Week 15 — Synthesis and the bridge to Mathematical Statistics II

Where this week starts

Fourteen weeks ago you were handed a family of distributions and asked what it was for. Since then you have built order statistics, normal theory, limit theorems, likelihoods, posteriors, sufficiency, unbiased optimality, information bounds, intervals, and an account of misspecification. Written out like that it reads as fourteen techniques in a row, and stored like that it will not survive the summer: a list has no structure to recover when one item goes missing.

It is not a list. It is one argument. A scientific question fixes an estimand; a model turns that estimand into a function of a parameter; the likelihood ranks candidate parameter values against the data; sufficiency says which part of the data carries that ranking; information measures how sharply it discriminates; risk prices the consequences of landing on the wrong value. Every interval you have built, every Bayes estimator, every optimality theorem is a statement about one of those objects or about the link between two of them.

So this page is one estimation problem carried the whole distance: twenty repair jobs, a gamma model, one rate, taken through sufficiency, likelihood, information, Rao-Blackwell, a prior, a risk comparison and four intervals, and then asked what survives when the model is wrong. Each step carries the week that supplied it; nothing new appears except the connective tissue.

By Thursday you should be able to take an unfamiliar estimation problem through every layer without being told which comes next, to state for any two objects above the theorem joining them, and to name for each major topic of Mathematical Statistics II the result from this course it continues.

Why this matters downstream

The concrete stake is diagnostic. A colleague sends you a fitted model, an estimate, a standard error and an interval, and asks whether the analysis is sound. An interval can be too short because the likelihood is skewed, or because a nuisance parameter was fixed at a guess, or because the model is wrong — three faults producing numbers that look identical on the page. They live in three different places in the structure above, so only someone who can move between estimand, likelihood, sufficiency, information and risk can separate them.

The forward stake is that the sequel assumes exactly that mobility. Neyman-Pearson theory opens by asking you to form a likelihood ratio and study its distribution; decision theory opens with a risk function already on the board. If those are still fourteen separate things in your head, the next course will spend three weeks rebuilding this one.

What you will be able to do

• Carry one estimation problem through model, estimand, sufficiency, likelihood, information, risk and interval, naming the result used at each step.
• Rao-Blackwellize a crude unbiased estimator and verify the variance reduction by direct computation rather than by appeal to the theorem.
• Show that a maximum likelihood estimator can be inadmissible, by exhibiting a competitor with smaller mean squared error at every parameter value.
• Explain why the Cramer-Rao bound is attained exactly in one parameterization and missed in another, and predict which one it will be.
• Form the likelihood ratio for two simple hypotheses, find the cut giving a stated size, and compute the power of the resulting test.
• Name, for each major topic of Mathematical Statistics II, the theorem from this course it extends.

Words worth owning

Term	What it means in this course
Estimand	The number the scientific question asks for, fixed before any estimator exists. A property of a distribution, not a procedure.
Power	$P_{\theta_1}(\mathrm{reject})$ for a test at a specific alternative $\theta_1$. Like risk, a function of the parameter and not one number.
Sufficient statistic	A function $T$ of the data such that the likelihood depends on the sample only through $T$.
Fisher information	$I(\theta) = \operatorname{Var}(U(\theta)) = -\mathbb{E}[\partial^2 \ell / \partial \theta^2]$, the expected sharpness of the log-likelihood.
Risk function	$R(\theta, \hat\theta) = \mathbb{E}_\theta[L(\theta, \hat\theta)]$. Under squared error, the mean squared error.
Admissible	No competitor has risk no larger at every $\theta$ and strictly smaller somewhere. When one does, the estimator is inadmissible.
Most powerful test	For two simple hypotheses at a fixed size, the test maximizing power at the alternative.
Monotone likelihood ratio	A family whose likelihood ratio between any two parameter values is monotone in one statistic. It promotes a most powerful test to a uniformly most powerful one.

The course as one argument

Start with the map, because the shape is easier to see than to describe.

The fifteen weeks drawn as a dependency graph, with an arrow from each result to the later result that uses it.

Eleven results and the arrows between them

Eleven boxes, fourteen arrows, and the arrows are the content. Each runs from a result to a later result that could not be stated without it, which is why none run backwards.

