Lecture notes. Written from the material of Berkeley Stat 210A, licensed CC BY 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.

82. Nuisance Parameters¶

What this covers¶

Most testing problems have more unknown parameters than the hypothesis is actually about. This chapter asks how to test a hypothesis on a parameter of interest $\theta$ when the model also carries an unknown nuisance parameter $\lambda$ that can shift the size or power of any test. It develops the main tool — conditioning on a statistic that is sufficient for $\lambda$ — states the resulting optimal (UMPU) test for exponential families, sketches why it is optimal, and shows the same conditioning idea surviving even when there is no exponential family at all, in permutation tests. It assumes UMP and UMP-unbiased testing for one-parameter exponential families, sufficiency and completeness, and the natural-parameter form of an exponential family density.

Nuisance parameters: the setup¶

One-parameter families are the exception in statistics, not the rule. Formally, we observe $X$ from a model

\[ \mathcal P = \{P_{\theta,\lambda} : (\theta,\lambda) \in \Omega\}, \]

where $\theta$ and $\lambda$ may be real vectors, or may represent infinite-dimensional objects such as unknown distributions. The hypotheses concern $\theta$ alone — $H_0:\theta\in\Theta_0$ vs $H_1:\theta\in\Theta_1$ — and $\lambda$ is the nuisance parameter: unknown, not of interest, but capable of changing the type-I error or the power of a test we might otherwise like to use.

Three examples fix the pattern.

Two-sample Gaussian. $X_1,\dots,X_n\stackrel{\text{iid}}\sim N(\mu,\sigma^2)$, $Y_1,\dots,Y_m\stackrel{\text{iid}}\sim N(\nu,\sigma^2)$, all three of $\mu,\nu,\sigma^2$ unknown, testing $H_0:\mu=\nu$ vs $H_1:\mu\neq\nu$. Here $\theta=\mu-\nu$ is the parameter of interest, and $\lambda=(\mu+\nu,\sigma^2)$ (or $(\mu,\sigma^2)$, or any other coordinatization of what remains) is the nuisance parameter.
Comparing two Poissons. $X\sim\mathrm{Pois}(\mu)$, $Y\sim\mathrm{Pois}(\nu)$, testing $H_0:\mu\le\nu$ vs $H_1:\mu>\nu$. One could use $\mu-\nu$ as the parameter of interest, but it turns out to be more natural to use the ratio $\mu/\nu$, or $\theta=\mu/(\mu+\nu)$.
Two-sample binomial. $X_1\sim\mathrm{Binom}(n_1,\pi_1)$, $X_2\sim\mathrm{Binom}(n_2,\pi_2)$, with $n_1,n_2$ known (so not nuisance parameters), testing $H_0:\pi_1\le\pi_2$ vs $H_1:\pi_1>\pi_2$. Again $\pi_1-\pi_2$ is an option, but the natural quantity turns out to be the odds ratio $\rho = \frac{\pi_1/(1-\pi_1)}{\pi_2/(1-\pi_2)}$, whose log is the difference between the two natural (logit) parameters.

In each case the question is the same: how do we test $\theta$ without the unknown $\lambda$ contaminating the answer?

Conditioning as a strategy¶

The central idea is to condition on a statistic $U$ that carries all the information about $\lambda$, so that the conditional distribution of the rest of the data no longer depends on $\lambda$ at all.

A warm-up: binomial counts with a random sample size. Suppose $N$ is a sample size and, given $N=n$, we observe $X\sim\mathrm{Binom}(n,\theta)$ — the situation, for instance, of a survey with a random number of respondents. First suppose the distribution of $N$ is known, say $N\sim\mathrm{Pois}(10)$, and we want to test $H_0:\theta\le\tfrac12$ vs $H_1:\theta>\tfrac12$. The full likelihood is

\[ p_\theta(n,x) = \frac{10^n e^{-10}}{n!}\binom nx \theta^x(1-\theta)^{n-x}, \qquad 0\le x\le n. \]

There is no UMP test here: the likelihood ratio between $\theta=\tfrac12$ and any alternative $\theta_1>\tfrac12$,

\[ \log\frac{p_{\theta_1}}{p_{1/2}}(n,x) = x\log\frac{\theta_1}{1-\theta_1} - n\log\big(2(1-\theta_1)\big), \]

is a different linear combination of $x$ and $n$ for each $\theta_1$, so no single rejection region can be best against every alternative at once. Two natural tests are available instead.

