑i=1nxi−nc1+θc=0\frac{2n}{\theta}-\sum_{i=1}^n x_i-\frac{nc}{1+\theta c}=0
  3. 2nθ−∑i=1n(xi−c)+nc1+θc=0\frac{2n}{\theta}-\sum_{i=1}^n(x_i-c)+\frac{nc}{1+\theta c}=0
  4. nθ−∑i=1n(xi−c)−nc1+θc=0\frac{n}{\theta}-\sum_{i=1}^n(x_i-c)-\frac{nc}{1+\theta c}=0
Explanation: When dealing with truncated data, the key insight is that you're not working with the ordinary likelihood — you must condition on the event that each observation exceeds the threshold cc. This means the conditional density for each observed xi>cx_i > c is f(x;θ)/P(X>c;θ)f(x;\theta)/P(X > c;\theta), making the log-likelihood ℓ(θ)=∑log⁡f(xi;θ)−nlog⁡P(X>c;θ)\ell(\theta) = \sum \log f(x_i;\theta) - n\log P(X>c;\theta). For the Gamma(2, θ\theta) density f(x;θ)=θ2xe−θxf(x;\theta) = \theta^2 x e^{-\theta x}, the log contribution per observation is 2log⁡θ+log⁡xi−θxi2\log\theta + \log x_i - \theta x_i. The survival function is P(X>c;θ)=e−θc(1+θc)P(X>c;\theta) = e^{-\theta c}(1+\theta c), so log⁡P(X>c;θ)=−θc+log⁡(1+θc)\log P(X>c;\theta) = -\theta c + \log(1+\theta c). Differentiating the full log-likelihood with respect to θ\theta: dℓdθ=2nθ−∑xi−n(−c+c1+θc)=2nθ−∑xi+nc−nc1+θc\frac{d\ell}{d\theta} = \frac{2n}{\theta} - \sum x_i - n\left(-c + \frac{c}{1+\theta c}\right) = \frac{2n}{\theta} - \sum x_i + nc - \frac{nc}{1+\theta c} Rewriting −∑xi+nc=−∑(xi−c)-\sum x_i + nc = -\sum(x_i - c) confirms the likelihood equation is 2nθ−∑(xi−c)−nc1+θc=0\frac{2n}{\theta} - \sum(x_i - c) - \frac{nc}{1+\theta c} = 0, which is answer A. Answer B incorrectly uses ∑xi\sum x_i instead of ∑(xi−c)\sum(x_i - c), forgetting to incorporate the +nc+nc term from differentiating −n(−θc)-n(-\theta c). Answer C has the wrong sign on the last term, which would arise from differentiating incorrectly. Answer D uses n/θn/\theta instead of 2n/θ2n/\theta, reflecting a shape parameter of 1 rather than 2. Your strategy: always write out log⁡P(X>c;θ)\log P(X>c;\theta) explicitly and differentiate it carefully — sign errors and forgotten terms from the survival function account for most mistakes on truncated-likelihood problems.

Question 5

Independent responses satisfy Yi∼N(μ,σ2/wi)Y_i\sim N(\mu,\sigma^2/w_i), where all weights wi>0w_i>0 are known and both μ\mu and σ2\sigma^2 are unknown. Which pair gives their joint maximum likelihood estimators?