Follow the darker spine, the path this page walks. Model and estimand (Weeks 0 to 3) feeds likelihood and score (Weeks 7 and 8): you cannot write $L(\theta)$ before you have a family $\{P_\theta : \theta \in \Theta\}$. Likelihood feeds sufficiency (Week 10) very literally — the Fisher-Neyman factorization $f(x \mid \theta) = g(T(x), \theta)h(x)$ says that $\ell(\theta)$ depends on the sample only through $T$. Sufficiency feeds UMVU estimation (Week 11), since Rao-Blackwell conditions on a sufficient statistic and Lehmann-Scheffe needs it complete. Likelihood also feeds information bounds (Week 12) along a second edge, because $I(\theta)$ is the variance of the score. The spine closes with limits of the model (Week 14) feeding this week.

The lighter arrows carry the rest: sampling behaviour (Weeks 4 and 5) turns information into standard errors, and risk (Week 6) joins prior and posterior (Week 9) to make Bayes estimators, which feed interval estimation (Week 13). The map has no box for “review”.

Five views of one structure

The same claim without the picture. Let $X_1, \dots, X_n$ be independent with density $f(x \mid \theta)$, write $\ell(\theta) = \sum_i \log f(X_i \mid \theta)$ and $U(\theta) = \ell'(\theta)$, and let $T$ be a statistic. Each pair below is joined by a theorem you have proved.

Estimand and likelihood meet through invariance: if the estimand is $g(\theta)$ then $\widehat{g(\theta)} = g(\hat\theta)$, because reparameterizing relabels the horizontal axis of $\ell$ without moving its maximum.

Likelihood and sufficiency. Factorization says $T$ is sufficient exactly when $\ell(\theta)$ is, up to a $\theta$-free additive term, a function of $T$ alone. Sufficiency is not an extra property of the model; it describes the likelihood’s shape.

Likelihood and information. $I(\theta) = \operatorname{Var}_\theta(U(\theta)) = -\mathbb{E}_\theta[\ell''(\theta)]$ under the Week 7 regularity conditions. Information is curvature.

Information and risk. Cramer-Rao gives $\operatorname{Var}(\hat\theta) \ge 1/(nI_1(\theta))$ for unbiased $\hat\theta$, and maximum likelihood meets that floor asymptotically.

Risk and the posterior. A Bayes estimator minimizes posterior expected loss pointwise, and so minimizes risk averaged against the prior. Frequentist and Bayes risk are one integrand read in two orders.

Note

The audit habit, one last time. Every arrow carries conditions, and the standing counterexample breaks several at once: for $X_1, \dots, X_n$ uniform on $(0, \theta)$ the support depends on $\theta$, so the score identity fails, $I(\theta)$ is not what the formula says, Cramer-Rao does not apply, and $X_{(n)}$ has variance of order $n^{-2}$ rather than $n^{-1}$. Factorization survives intact. Check which side of an arrow your model is on before using it.

What the sequel adds and why it needs this course

Mathematical Statistics II is usually advertised as “testing, decision theory, and resampling”, which is accurate and useless, because it makes the topics sound new. Each continues a result you own.

Testing is the likelihood ratio, calibrated

Week 7 built $L(\theta)$ and Week 13 cut it horizontally to make an interval. Neyman-Pearson reads the same object as a ranking of samples instead of parameter values: for simple $H_0: \theta = \theta_0$ against simple $H_1: \theta = \theta_1$, the test rejecting when $L(\theta_1)/L(\theta_0) \ge k$ is most powerful at its own size. The second worked example builds exactly that test.

Uniformly most powerful tests need one more ingredient, and Week 3 supplied it. If the family has a monotone likelihood ratio in a statistic $T$ — as every one-parameter exponential family does — the cut Neyman-Pearson produces for one alternative is the cut for every alternative on that side, so one test is most powerful against all of them. That is the Karlin-Rubin theorem, and it is why exponential families were worth a week in September.