Marginal test. Marginally $X\sim\mathrm{Pois}(10\theta)$, which has monotone likelihood ratio in $X$; reject when $X$ exceeds the upper-$\alpha$ quantile of $\mathrm{Pois}(5)$. Call this $\phi_1$; it certainly satisfies $\mathbb E_\theta[\phi_1]\le\alpha$ for $\theta\le\tfrac12$.
Conditional test. Conditionally, $X\mid N=n \sim \mathrm{Binom}(n,\theta)$, which also has MLR in $x$ for fixed $n$; reject when $X$ exceeds the upper-$\alpha$ quantile of $\mathrm{Binom}(n,\tfrac12)$. Call this $\phi_2$; it satisfies $\mathbb E_\theta[\phi_2\mid N=n]\le\alpha$ for every $n$ and every $\theta\le\tfrac12$, and hence $\mathbb E_\theta[\phi_2]\le\alpha$ as well.

The test $\phi_2$ is UMP in the conditional model $\mathcal Q_n=\{q_\theta(x\mid n):\theta\}$, the family of conditional distributions of $X$ given $N=n$. This example illustrates a fact that holds in general: any level-$\alpha$ (or unbiased) test in a conditional model is automatically level-$\alpha$ (or unbiased) in the marginal model, because marginal power is an average of conditional powers, and an average of numbers each $\le\alpha$ is itself $\le\alpha$. The same holds for confidence sets.

The conditionality principle says that, since $N$ here is ancillary (its distribution does not involve $\theta$), we ought to condition on it: the observed value of $N$ tells us nothing about $\theta$ on its own, so it should be treated as fixed rather than as informative data. Whether or not one accepts this principle as a general rule, conditioning is forced on us in a more serious case:

When the nuisance parameter is the distribution of $N$ itself. Suppose instead $N$ is drawn from an unknown distribution $P^N$ on $\{0,1,2,\dots\}$, with $X\mid N=n$ still $\mathrm{Binom}(n,\theta)$. Now $P^N$ is an infinite-dimensional nuisance parameter, and the marginal test $\phi_1$ is no longer even available — there is no fixed marginal distribution of $X$ to appeal to. But the conditional test $\phi_2$ is untouched: the conditional model $\mathcal Q_n$ never depended on $P^N$ in the first place. Conditioning on $N$ removes the nuisance parameter entirely.

Comparing two Poissons, revisited. Let $N=X+Y\sim\mathrm{Pois}(\mu+\nu)$. A short calculation (expand each Poisson pmf, divide, and use $\mathrm{Pois}(\mu)*\mathrm{Pois}(\nu)=\mathrm{Pois}(\mu+\nu)$ for the sum) gives

\[ X \mid X+Y=n \;\sim\; \mathrm{Binom}(n,\theta), \qquad \theta = \frac{\mu}{\mu+\nu}. \]

Since $Y=N-X$, the pair $(N,X)$ carries the same information as $(X,Y)$, and conditioning on $N$ removes the nuisance parameter $\lambda=\mu+\nu$ altogether, leaving exactly the hypothesis $H_0:\theta\le\tfrac12 \iff \mu\le\nu$ against $H_1:\theta>\tfrac12$. This reduction is not a coincidence — it is forced by the exponential family structure of the model, which is the general mechanism behind conditioning.