  1. μ^=∑iwiYi∑iwi,σ^2=1n∑iwi(Yi−μ^)2\widehat\mu=\frac{\sum_iw_iY_i}{\sum_iw_i},\qquad \widehat\sigma^2=\frac{1}{n}\sum_iw_i(Y_i-\widehat\mu)^2 (correct answer)
  2. μ^=1n∑iYi,σ^2=1n∑iwi(Yi−μ^)2\widehat\mu=\frac{1}{n}\sum_iY_i,\qquad \widehat\sigma^2=\frac{1}{n}\sum_iw_i(Y_i-\widehat\mu)^2
  3. μ^=∑iwiYi∑iwi,σ^2=1n−1∑iwi(Yi−μ^)2\widehat\mu=\frac{\sum_iw_iY_i}{\sum_iw_i},\qquad \widehat\sigma^2=\frac{1}{n-1}\sum_iw_i(Y_i-\widehat\mu)^2
  4. μ^=∑iYi/wi∑i1/wi,σ^2=1n∑i(Yi−μ^)2wi\widehat\mu=\frac{\sum_iY_i/w_i}{\sum_i1/w_i},\qquad \widehat\sigma^2=\frac{1}{n}\sum_i\frac{(Y_i-\widehat\mu)^2}{w_i}
Explanation: When you see a weighted normal model like this, your instinct should be to derive the MLEs directly from the log-likelihood rather than memorizing formulas — the structure of the variance tells you everything. Since Yi∼N(μ,σ2/wi)Y_i \sim N(\mu, \sigma^2/w_i), the log-likelihood is: ℓ(μ,σ2)=−n2log⁡σ2−12σ2∑iwi(Yi−μ)2+const\ell(\mu,\sigma^2) = -\frac{n}{2}\log\sigma^2 - \frac{1}{2\sigma^2}\sum_i w_i(Y_i - \mu)^2 + \text{const} Notice that wiw_i appears because the precision of each observation is wi/σ2w_i/\sigma^2. Setting ∂ℓ/∂μ=0\partial\ell/\partial\mu = 0 gives ∑iwi(Yi−μ)=0\sum_i w_i(Y_i - \mu) = 0, which yields the weighted mean μ^=∑iwiYi∑iwi\widehat\mu = \frac{\sum_i w_i Y_i}{\sum_i w_i}. Then setting ∂ℓ/∂σ2=0\partial\ell/\partial\sigma^2 = 0 gives σ^2=1n∑iwi(Yi−μ^)2\widehat\sigma^2 = \frac{1}{n}\sum_i w_i(Y_i - \widehat\mu)^2. This is exactly answer A, confirming it as correct. B uses the unweighted sample mean for μ^\widehat\mu, ignoring that observations with larger wiw_i are more precise and deserve more influence — a fundamental misread of the model. C uses the correct μ^\widehat\mu but divides by n−1n-1 instead of nn. That divisor appears in unbiased variance estimation (REML-style), not MLE — MLEs are known to be biased for σ2\sigma^2. D inverts the weighting scheme entirely, treating higher-weight observations as less informative, which contradicts the model where Var(Yi)=σ2/wi\text{Var}(Y_i) = \sigma^2/w_i. Study tip: Always write out the log-likelihood explicitly. The weight wiw_i divides the variance, so it multiplies the squared residual in the exponent — that asymmetry is what separates A from D every time.

Question 6

A sample comes from a Pareto distribution with density f(x;α,θ)=αθαx−(α+1)f(x;\alpha,\theta)=\alpha\theta^\alpha x^{-(\alpha+1)} for x≥θx\ge\theta, where both α>0\alpha>0 and θ>0\theta>0 are unknown. Which pair is the joint maximum likelihood estimator?

