Refresher: Probability and Statistics for MVA

Contents

• Introduction and overview
• 0. Review: random variables, laws and moments
• 1. Conditional expectation and Gaussian vectors
• 2. Modes of convergence and limit theorems
• 3. Concentration inequalities
• 4. Point estimation: likelihood, Fisher information, Cramér–Rao
• 5. Confidence intervals and hypothesis tests
• 6. Linear regression
• 7. Bayesian basics
• 8. Martingales and Markov chains
• 9. Research directions and open problems
• Exercises
• Projects and assessment
• References

Introduction and overview

Audience: a student entering the MVA with a good M1 in mathematics, computer science or physics, who wants to check, before the first lectures, the probability and statistics that the MVA courses use without recalling them.

Empirical first: a classifier answers 912 of 1,000 test images correctly, and the published baseline is 90 %. Is the new model better? The central limit theorem gives the 95 % interval [0.894, 0.930] for its true accuracy. Hoeffding’s inequality gives the wider, non-asymptotic interval [0.869, 0.955]. Both intervals contain 0.90: on this test set, a gain of 1.2 points is within the noise. Chapters 2, 3 and 5 explain the two intervals, and ch. 7 gives the Bayesian reading, a posterior probability of 0.89 that the accuracy exceeds 90 %.

This page is the probability and statistics refresher of these MVA notes. It reviews random variables and conditional expectation, Gaussian vectors, the limit theorems, the concentration inequalities of learning theory, estimation by maximum likelihood, confidence intervals and tests, linear regression, Bayesian inference, martingales and Markov chains. Each tool is stated with its hypotheses, proved or sketched, computed on numbers and checked in Python.

It is one of three refreshers, with Linear Algebra, Analysis and Optimization and Scientific Python, NumPy and PyTorch. The table of the next section names, for each tool, the MVA courses that use it.

What you will be able to do

• Compute laws, moments and conditional expectations of discrete and continuous random variables, including after a change of variables.
• Condition a Gaussian vector on some of its coordinates, and read the result as a linear regression with a residual variance.
• Distinguish almost-sure, L^p, in-probability and in-law convergence, and apply the law of large numbers, the central limit theorem, Slutsky’s lemma and the delta method.
• Bound the deviations of an empirical mean with the Markov, Chebyshev, Hoeffding and Bernstein inequalities, and choose the bound that fits the variance at hand.
• Recognise a sub-Gaussian variable and bound the expected maximum of many of them.
• Derive maximum-likelihood estimators and their Fisher information, apply the Cramér–Rao bound and state the asymptotic normality of the maximum-likelihood estimator.
• Build confidence intervals and tests, read a p-value correctly and correct for multiple testing.
• Fit a linear regression, quantify the uncertainty of its coefficients, and connect ridge regression with a Gaussian prior.
• Compute a posterior in a conjugate model and sample it with the Metropolis–Hastings algorithm.
• Use martingales (optional stopping, the Azuma–Hoeffding inequality) and the convergence theorem for Markov chains.

Official objectives and syllabus (2026–2027)

The planning page of master-mva.com announces refresher courses (“cours de remise à niveau”) from 14 to 24 September 2026. They are optional and without registration, and mandatory for the I&I track. No programme, volume or instructor is published for them.

This page therefore reconstructs no official syllabus: it is independent support, written for the level that the MVA courses assume. The table below replaces the usual correspondence table. For each tool of the page, it names MVA courses of 2026–2027 that use it, according to their official descriptions or to their pages in these notes.

Beyond the level of a refresher, ch. 9 surveys research directions (self-study): heavy-tailed mean estimation, conformal prediction, anytime-valid inference, high-dimensional regression and Bayesian asymptotics.

Where this course fits in the programme

Official prerequisites: none published. The page assumes a first course in probability and statistics at bachelor level; ch. 0 reviews what it needs, and the rest is built from there.

• The other refreshers — Linear Algebra, Analysis and Optimization supplies the spectral theorem, projections and positive semidefinite matrices used for Gaussian vectors and regression. Scientific Python, NumPy and PyTorch supplies the NumPy and SciPy used by every code block.
• Semester 1 courses that rely on it — Introduction to Statistical Learning, Probabilistic Graphical Models, Computational Statistics, Stochastic Calculus in Machine Learning, Theoretical Foundations of Deep Learning, Statistical Learning with Extreme Values, Image Denoising and Machine Learning for Time Series.
• Semester 2 courses that rely on it — Bayesian Machine Learning, Reinforcement Learning, Sequential Learning, Kernel Methods, Large Random Matrices, Biostatistics, Detection Theory and Inverse Problems and Imaging.