Decision theory, resampling, and letting the model go

Week 6 defined risk and Week 9 defined Bayes risk; decision theory asks the global questions those definitions make possible. Is there an estimator no competitor beats everywhere — an admissible one? Is there one whose worst-case risk is smallest, a minimax one? The risk figure below settles the first question in the negative for maximum likelihood, in a line of arithmetic; Stein’s phenomenon, where the obvious estimator of a normal mean is inadmissible in three or more dimensions, is the sequel’s famous instance.

The bootstrap continues Week 13: when no pivot exists and the asymptotics are not trusted, resample, recompute, take percentiles — the next course supplies the theory saying when that is legitimate. Robust and nonparametric procedures continue Week 14, and the sandwich variance gets a chapter.

Worked example — twenty repair times, carried through every layer

Setting. A maintenance group records the time, in hours, to complete twenty independent service jobs of one type:

$$9.8,\ 3.2,\ 6.3,\ 4.5,\ 14.7,\ 2.1,\ 5.4,\ 7.7,\ 3.9,\ 6.8,\ 1.3,\ 11.9,\ 4.6,\ 8.0,\ 5.3,\ 2.3,\ 9.2,\ 3.0,\ 6.2,\ 3.8 .$$

These sum to $T = \sum_i x_i = 120.0$ hours, so $\bar{x} = 6.0$. Model them as independent $\mathrm{Gamma}(\alpha, \lambda)$ with known shape $\alpha = 3$ and unknown rate $\lambda \gt 0$:

$$f(x \mid \lambda) = \frac{\lambda^3 x^2 e^{-\lambda x}}{\Gamma(3)} = \frac{\lambda^3 x^2 e^{-\lambda x}}{2}, \qquad x \gt 0 .$$

Step one, the estimand (Week 1). The parameter $\lambda$ is a rate in repairs per hour of work. What the group cares about is the fraction of jobs overrunning an eight-hour shift, $\psi = P_\lambda(X \gt 8)$. Two objects joined by the model and nothing else; Step nine asks what happens to each when the family is wrong.

Step two, sufficiency (Week 10). The joint density factors as

$$\prod_{i=1}^{20} \frac{\lambda^3 x_i^2 e^{-\lambda x_i}}{2} = \underbrace{\lambda^{60} e^{-\lambda T}}_{g(T, \lambda)} \cdot \underbrace{2^{-20}\textstyle\prod_i x_i^2}_{h(x)},$$

so $T = \sum_i X_i$ is sufficient by Fisher-Neyman. It is complete as well: writing the density as $h(x)\exp\{\eta x - A(\eta)\}$ with $\eta = -\lambda$ and $A(\eta) = -3\log(-\eta)$ puts the model in a full-rank one-parameter exponential family. From here the twenty numbers matter only through $120.0$.

Step three, likelihood, score, information (Weeks 7 and 12). Up to an additive constant,

$$\ell(\lambda) = 60 \log \lambda - 120\lambda, \qquad U(\lambda) = \frac{60}{\lambda} - 120, \qquad \ell''(\lambda) = -\frac{60}{\lambda^2} .$$

The score vanishes at $\hat\lambda = 60/120 = 0.500$ per hour, and $\ell''$ is negative throughout, so that is the maximum. Since $\ell''$ contains no data, observed and expected information coincide: $J(\hat\lambda) = I_{20}(\hat\lambda) = 240$, with $I_1(\hat\lambda) = \alpha/\hat\lambda^2 = 12$, so the standard error is $240^{-1/2} = 0.06455$. By invariance the mean job length is estimated by $3/\hat\lambda = 6.0$ hours and the overrun fraction by

$$\hat\psi = e^{-4}\left(1 + 4 + \frac{4^2}{2}\right) = 13e^{-4} = 0.2381 .$$

Step four, a crude competitor improved (Week 11). Throw away nineteen observations. Since $E_\lambda[1/X_1] = \lambda/(\alpha - 1) = \lambda/2$, the statistic $\delta_0 = 2/X_1$ is unbiased for $\lambda$; here it gives $2/9.8 = 0.204$, badly wrong. Its variance is $4\{\lambda^2/[(\alpha-1)(\alpha-2)] - \lambda^2/(\alpha-1)^2\} = 4(\lambda^2/2 - \lambda^2/4) = \lambda^2$. Now condition on the sufficient statistic. Because $X_1 \sim \mathrm{Gamma}(3, \lambda)$ and $T - X_1 \sim \mathrm{Gamma}(57, \lambda)$ are independent, $X_1/T \sim \mathrm{Beta}(3, 57)$ independently of $T$; and $E[1/B] = (a + b - 1)/(a-1)$ for $B \sim \mathrm{Beta}(a,b)$, so