  1. θ^=X(1),α^=n∑ilog⁡Xi\widehat\theta=X_{(1)},\qquad \widehat\alpha=\frac{n}{\sum_i\log X_i}
  2. θ^=X(n),α^=n∑ilog⁡(X(n)/Xi)\widehat\theta=X_{(n)},\qquad \widehat\alpha=\frac{n}{\sum_i\log(X_{(n)}/X_i)}
  3. θ^=X(1),α^=n∑ilog⁡(Xi/X(1))\widehat\theta=X_{(1)},\qquad \widehat\alpha=\frac{n}{\sum_i\log(X_i/X_{(1)})} (correct answer)
  4. θ^=exp⁡ ⁣(1n∑ilog⁡Xi),α^=n∑ilog⁡(Xi/θ^)\widehat\theta=\exp\!\left(\frac{1}{n}\sum_i\log X_i\right),\qquad \widehat\alpha=\frac{n}{\sum_i\log(X_i/\widehat\theta)}
Explanation: When both parameters of a distribution are unknown and appear in the support, you must be especially careful: the likelihood is not always differentiable with respect to every parameter, so blind calculus can mislead you. For this Pareto distribution, the log-likelihood is ℓ(α,θ)=nlog⁡α+nαlog⁡θ−(α+1)∑ilog⁡xi,\ell(\alpha,\theta) = n\log\alpha + n\alpha\log\theta - (\alpha+1)\sum_i\log x_i, but this is only valid when θ≤x(1)\theta \le x_{(1)} (the sample minimum), since every observation must satisfy xi≥θx_i \ge \theta. For fixed α>0\alpha > 0, the term nαlog⁡θn\alpha\log\theta is strictly increasing in θ\theta, so you push θ\theta as high as possible without violating the constraint — giving θ^=X(1)\widehat{\theta} = X_{(1)}. With θ\theta fixed at X(1)X_{(1)}, differentiate with respect to α\alpha and set equal to zero: ∂ℓ∂α=nα+nlog⁡X(1)−∑ilog⁡Xi=0,\frac{\partial\ell}{\partial\alpha} = \frac{n}{\alpha} + n\log X_{(1)} - \sum_i\log X_i = 0, which yields α^=n∑ilog⁡(Xi/X(1)).\widehat{\alpha} = \frac{n}{\sum_i\log(X_i/X_{(1)})}. This is Choice C, the correct answer. Choice A uses ∑ilog⁡Xi\sum_i \log X_i in the denominator without subtracting nlog⁡X(1)n\log X_{(1)}, which is what you'd get if you incorrectly treated θ\theta as known or equal to 1. Choice B sets θ^=X(n)\widehat{\theta} = X_{(n)}, the sample maximum — this violates the support constraint since X(n)≥xiX_{(n)} \ge x_i for all ii is impossible unless all observations are equal. Choice D uses a geometric-mean formula for θ^\widehat{\theta}, which has no MLE justification here and also fails to respect the boundary constraint. The key study tip: whenever a parameter appears in the boundary of the support, recognize that the MLE for that parameter is found by a boundary argument, not calculus. Then condition on that estimate before differentiating for the remaining parameters.

Question 7

Independent counts originally follow a Poisson distribution with mean λ\lambda, but observations equal to zero are not recorded. For the resulting zero-truncated sample, let Xˉ\bar X denote the sample mean. Which equation characterizes an interior maximum likelihood estimate of λ\lambda?

  1. Xˉ=λ\bar X=\lambda
  2. Xˉ=λ1−e−λ\bar X=\frac{\lambda}{1-e^{-\lambda}} (correct answer)
  3. Xˉ=λ(1−e−λ)\bar X=\lambda(1-e^{-\lambda})
  4. Xˉ=λeλ−1\bar X=\frac{\lambda}{e^{\lambda}-1}
Explanation: When sampling from a truncated distribution, you must re-normalize the likelihood to reflect only the observable outcomes. Here, zeros are excluded, so the zero-truncated Poisson assigns probability P(X=k∣X>0)=e−λλk/k!1−e−λP(X=k \mid X>0) = \frac{e^{-\lambda}\lambda^k/k!}{1-e^{-\lambda}} for k=1,2,3,…k=1,2,3,\ldots The log-likelihood for a sample of size nn is ℓ(λ)=nXˉln⁡λ−nλ−nln⁡(1−e−λ)+const\ell(\lambda) = n\bar{X}\ln\lambda - n\lambda - n\ln(1-e^{-\lambda}) + \text{const}. Setting the derivative with respect to λ\lambda equal to zero gives Xˉλ−1−e−λ1−e−λ=0\frac{\bar{X}}{\lambda} - 1 - \frac{e^{-\lambda}}{1-e^{-\lambda}} = 0. Rearranging, Xˉ=λ(1+e−λ1−e−λ)=λ1−e−λ\bar{X} = \lambda\left(1 + \frac{e^{-\lambda}}{1-e^{-\lambda}}\right) = \frac{\lambda}{1-e^{-\lambda}}, confirming answer B. This equation also has a beautiful intuitive interpretation: λ1−e−λ\frac{\lambda}{1-e^{-\lambda}} is exactly the mean of the zero-truncated Poisson distribution, so the MLE simply equates the sample mean to the truncated population mean — a standard MLE moment-matching result. A sets Xˉ=λ\bar{X}=\lambda, which would be correct for an untruncated Poisson, ignoring the re-normalization entirely. C inverts the relationship incorrectly, placing the denominator 1−e−λ1-e^{-\lambda} in the numerator instead. D rewrites λeλ−1\frac{\lambda}{e^\lambda - 1}, which equals λe−λ1−e−λ\frac{\lambda e^{-\lambda}}{1-e^{-\lambda}} — a structurally plausible but incorrect expression that confuses the truncation adjustment. As a study strategy: whenever you encounter truncated or censored distributions on exams, always re-derive the likelihood from scratch using the conditional probability. The MLE scoring equation will differ from the standard case precisely because of the normalizing denominator.