Multiparameter exponential families: the general recipe¶

Take a generic exponential family in which the natural parameter splits into a parameter of interest $\theta\in\mathbb R^s$ and a nuisance parameter $\lambda\in\mathbb R^r$:

\[ p_{\theta,\lambda}(x) = e^{\theta\cdot T(x) + \lambda\cdot U(x) - A(\theta,\lambda)}\, h(x). \]

We want to test $H_0:\theta\in\Theta_0$ vs $H_1:\theta\in\Theta_1$, treating $\lambda$ as a nuisance parameter. The recipe is: condition on $U(X)$, and base the test on the conditional model for $T(X)$, which will depend on $\theta$ alone.

No generality is lost in reducing to the sufficient statistic $(T,U)$, since any test $\phi(X)$ has exactly the same power function as $\psi(T(X),U(X)) := \mathbb E[\phi(X)\mid T,U]$. The joint density of $(T,U)$ has the same exponential-family form,

\[ (T,U) \sim p_{\theta,\lambda}(t,u) = e^{\theta\cdot t + \lambda\cdot u - A(\theta,\lambda)}\,g(t,u), \]

and when we form the conditional density of $T$ given $U=u$, the factor $e^{\lambda\cdot u - A(\theta,\lambda)}$ appears in both numerator and denominator and cancels:

\[ q_\theta(t\mid u) = \frac{p_{\theta,\lambda}(t,u)}{\int p_{\theta,\lambda}(z,u)\,dz} = \frac{e^{\theta\cdot t}\,g(t,u)}{\int e^{\theta\cdot z}\,g(z,u)\,dz} = e^{\theta\cdot t - B_u(\theta)}\, g(t,u), \qquad B_u(\theta)=\log\!\int e^{\theta\cdot z}g(z,u)\,dz. \]

The nuisance parameter $\lambda$ has vanished completely: $q_\theta(t\mid u)$ is an $s$-parameter exponential family in $\theta$ with sufficient statistic $T$, for every fixed $u$. If $s=1$, this conditional family automatically has monotone likelihood ratio in $T$, so ordinary one-parameter theory (UMP one-sided tests, UMPU two-sided tests) applies directly, conditionally on $u$. Even when $s>1$, the nuisance parameter is still gone, which is already useful. Note also that the base measure $h$ never has to be dealt with separately: it is absorbed into $g(t,u)$ by the sufficiency reduction to $(T,U)$, and plays no further role.

Poisson comparison, done via the recipe. Write the joint pmf of $(X,Y)$ and split the exponent by adding and subtracting $x\log\nu$:

\[ p_{\mu,\nu}(x,y) = \exp\{x\log\mu + y\log\nu - (\mu+\nu)\}\frac{1}{x!\,y!} = \exp\Big\{ x\log\frac\mu\nu + (x+y)\log\nu - (\mu+\nu)\Big\}\frac1{x!\,y!}. \]

This is exactly the multiparameter exponential-family form with $\theta=\log(\mu/\nu)$, $T(x,y)=x$, $\lambda=\log\nu$, and $U(x,y)=x+y$. So the general strategy again tells us to condition on $X+Y=u$, and the conditional family in $T=X$ is one-parameter with MLR — reconciling this with the earlier computation: writing out $P_\theta(X=x_1\mid X_1+X_2=u)$ explicitly gives

\[ P_\theta(X_1=x_1\mid X_1+X_2=u) = \frac{e^{\theta x_1}}{\sum_{i=0}^u e^{\theta i}} = \binom{u}{x_1}\Big(\frac{e^\theta}{1+e^\theta}\Big)^{x_1}\Big(\frac1{1+e^\theta}\Big)^{u-x_1}, \]

i.e. $X_1\mid X_1+X_2=u \sim \mathrm{Binom}\big(u,\,e^\theta/(1+e^\theta)\big)$. The success probability $e^\theta/(1+e^\theta)$ is exactly $\mu/(\mu+\nu)$ written on the log-odds scale — the same conclusion as before, just derived structurally rather than by direct calculation. In the end, testing $H_0:\theta=0$ vs $H_1:\theta\ne 0$ reduces to an ordinary two-sided binomial test. The two-sample binomial problem from the first section is another instance of exactly this pattern: conditioning $X_1$ on $X_1+X_2$ leaves a conditional distribution that depends only on the log odds ratio $\rho$, and not on the nuisance parameter $\pi_2$.