$$E[\delta_0 \mid T = t] = \frac{2}{t}\,E\!\left[\frac{1}{B}\right] = \frac{2}{t}\cdot\frac{59}{2} = \frac{59}{t} .$$

The improved estimator is $59/T = 0.4917$, with variance $59^2E[T^{-2}] - \lambda^2 = \lambda^2\{59/58 - 1\} = \lambda^2/58$. Conditioning cut the variance by a factor of $58$, and because $T$ is complete, Lehmann-Scheffe upgrades $59/T$ from “better” to UMVU.

Step five, the bound (Week 12). For unbiased estimators of $\lambda$, Cramer-Rao gives $\operatorname{Var} \ge 1/(nI_1(\lambda)) = \lambda^2/60$; the UMVU estimator sits at $\lambda^2/58$, so the bound is real but unreached. Change the estimand and that reverses: for $\mu = 3/\lambda$ the bound is $(3/\lambda^2)^2\lambda^2/60 = 0.15/\lambda^2$, and $\bar{X}$ is unbiased for $\mu$ with variance $(3/\lambda^2)/20 = 0.15/\lambda^2$ exactly. Attainment needs the score to be proportional to $\hat g - g(\lambda)$, and $U(\lambda) = -60\,(T/60 - 1/\lambda)$ delivers that for $1/\lambda$ and its affine images, never for $\lambda$ itself.

Step six, the posterior (Week 9). Put a $\mathrm{Gamma}(8, 20)$ prior on $\lambda$: prior mean $0.4$, prior standard deviation $\sqrt{8}/20 = 0.141$ — a loosely held belief that jobs run near seven and a half hours. Conjugacy gives

$$\pi(\lambda \mid x) \propto \lambda^{7}e^{-20\lambda}\cdot\lambda^{60}e^{-120\lambda} = \lambda^{67}e^{-140\lambda}, \qquad \lambda \mid x \sim \mathrm{Gamma}(68, 140).$$

The posterior mean $68/140 = 0.4857$ is the precision-weighted average $(20/140)(0.4) + (120/140)(0.5)$, one part prior to six parts data. The posterior predictive mean for the next job is $E[3/\lambda \mid x] = 3 \times 140/67 = 6.27$ hours with standard deviation $3.73$ against the plug-in $3.57$; the extra spread is parameter uncertainty.

Step seven, risk (Week 6). The two estimators of $\lambda$ built so far from the sufficient statistic — the maximum likelihood $60/T$ and the UMVU $59/T$ — both have the form $c/T$ with $T \sim \mathrm{Gamma}(60, \lambda)$, and $E[T^{-1}] = \lambda/59$, $E[T^{-2}] = \lambda^2/(59\cdot 58)$, so

$$\frac{\mathrm{MSE}(c/T)}{\lambda^2} = \frac{c^2}{59 \cdot 58} - \frac{2c}{59} + 1 ,$$

a parabola in $c$ minimized at $c = 58$ with value $1 - 58/59 = 1/59$. At $c = 60, 59, 58$ it equals $62/3422 = 0.01812$, $1/58 = 0.01724$ and $1/59 = 0.01695$.

Scaled mean squared error of four estimators of the repair rate, plotted against the true rate.