Question 8

A multinomial model has three cell probabilities p1=θp_1=\theta, p2=θ(1−θ)p_2=\theta(1-\theta), and p3=(1−θ)2p_3=(1-\theta)^2, where 0<θ<10<\theta<1. The observed cell counts are n1=12n_1=12, n2=8n_2=8, and n3=10n_3=10. What is the maximum likelihood estimate of θ\theta?

  1. θ^=n1+2n22(n1+n2+n3)=715\widehat\theta=\frac{n_1+2n_2}{2(n_1+n_2+n_3)}=\frac{7}{15}
  2. θ^=n1+n22(n1+n2+n3)=13\widehat\theta=\frac{n_1+n_2}{2(n_1+n_2+n_3)}=\frac{1}{3}
  3. θ^=n1n1+n2+n3=25\widehat\theta=\frac{n_1}{n_1+n_2+n_3}=\frac{2}{5}
  4. θ^=n1+n2n1+2n2+2n3=512\widehat\theta=\frac{n_1+n_2}{n_1+2n_2+2n_3}=\frac{5}{12} (correct answer)
Explanation: When working with multinomial MLE problems, your first instinct should be to write down the log-likelihood and differentiate. The multinomial log-likelihood (ignoring constants) is: ℓ(θ)=n1ln⁡p1+n2ln⁡p2+n3ln⁡p3\ell(\theta) = n_1\ln p_1 + n_2\ln p_2 + n_3\ln p_3 Substituting the cell probabilities gives: ℓ(θ)=n1ln⁡θ+n2[ln⁡θ+ln⁡(1−θ)]+n3[2ln⁡(1−θ)]\ell(\theta) = n_1\ln\theta + n_2[\ln\theta + \ln(1-\theta)] + n_3[2\ln(1-\theta)] =(n1+n2)ln⁡θ+(n2+2n3)ln⁡(1−θ)= (n_1+n_2)\ln\theta + (n_2+2n_3)\ln(1-\theta) Differentiating with respect to θ\theta and setting equal to zero: dℓdθ=n1+n2θ−n2+2n31−θ=0\frac{d\ell}{d\theta} = \frac{n_1+n_2}{\theta} - \frac{n_2+2n_3}{1-\theta} = 0 Cross-multiplying and solving: (n1+n2)(1−θ)=(n2+2n3)θ(n_1+n_2)(1-\theta) = (n_2+2n_3)\theta, which gives: θ^=n1+n2n1+2n2+2n3=2048=512\widehat\theta = \frac{n_1+n_2}{n_1+2n_2+2n_3} = \frac{20}{48} = \frac{5}{12} This confirms D is correct. Choice A doubles n2n_2 in the numerator, which would arise if you incorrectly counted each appearance of θ\theta in p2p_2 twice rather than tracking contributions to the θ\theta and (1−θ)(1-\theta) terms separately. Choice B forgets that p2p_2 also contributes a factor of (1−θ)(1-\theta), losing the n2n_2 term in the denominator's coefficient. Choice C naively uses only n1n_1, as if θ\theta appeared only in p1p_1, ignoring that p2p_2 also contains θ\theta. Your study tip: always expand each ln⁡pi\ln p_i fully before differentiating. Collecting coefficients on ln⁡θ\ln\theta versus ln⁡(1−θ)\ln(1-\theta) separately prevents the algebra errors that generate the wrong distractors here.