The UMPU test in exponential families¶

Putting the recipe together with one-parameter exponential family theory gives an exact optimal test whenever $\theta$ is scalar ($s=1$). Let $\mathcal P$ be a full-rank exponential family $p_{\theta,\lambda}(x) = e^{\theta T(x) + \lambda\cdot U(x) - A(\theta,\lambda)}h(x)$ with $\theta\in\mathbb R$, $\lambda\in\mathbb R^r$.

Two-sided case. To test $H_0:\theta=\theta_0$ vs $H_1:\theta\ne\theta_0$, there is a UMPU test of the form $\phi(x)=\psi(T(x),U(x))$, where

\[ \psi(t,u) = \begin{cases} 1 & t>c_2(u)\\ \gamma_2(u) & t=c_2(u)\\ 0 & c_1(u)<t<c_2(u)\\ \gamma_1(u) & t=c_1(u)\\ 1 & t<c_1(u) \end{cases} \]

with $c_1,c_2,\gamma_1,\gamma_2$ chosen so that $\mathbb E_{\theta_0}[\phi]=\alpha$ and $\mathbb E_{\theta_0}[T\phi] = \alpha\,\mathbb E_{\theta_0}[T]$ (the second condition is what makes $\theta_0$ a stationary point of the power function, the unbiasedness requirement).

One-sided case. To test $H_0:\theta\le\theta_0$ vs $H_1:\theta>\theta_0$, there is a UMP test $\phi(x)=\psi(T(x),U(x))$ with

\[ \psi(t,u) = \begin{cases} 1 & t>c(u)\\ \gamma(u) & t=c(u)\\ 0 & t<c(u) \end{cases} \]

and $c(u),\gamma(u)$ chosen so that $\mathbb E_{\theta_0}[\phi]=\alpha$.

In both cases, the thresholds and randomization probabilities depend on $u$: the test conditions on $U(X)=u$, applies the ordinary optimal one-parameter test inside the conditional family $q_\theta(\cdot\mid u)$, and lets $u$ vary freely across its whole range. Applied to the Poisson comparison example, this is the two-sided binomial test derived above: reject when $X_1$, conditional on $X_1+X_2=u$, is either far below or far above what $H_0$ predicts.

Why conditioning is optimal: a proof sketch¶

The claim is that among all unbiased tests of $H_0:\theta=\theta_0$ — not just tests built by conditioning — $\psi$ has the highest possible power at every alternative. The argument has three steps.

1. Unbiasedness pins down a derivative condition at $\theta_0$. Let $\phi$ be any unbiased test, with power function $\beta(\theta)=\mathbb E_\theta[\phi]$. Unbiasedness forces $\beta(\theta_0)=\alpha$, and since $\beta(\theta)\ge\alpha$ for $\theta\ne\theta_0$ while $\beta(\theta_0)=\alpha$, the point $\theta_0$ is a minimum of $\beta$, so $\beta'(\theta_0)=0$ (the power function of an exponential-family test is infinitely differentiable, and one may differentiate under the integral sign). Since $\partial_\theta\mathbb E_\theta[\phi]=\mathbb E_\theta[T\phi]-\mathbb E_\theta[T]\,\mathbb E_\theta[\phi]$ for an exponential family, this translates into exactly the two moment conditions used to define $\psi$ above: $\mathbb E_{\theta_0}[\phi]=\alpha$ and $\mathbb E_{\theta_0}[T\phi]=\alpha\,\mathbb E_{\theta_0}[T]$.