Two readings. First, the flat lines never cross, because each risk is a constant multiple of $\lambda^2$; so $58/T$ beats $60/T$ at every $\lambda$, making the maximum likelihood estimator inadmissible under squared-error loss. The UMVU estimator is dominated too — optimality within the unbiased class is not optimality. Second, the posterior mean is no flat line: its scaled risk dips to $0.0123$ near $\lambda = 0.5$, below the Cramer-Rao value $1/60 = 0.01667$, which is legal precisely because it is biased. It beats the best flat competitor between $\lambda = 0.31$ and $\lambda = 0.78$, and outside that band it climbs steeply — but to a ceiling, not past one. Because $68/(20 + T) \lt 68/20 = 3.4$ for every $T \gt 0$, the estimator simply stops following a large $\lambda$, so beyond its minimum the scaled risk rises monotonically to the limit $1$: it is $0.027$ at $\lambda = 1$, $0.330$ at $5$, $0.876$ at $50$ and $0.935$ at $100$. That limit is exactly the scaled risk $1.0$ of the crude $\delta_0$, which sits off the plot — however far the truth strays from the prior, the posterior mean approaches the estimator Step four called badly wrong and never becomes worse than it. Its raw mean squared error still grows like $\lambda^2$, as every risk here does.

Step eight, four intervals (Weeks 4, 12 and 13). With $z = 1.96$ and $\chi^2_{1,0.95} = 3.84146$:

• Wald: $0.500 \pm 1.96(0.06455) = (0.3735,\ 0.6265)$.
• Likelihood ratio: with $u = \lambda/\hat\lambda$ the deviance is $D(\lambda) = 120[u - 1 - \log u]$; setting it equal to $3.84146$ gives $u = 0.76785$ and $u = 1.27481$, hence $(0.3839,\ 0.6374)$.
• Exact: $2\lambda T \sim \chi^2_{120}$ whatever $\lambda$ is, so $(91.573/240,\ 152.211/240) = (0.3816,\ 0.6342)$.
• Posterior, central: the $0.025$ and $0.975$ quantiles of $\mathrm{Gamma}(68, 140)$, namely $(0.3772,\ 0.6078)$.

Everything the twenty repair times say about the rate: the relative likelihood, the posterior, and four intervals on one axis.

The three frequentist intervals differ by about $0.01$ at each end, the mild right skew of a rate likelihood appearing where Week 13 said it would: the symmetric Wald interval sits lowest at both ends, and the two shape-aware intervals agree closely. The posterior interval is shorter, pulled toward $0.4$, and alone has endpoints that are quantiles of a distribution for $\lambda$.

Step nine, what breaks if the model is wrong (Week 14). Three checks first, all using information $T$ discarded: the sample standard deviation is $3.44$ against the model’s $\sqrt{3}/\hat\lambda = 3.46$; the sample median is $5.35$ against a model median of $5.348$; four of twenty jobs exceeded eight hours against $20\hat\psi = 4.8$. The shape-three gamma is doing well — and you had to look at the data rather than at $T$ to learn that.

Now suppose the shape were really one, an exponential of the same mean. Then $\hat\lambda = 3/\bar{X}$ still converges, to the Kullback-Leibler projection $3/E[X]$, which is $0.5$ again; consistency survives. The standard error does not. Under the exponential $\operatorname{Var}(X) = 36$ rather than $12$, so the delta-method variance of $3/\bar{X}$ triples and the honest standard error is $\sqrt{3} \times 0.06455 = 0.1118$: every interval in Step eight is too short by a factor near $1.73$, and more data will not repair it. And $\hat\psi$ is aimed at the wrong number, since $P(X \gt 8)$ under that exponential is $e^{-4/3} = 0.264$, not $0.238$.

The same reasoning, transferred

Change the model, keep the structure. A service desk logs whether each of $n = 40$ independently handled machines needed a second visit; $x = 12$ did. Model the indicators as $\mathrm{Bernoulli}(p)$, with estimand $p$.

Same. $T = \sum_i X_i = 12$ is sufficient and complete by factorization in the exponential family, reducing forty numbers to one. The score $\ell'(p) = 12/p - 28/(1-p)$ vanishes at $\hat p = 0.300$, and $I_1(p) = 1/\{p(1-p)\}$ gives $I_{40}(\hat p) = 40/0.21 = 190.48$ and $\mathrm{se} = 0.07246$. Rao-Blackwell runs identically: $X_1$ is unbiased with variance $p(1-p)$, and $E[X_1 \mid T = t] = t/40 = \bar{X}$, a variance cut by a factor of forty. So does conjugacy: a $\mathrm{Beta}(2,2)$ prior gives a $\mathrm{Beta}(14, 30)$ posterior with mean $14/44 = 0.3182 = (4/44)(0.5) + (40/44)(0.3)$, the same weighting in new clothes. The Wald interval is $(0.1580,\ 0.4420)$, the likelihood-ratio interval $(0.1735,\ 0.4514)$, the credible interval $(0.1908,\ 0.4613)$.