Question 9

Suppose independent counts satisfy Yi∼Poisson⁡(bi+θei)Y_i\sim\operatorname{Poisson}(b_i+\theta e_i), where bi>0b_i>0 and ei>0e_i>0 are known and θ≥0\theta\ge 0 is an unknown signal strength. Which statement correctly characterizes the maximum likelihood estimator?

  1. It is zero when ∑iei(Yi/bi−1)≤0\sum_i e_i(Y_i/b_i-1)\le 0; otherwise it is the positive root of ∑iei{Yi/(bi+θei)−1}=0\sum_i e_i\{Y_i/(b_i+\theta e_i)-1\}=0. (correct answer)
  2. It is zero when ∑iei(Yi/bi−1)≥0\sum_i e_i(Y_i/b_i-1)\ge 0; otherwise it is the positive root of ∑iei{Yi/(bi+θei)−1}=0\sum_i e_i\{Y_i/(b_i+\theta e_i)-1\}=0.
  3. It always equals max⁡{0,(∑iYi−∑ibi)/∑iei}\max\{0,(\sum_iY_i-\sum_i b_i)/\sum_i e_i\} because total observed and fitted counts must agree.
  4. It is zero when ∑i(Yi−bi)≤0\sum_i(Y_i-b_i)\le 0; otherwise it is the positive root of ∑i{Yi/(bi+θei)−1}=0\sum_i\{Y_i/(b_i+\theta e_i)-1\}=0.
Explanation: When maximizing a likelihood over a constrained parameter space like θ≥0\theta \ge 0, you must check whether the unconstrained optimum respects the boundary. Here, the log-likelihood is ℓ(θ)=∑i[Yilog⁡(bi+θei)−(bi+θei)]\ell(\theta) = \sum_i [Y_i \log(b_i + \theta e_i) - (b_i + \theta e_i)], and its derivative is ℓ′(θ)=∑iei{Yi/(bi+θei)−1}\ell'(\theta) = \sum_i e_i\{Y_i/(b_i + \theta e_i) - 1\}. Because ℓ\ell is concave in θ\theta, the MLE is θ^=0\hat\theta = 0 when ℓ′(0)≤0\ell'(0) \le 0, and otherwise it's the unique positive root of ℓ′(θ)=0\ell'(\theta) = 0. Evaluating at θ=0\theta = 0 gives ℓ′(0)=∑iei(Yi/bi−1)\ell'(0) = \sum_i e_i(Y_i/b_i - 1). So the MLE is zero precisely when this quantity is non-positive — which is exactly what A states, making it correct. B flips the inequality sign, claiming the MLE is zero when ℓ′(0)≥0\ell'(0) \ge 0. This is backwards: a non-negative derivative at zero means the likelihood is still increasing, so the optimum lies in the interior, not at the boundary. C proposes a simple closed-form formula analogous to a homogeneous Poisson model. This would only be valid if all eie_i were equal and bib_i were a common constant — in general, the score equation doesn't collapse to a single balance of totals. D uses an unweighted sum ∑i(Yi−bi)\sum_i(Y_i - b_i) for the boundary condition, omitting the essential eie_i weights. It also drops the eie_i weights inside the score equation itself, making neither condition match the true derivative. Study tip: For constrained MLEs, always evaluate the score function at the boundary — the sign tells you whether to report the boundary value or solve the interior equation.