2. These conditions hold for every value of the nuisance parameter, hence almost surely given $U$. At $\theta=\theta_0$, the family reduces to the boundary submodel $\mathcal P_0=\{p_{\theta_0,\lambda}:\lambda\in\mathbb R^r\}$, itself a full-rank $r$-parameter exponential family in $\lambda$ with complete sufficient statistic $U(X)$. Both moment conditions above hold for every $\lambda$, since $\phi$ is unbiased against the whole boundary. But two functions of $U$ with the same expectation for every $\lambda$ in a complete family must agree almost surely — this is exactly what completeness buys us — so the conditions strengthen from "on average" to "conditionally, given $U$, almost surely": $\mathbb E_{\theta_0}[\phi\mid U]=\alpha$ a.s., and (two-sided case) $\mathbb E_{\theta_0}[T\phi\mid U]=\alpha\,\mathbb E_{\theta_0}[T\mid U]$ a.s.

3. Conditionally on each value of $U=u$, $\phi$ is competing inside a one-parameter exponential family, where the optimal test is already known. For any fixed $u$, the conditional model $\{p_\theta(\cdot\mid u)\}$ is a one-parameter exponential family in $T$ with density $e^{\theta t}g(t,u)$. Step 2 says that $\phi$'s conditional expectation given $(T,U)=(t,u)$, $g(t,u) := \mathbb E_{\theta_0}[\phi\mid T=t,U=u]$, defines a test of level (or unbiasedness) exactly $\alpha$ inside this one-parameter conditional family. But $\psi(t,u)$ is, by construction, the UMP test (one-sided case) or UMPU test (two-sided case, via the analogous one-parameter theorem) in precisely this conditional family. So $\psi$ has conditional power at least as large as $\phi$'s conditional power, for every $\theta$ and (almost) every $u$. Averaging over $u$,

\[ \mathbb E_\theta[\phi] = \mathbb E_\theta\big[\mathbb E[\phi\mid T,U]\big] \le \mathbb E_\theta\big[\mathbb E[\psi\mid T,U]\big] = \mathbb E_\theta[\psi] \]

for every $\theta\ne\theta_0$ and every unbiased $\phi$ — which is exactly UMPU optimality of $\psi$. (The underlying one-parameter facts used here — the derivative characterization of unbiasedness, and the two-sided UMPU theorem for one-parameter exponential families — are standard results referred to in the lecture as Keener's Statistical Theory, Theorems 12.4 and 12.22.)

Example: normal mean, unknown variance¶

Let $X=(X_1,\dots,X_n)\sim N(\mu\mathbf 1,\sigma^2 I_n)$ with $\mu\in\mathbb R$ the parameter of interest and $\sigma^2>0$ unknown, and test $H_0:\mu=0$ vs $H_1:\mu\ne0$. Writing the density in natural-parameter form gives $T=\bar X/\|X\|$ and $U=\|X\|^2$, and the recipe says the optimal test rejects when $\bar X$ is conditionally extreme given $\|X\|^2$.

Here is where the geometry does the work. Decompose $\mathbb R^n$ into the line spanned by $\mathbf 1=(1,\dots,1)$ — the direction along which the mean $\mu$ moves the data — and its orthogonal complement $\mathbf 1^\perp$. Any Gaussian vector splits into independent projections onto these two orthogonal subspaces, so $\mathrm{Proj}_{\mathbf 1}(X)$ and $\mathrm{Proj}_{\mathbf 1^\perp}(X)$ are independent; the first carries all the information about $\mu$ (its length is $\sqrt n\,\bar X$), while the second carries none — it is exactly $N(0,\sigma^2 P)$ regardless of $\mu$, so it is what informs us about $\sigma^2$ alone.