Changed. The Cramer-Rao bound is now attained exactly: $\operatorname{Var}(\bar{X}) = p(1-p)/40 = 1/I_{40}(p)$, which is $0.00525$ at $\hat p$, so $\bar{X}$ is efficient at every $n$. Resist the inference that therefore nothing beats it. Efficiency settles optimality inside the unbiased class and, as Step seven just showed with $59/T$, settles nothing outside it, so the shrinkage question has to be asked directly. For $c\bar{X}$,

$$\mathrm{MSE}(c\bar{X}) = \frac{c^2 p(1-p)}{40} + p^2(c-1)^2 ,$$

minimized at $c = 40p/(40p + 1 - p)$, which is $0.945$ at $p = 0.3$ and $0.678$ at $p = 0.05$. The best multiplier depends on the unknown $p$, so no single choice of $c$ can be made in advance; and near $p = 1$ every choice but one is a mistake, since $\mathrm{MSE}(c\bar{X}) - \mathrm{MSE}(\bar{X}) = (c^2-1)p(1-p)/40 + p^2(c-1)^2 \to (c-1)^2$, strictly positive for $c \ne 1$. No constant multiple of $\bar{X}$ dominates it, and the stronger statement, that $\bar{X}$ is admissible here, is one the sequel’s decision theory proves. In the gamma case the minimizing constant was $n\alpha - 2$, free of $\lambda$ — which is exactly why $58/T$ could dominate there and why nothing of that shape can here. Meanwhile the exact row of Step eight has no counterpart: the data are discrete, no function of $(X, p)$ has a $p$-free distribution, and every interval here is an approximation whose coverage sawtooths with $p$. One model gave an unreachable bound and an exact pivot; the other gives the reverse.

A computation you can run

Nothing above needs more than a dozen lines of R, and typing them beats trusting the page.

hours <- c(9.8, 3.2, 6.3, 4.5, 14.7, 2.1, 5.4, 7.7, 3.9, 6.8,
           1.3, 11.9, 4.6, 8.0, 5.3, 2.3, 9.2, 3.0, 6.2, 3.8)
n <- length(hours); alpha <- 3; tot <- sum(hours); K <- n * alpha   # 20, 120, 60

lhat <- K / tot; se <- lhat / sqrt(K)                # 0.5000   0.06455
lhat + c(-1, 1) * 1.959964 * se                      # 0.3735   0.6265  Wald
dev <- function(l) 2 * K * (l / lhat - 1 - log(l / lhat))
c(uniroot(function(l) dev(l) - qchisq(0.95, 1), c(1e-6, lhat))$root,
  uniroot(function(l) dev(l) - qchisq(0.95, 1), c(lhat, 5))$root)   # 0.3839  0.6374
qchisq(c(0.025, 0.975), 2 * K) / (2 * tot)           # 0.3816   0.6342  exact
qgamma(c(0.025, 0.975), 8 + K, 20 + tot)             # 0.3772   0.6078  posterior

The risk claims are simulation checks, each of this shape: draw the sufficient statistic from its own distribution, apply each estimator, average the squared error.

tsim <- rgamma(2e5, K, rate = 0.5)
mean((K / tsim         - 0.5)^2) / 0.25              # about 0.0181   the MLE
mean((58 / tsim        - 0.5)^2) / 0.25              # about 0.0169   best flat
mean((68 / (20 + tsim) - 0.5)^2) / 0.25              # about 0.0123   posterior mean

Agreement confirms only that the algebra was transcribed correctly, and at this many draws each figure is pinned to about the third decimal place. Week 8’s rule stands.