Question 10

Observations are independent from a two-component mixture with density g(x;θ)=θf1(x)+(1−θ)f0(x)g(x;\theta)=\theta f_1(x)+(1-\theta)f_0(x), where f0f_0 and f1f_1 are known and 0<θ<10<\theta<1. Define ri(θ)=θf1(xi)/g(xi;θ)r_i(\theta)=\theta f_1(x_i)/g(x_i;\theta). Which equation must any interior maximum likelihood estimate satisfy?

  1. θ1−θ={∏i=1nf1(xi)f0(xi)}1/n\frac{\theta}{1-\theta}=\left\{\prod_{i=1}^n\frac{f_1(x_i)}{f_0(x_i)}\right\}^{1/n}
  2. θ=1n∑i=1nf1(xi)f1(xi)+f0(xi)\theta=\frac{1}{n}\sum_{i=1}^n\frac{f_1(x_i)}{f_1(x_i)+f_0(x_i)}
  3. θ=∑i=1nri(θ)f1(xi)∑i=1nf1(xi)\theta=\frac{\sum_{i=1}^n r_i(\theta)f_1(x_i)}{\sum_{i=1}^n f_1(x_i)}
  4. θ=1n∑i=1nri(θ)\theta=\frac{1}{n}\sum_{i=1}^n r_i(\theta) (correct answer)
Explanation: Mixture model MLE questions hinge on one core move: differentiating the log-likelihood and setting the score to zero. The log-likelihood is ℓ(θ)=∑i=1nlog⁡[θf1(xi)+(1−θ)f0(xi)]\ell(\theta) = \sum_{i=1}^n \log[\theta f_1(x_i) + (1-\theta)f_0(x_i)]. Differentiating with respect to θ\theta gives dℓdθ=∑i=1nf1(xi)−f0(xi)g(xi;θ)\frac{d\ell}{d\theta} = \sum_{i=1}^n \frac{f_1(x_i) - f_0(x_i)}{g(x_i;\theta)}. Setting this equal to zero and rearranging: ∑i=1nf1(xi)g(xi;θ)=∑i=1nf0(xi)g(xi;θ)\sum_{i=1}^n \frac{f_1(x_i)}{g(x_i;\theta)} = \sum_{i=1}^n \frac{f_0(x_i)}{g(x_i;\theta)}. Now notice that ri(θ)=θf1(xi)/g(xi;θ)r_i(\theta) = \theta f_1(x_i)/g(x_i;\theta), so f1(xi)/g(xi;θ)=ri(θ)/θf_1(x_i)/g(x_i;\theta) = r_i(\theta)/\theta. Substituting and multiplying both sides by θ/n\theta/n yields θ=1n∑i=1nri(θ)\theta = \frac{1}{n}\sum_{i=1}^n r_i(\theta), confirming D is the correct fixed-point equation for any interior MLE. Choice A resembles a geometric-mean likelihood-ratio condition — it has no derivation from the score equation and conflates the MLE with a simple ratio test statistic. Choice B is tempting because it looks like a posterior mean, but it replaces g(xi;θ)g(x_i;\theta) in the denominator with f1(xi)+f0(xi)f_1(x_i)+f_0(x_i), which ignores θ\theta entirely and destroys the self-consistency structure. Choice C introduces an extra f1(xi)f_1(x_i) factor in both numerator and denominator with no basis in the score equation — it's a plausible-looking distortion of D. A useful mental anchor: the MLE fixed-point equation for a mixture always has the form "parameter equals average posterior weight," since ri(θ)r_i(\theta) is exactly the posterior probability that observation ii came from component f1f_1. Recognizing this EM-style self-consistency structure will help you immediately identify the correct score equation on mixture problems.