Under $H_0:\mu=0$, the whole vector $X$ is spherically symmetric, so $X/\|X\|$ is uniformly distributed on the sphere $S^{n-1}$ and is independent of $\|X\|$. This means the optimal test rejects for conditionally extreme $\bar X$ given $\|X\|^2$, and — because the null distribution of the angle doesn't depend on $u$ at all — for marginally extreme $\bar X/\|X\|$: the two descriptions coincide. Squaring and rearranging gives a familiar statistic:

\[ T^2 = \frac{(\sum X_i)^2}{\sum X_i^2 - \tfrac1n(\sum X_i)^2} = \frac{n\bar X^2}{\|X\|^2-n\bar X^2} = \frac{n\bar X^2}{S^2}, \qquad S^2=\frac1{n-1}\sum(X_i-\bar X)^2, \]

which, written in terms of the two projections, is

\[ T^2 = \frac{\|\mathrm{Proj}_{\mathbf 1}X\|^2}{\|\mathrm{Proj}_{\mathbf 1^\perp}X\|^2}\cdot\frac{n-1}n. \]

This is (the square of) the ordinary one-sample $t$-statistic, arrived at here as a ratio of the squared length along the "signal" direction to the squared length along the "noise" directions — exactly the picture that makes ratios of projections the right way to think about tests with an unknown variance nuisance parameter, a theme that recurs beyond this one example.

Fixing $\|X\|^2=u$ places $X$ on a sphere. $X$ splits into an independent projection along $\mathbf 1$ (carrying $\mu$) and onto $\mathbf 1^\perp$ (carrying only $\sigma^2$); the two-sided test rejects when $X$'s angle from $\mathbf 1^\perp$ is extreme in either direction (shaded wedges), which is both the conditional-given-$u$ test and, since the null angle distribution does not depend on $u$, the marginal test.

Beyond exponential families: permutation tests¶

Conditioning on a statistic that is sufficient under the null remains useful even when there is no exponential family — and no UMPU test — in sight.

Two-sample permutation test. Observe independent samples $X_1,\dots,X_n\stackrel{\text{iid}}\sim P$ and $Y_1,\dots,Y_m\stackrel{\text{iid}}\sim Q$, and test $H_0:P=Q$ vs $H_1:P\ne Q$. This model is not an exponential family, but under $H_0$ all $n+m$ observations are iid from the common distribution $P$. Pool them into $Z=(Z_1,\dots,Z_{n+m})=(X_1,\dots,X_n,Y_1,\dots,Y_m)$. Under $H_0$, the vector of pooled order statistics $U(Z)=(Z_{(1)},\dots,Z_{(n+m)})$ is complete sufficient for $Z$, and the conditional distribution of $Z$ given $U(Z)=u$ is uniform over all $(n+m)!$ permutations of $u$:

\[ Z \mid U(Z)=u \ \stackrel{H_0}{\sim}\ \mathrm{Unif}\{\pi u : \pi \in \mathcal S_{n+m}\}. \]

Conditioning on $U$ makes $H_0$ a simple null in the conditional model — there is exactly one conditional distribution to compare against — even though the alternative remains highly composite (a stochastically larger $Q$, a more variable $Q$, and so on are all different ways $H_1$ could hold, with no single best direction to test against). Because there is no generically optimal statistic, we are free to plug in any test statistic $T(X,Y)$ and still get an exact conditional test by conditioning on $U(X,Y)$. Rejecting when $T$ exceeds its conditional $\alpha$-quantile is equivalent to rejecting for a small $p$-value

\[ p(x,y\mid u) = \mathbb P_{H_0}\big(T(X,Y)\ge T(x,y) \mid U(X,Y)=u\big) = \frac1{(n+m)!}\sum_{\pi\in\mathcal S_{n+m}} \mathbb 1\{T(\pi u)\ge T(x,y)\}. \]

Enumerating all $(n+m)!$ permutations is rarely feasible, so in practice one samples $\pi_1,\dots,\pi_B\stackrel{\text{iid}}\sim\mathrm{Unif}(\mathcal S_{n+m})$ and forms the Monte Carlo $p$-value

\[ p = \frac1{B+1}\Big(1+\sum_{b=1}^B \mathbb 1\{T(\pi_b u)\ge T(x,y)\}\Big). \]