Second worked example — the most powerful test for two simple rates

Setting. Same twenty jobs, same gamma model with $\alpha = 3$. A contract turns on one comparison: the incumbent’s advertised rate $\lambda_0 = 0.4$ per hour, a mean job of $3/0.4 = 7.5$ hours, against a challenger’s claimed $\lambda_1 = 0.6$, five hours. Both hypotheses are simple, which is exactly what Neyman-Pearson governs. Fix the size at $0.05$.

Step one, the ratio. Everything free of $\lambda$ cancels:

$$\Lambda(x) = \frac{L(\lambda_1)}{L(\lambda_0)} = \left(\frac{\lambda_1}{\lambda_0}\right)^{60} e^{-(\lambda_1 - \lambda_0)T} = 1.5^{60}\,e^{-0.2T} .$$

Two facts fall out. The ratio depends on the sample only through $T$ — sufficiency again, now inside a testing problem — and it is strictly decreasing in $T$. So “reject when $\Lambda \ge k$” is the rule “reject when $T \le c$”: the most powerful test rejects the advertised rate when the total is small.

Step two, the cut. Under $H_0$, $T \sim \mathrm{Gamma}(60, 0.4)$, equivalently $0.8\,T \sim \chi^2_{120}$. Choosing $c$ with $P_{0.4}(T \le c) = 0.05$ gives $c = \chi^2_{120,\,0.05}/0.8 = 95.705/0.8 = 119.63$ hours, and the equivalent cut on the ratio is $k = 1.5^{60}e^{-0.2(119.63)} = \exp(24.3279 - 23.9262) = 1.494$.

Step three, power. Under $H_1$, $T \sim \mathrm{Gamma}(60, 0.6)$ and $P_{0.6}(T \le 119.63) = 0.930$: the test detects the challenger’s rate about $93$ times in a hundred. What made that computable was not the test but Week 3’s fact that independent gammas sharing a rate add to a gamma.

The total of the twenty repair times under two simple hypotheses, with the most powerful test’s rejection region shaded.

Step four, the verdict. The observed total is $120.0$, exceeding $119.63$ by four tenths of an hour, so the test does not reject $H_0$. The observed ratio is $\Lambda = 1.5^{60}e^{-24} = 1.388$, just under the cut of $1.494$: the data do favour the challenger, by $1.4$ to one, and the procedure still declines to act. That is not a defect to argue away. A size-$0.05$ test is a commitment made before the data arrive about how much evidence is enough, and honouring it in awkward cases is the whole content of the commitment. The attained tail probability $P_{0.4}(T \le 120) = 0.052$ reports how awkward this case was.

What it buys and what it assumed. The lemma delivers optimality only against $\lambda_1 = 0.6$. But the ratio is monotone in $T$ for every pair $\lambda_1 \gt \lambda_0$, and the cut $119.63$ never mentions $\lambda_1$ — so the same test is uniformly most powerful for $H_0: \lambda \le 0.4$ against $H_1: \lambda \gt 0.4$. Karlin-Rubin, visible as a property of the algebra rather than a new theorem. It assumed independence, the shape $\alpha = 3$, and a pre-specified size. The exact law of $\Lambda$ under $H_0$, available only because this model is friendly, is what the sequel studies when it is not: Wilks’ theorem exists because $\Lambda$ usually has no tractable exact distribution.

The misreading to avoid

The sentence that arrives in the last week, in something close to these words:

Note

“So the course was fourteen separate techniques and a review. I will memorize the formulas that get tested and let the rest go.”

Two things are wrong with it: one about the material, one about what memorizing will cost.

The material first. Take the integrated example and change a single thing — let $\alpha = 1$, so the jobs are exponential. Every step moves at once. The sufficient statistic is still $T$; the maximum likelihood estimator becomes $20/T$; the UMVU estimator $19/T$; the best estimator of the form $c/T$ becomes $18/T$; the Cramer-Rao value becomes $\lambda^2/20$; the exact pivot becomes $2\lambda T \sim \chi^2_{40}$; the deviance becomes $40[u - 1 - \log u]$; the conjugate prior is still gamma. Each is the same formula with $K = n\alpha$ changed from $60$ to $20$, because each was a statement about the distribution of the sufficient statistic, and the model entered only through $K$. There were never eight formulas. There was one structure with a parameter in it, and whoever learned the eight separately now has eight more to learn.