This is exact, not approximate: if $\pi_0$ is the permutation with $\pi_0 u = (x,y)$, then under $H_0$ the permutations $\pi_0,\pi_1,\dots,\pi_B$ are themselves iid draws from $\mathcal S_{n+m}$, so the observed data has exactly the same chance as any of the $B$ resampled versions of producing the largest value of $T$ — giving a valid $p$-value with no asymptotics needed. (In practice one usually permutes $(x,y)$ directly rather than permuting $u$ and reapplying $\pi$; the two are equivalent, since $\tilde\pi_b=\pi_b\circ\pi_0^{-1}$ are again iid uniform on $\mathcal S_{n+m}$.) This idea — sampling from the exact null distribution rather than computing its quantiles analytically — is the Monte Carlo test, of which the permutation test is one instance; the same principle extends to settings where Markov chain Monte Carlo is needed to sample from the null when direct sampling isn't available.

It matters which statistic we condition on. $U(Z)$, the pooled order statistics, is sufficient for $Z$ under $H_0$ only — it is not sufficient once $P$ and $Q$ are allowed to differ, which is exactly why conditioning on it still leaves room to detect a difference. Compare this with conditioning on the order statistics of each sample separately, $V(X,Y)=(X_{(1)},\dots,X_{(n)},Y_{(1)},\dots,Y_{(m)})$: given $V$, the distribution of $(X,Y)$ would be completely known whether or not $H_0$ holds, collapsing the entire model — null and alternative alike — to a single point and destroying all information in the data. Conditioning on the right statistic removes the nuisance parameter (here, the unknown common distribution under $H_0$) without also erasing the distinction between the hypotheses; conditioning on too much erases everything.

Sources¶

Berkeley STAT 210A course reader, "Testing with Nuisance Parameters," as it appears across three offerings of the course (CC BY 4.0 throughout, converted 2026-09-18):

Fall 2024: reader/testing-nuisance.qmd, split into 01-nuisance-parameters.md, 02-theorem-informal.md, 03-proof-sketch.md. This is the source for the formal statement of the UMPU/UMP theorem, the proof-sketch steps (including the reference to Keener's Statistical Theory, Theorems 12.4 and 12.22, which the lecture cites but which is not itself part of the supplied material), and the normal-mean-with-unknown-variance example.
Fall 2025: reader/testing-nuisance.html and the duplicate units/reader/testing-nuisance.html (same content, split into 01-1-nuisance-parameters.md, 02-2-theorem-informal.md, 03-3-proof-sketch.md). This is the source for the fuller prose motivation: the definition of nuisance parameters, the three opening examples, the random-sample-size conditioning examples, and the conditionality principle.
Fall 2026: reader/testing-nuisance.qmd, split into 01-introduction.md, 02-conditional-testing.md, 03-multiparameter-exponential-families.md, 04-permutation-tests.md. Same content as the fall-2025 prose version, reorganized into shorter pages; used here for the multiparameter exponential family derivation and the permutation-test section.

Two things the lecture pointed to but the supplied material does not contain: the fall-2024 and fall-2025 "Geometric Picture" subsections mark the spot of a figure with the placeholder "[Insert geometric picture here]" and no image is present in the converted source — the diagram above reconstructs the projection argument described in the surrounding text, not the original figure. Separately, the fall-2025/fall-2026 pages embed an interactive Observable JS widget that lets a reader vary $n$, $\pi_2$, and the odds ratio $\rho$ to see how the conditional distribution of $X_1$ given $X_1+X_2$ depends only on $\rho$; the widget's code is in the source but is not reproducible here and is described in prose instead, in the two-sample binomial remark. Finally, the two-sample binomial comparison is noted in the fall-2025/2026 text as "shown in the problem set" — that problem set was not included among the supplied material, so no exercise is given for it here.

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