The second problem matters more. Formulas fail silently; structure does not. Remember only “$\mathrm{se} = 1/\sqrt{nI_1}$” and you will compute a standard error for the uniform on $(0, \theta)$, get a number, and never be told it is wrong. Remember instead that the formula is a claim about the curvature of a log-likelihood assumed differentiable at an interior maximum, and you look at the uniform likelihood first. It is zero for $\theta \lt x_{(n)}$ and $\theta^{-n}$ for $\theta \ge x_{(n)}$: it jumps from nothing to its maximum $x_{(n)}^{-n}$ at $\theta = x_{(n)}$ and decays from there, so it is discontinuous exactly at its maximizer, with no interior stationary point and no second derivative to read curvature from. Then you stop before computing. Every counterexample in this course has that shape: the technique runs, returns a plausible number, and is silently inapplicable.

One concession: the weeks are separable enough that you can compute a posterior without having proved Rao-Blackwell, which is why the misreading survives a semester of homework. It does not survive an unfamiliar problem, which will not tell you which technique it wants; only the map can.

Practice on your own

1. The integrated pass on a new model. Repeat all nine steps of the first worked example for $X_1, \dots, X_n$ independent $\mathrm{Poisson}(\lambda)$ with $n = 30$ and $\sum_i x_i = 45$: the sufficient statistic, the maximum likelihood estimator, $I_1(\lambda)$, a Rao-Blackwellized improvement of $X_1$, a gamma posterior, and three intervals. Say where your derivation stops matching the gamma case, and why.

2. An admissibility hunt. Show that $c = 58$ minimizes $c^2/(59\cdot 58) - 2c/59 + 1$ with minimum $1/59$. Then show that for a general known shape the minimizing constant is $n\alpha - 2$, and say why the maximum likelihood constant $n\alpha$ sits on the wrong side of the parabola.

3. A counterexample hunt. Find a model in which the Rao-Blackwellized estimator is not UMVU, and name which hypothesis of Lehmann-Scheffe your example violates. An incomplete sufficient statistic is one route; a curved exponential family is another.

4. An audit. A draft report states: “The posterior mean has smaller mean squared error than the maximum likelihood estimator, so it is the better estimator, and since it beats the Cramer-Rao bound the bound must be wrong.” Identify the three separate errors, and give a value of $\lambda$ from the risk figure at which the first claim is false.

5. A simulation to describe. Design, in words and pseudocode, a study checking whether the exact interval of Step eight holds coverage $0.95$ at $n = 5$ while the Wald interval does not. Work out how many replicates put the Monte Carlo standard error of a coverage near $0.95$ below $0.002$, and say what would falsify the claim.

Where to read more

• MIT OpenCourseWare 18.655, Mathematical Statistics for the decision-theoretic framing that unifies this page, and for admissibility.
• MIT OpenCourseWare 18.650, Statistics for Applications for the Neyman-Pearson lemma at the level the next course assumes.
• Penn State STAT 414 for gamma and beta facts, including the beta-gamma relationship of the Rao-Blackwell step.
• The R Project for qgamma, qchisq, uniroot and rgamma.
• Quarto, the format these pages use.
• Hogg, McKean and Craig treat sufficiency and completeness in their seventh chapter and Neyman-Pearson theory in their eighth.
• Course pages: the syllabus, the schedule, and the resources page.

Where this goes next

There is no Week 16. The course ends here, and Mathematical Statistics II begins where the second worked example stopped: with the distribution of a likelihood ratio and what to do with it. Every box on the map is a prerequisite for something there, and the honest way to prepare is not to reread these notes but to run an integrated pass on a model appearing nowhere in them — a two-parameter normal, a negative binomial, a censored exponential — and notice which step you cannot complete. That step is what to reread.

The last connection is the one this course has been making since Week 2. The posterior in Step six was not an alternative to the frequentist analysis: it used the same likelihood, the same sufficient statistic and the same information, differing only in what it treats as random. A Bayesian workflow — prior, posterior, predictive check, decision under loss — is assembled from parts built here. Week 14 is the way back, the notes index lists every unit in order, and the schedule records where each of them fell.