Course outline

• 0. Review: random variables, laws and moments — probability spaces, laws, expectation, changes of variables, independence, Jensen, Markov and Chebyshev.
• 1. Conditional expectation and Gaussian vectors — conditional expectation as an orthogonal projection, the law of total variance, Gaussian vectors and the conditioning formula.
• 2. Modes of convergence and limit theorems — the four modes and their implications, the laws of large numbers, the central limit theorem, Slutsky’s lemma and the delta method.
• 3. Concentration inequalities — the Chernoff method, sub-Gaussian variables, Hoeffding’s and Bernstein’s inequalities, maxima of sub-Gaussian variables.
• 4. Point estimation — bias and variance, maximum likelihood, score and Fisher information, the Cramér–Rao bound and the asymptotic normality of the maximum-likelihood estimator.
• 5. Confidence intervals and hypothesis tests — pivots, Student intervals, the Neyman–Pearson lemma, p-values, the likelihood-ratio test and multiple testing.
• 6. Linear regression — least squares as a projection, the Gauss–Markov theorem, inference in the Gaussian linear model, ridge regression.
• 7. Bayesian basics — priors and posteriors, conjugate models, Bayes estimators, credible intervals, the Bernstein–von Mises theorem.
• 8. Martingales and Markov chains — optional stopping, the Azuma–Hoeffding and McDiarmid inequalities, invariant laws, the ergodic theorem, Metropolis–Hastings.
• 9. Research directions and open problems — where the classical tools stop and current research continues (self-study).

Session plan

12 h of guided self-study = 4 steps of 3 h. The master-mva.com planning announces the refresher fortnight (14–24 September 2026) without publishing its content or volume; the 12 h volume comes from the catalogue of these notes, and the split into four steps is a proposal. The four steps cover the tools that almost every MVA course assumes; the later chapters are complements, to read when a course of the table above needs them.

Level 1 = foundations (reviews, definitions, first computations), 2 = core results, 3 = advanced or research-level material; the level never goes down from one step to the next.

Self-study, outside the 12 h (about 12 h in total, to read on demand): ch. 5 Confidence intervals and tests (≈ 3 h, Ex. 10, 13) · ch. 6 Linear regression (≈ 2.5 h, Ex. 11) · ch. 7 Bayesian basics (≈ 2 h, Ex. 12) · ch. 8 Martingales and Markov chains (≈ 3 h, Ex. 14–15) · ch. 9 Research directions and open problems (≈ 1.5 h, Ex. 16–17).

How to study this module

• Start with the diagnostic self-check. The question numbers point to chapters: skip a chapter whose questions you answer in a minute, and work through the others.
• Read each chapter with pen and paper: redo the proofs of the boxed theorems before reading them, then compare.
• Run every code block, then change one parameter (a sample size, a variance, a seed) and predict the new output before running it again.
• After each step of the session plan, do the exercises listed in its Practice column; the solutions follow each statement.
• Read the self-study chapters when a course needs them: ch. 5 before a course with tests, ch. 7–8 before Bayesian or sampling courses.
• The code needs Python 3 with NumPy and SciPy only; it runs on a laptop in a few seconds.

0. Review: random variables, laws and moments

Every later chapter handles random variables, their laws and their moments without recalling the definitions. This review fixes the notation and proves the few facts used everywhere else: changes of variables, independence, and the inequalities of Jensen, Markov and Chebyshev.

The laws used most often on this page:

For independent copies X_1, \dots, X_n of a variable with mean \mu and variance \sigma^2, the empirical mean \bar X_n = \frac1n \sum_i X_i has variance \sigma^2 / n. Chebyshev’s inequality then gives the first deviation bound of the page.

The code below samples the exponential law by inversion (Example 0.4) and compares its true tail with the bounds of Theorem 0.8. Both bounds hold, and both are far from the true value e^{-4} \approx 0.018: ch. 3 explains how to do better.

import numpy as np
rng = np.random.default_rng(0)
lam = 2.0
x = -np.log(rng.uniform(size=1_000_000)) / lam   # inverse transform: Exp(lam)
print(f"mean {x.mean():.4f} (exact {1/lam}), variance {x.var():.4f} (exact {1/lam**2})")
t = 2.0
print(f"P(X >= {t}): empirical {np.mean(x >= t):.4f}, exact {np.exp(-lam*t):.4f}")
print(f"Markov bound {(1/lam)/t:.4f}, Chebyshev bound {(1/lam**2)/(t - 1/lam)**2:.4f}")
# mean 0.4999 (exact 0.5), variance 0.2502 (exact 0.25)
# P(X >= 2.0): empirical 0.0183, exact 0.0183
# Markov bound 0.2500, Chebyshev bound 0.1111

1. Conditional expectation and Gaussian vectors

Predicting Y from X in mean square, denoising an image or filtering a trajectory all come down to computing \mathbb{E}[Y \mid X]. This chapter defines the conditional expectation in general, gives its rules of calculation, and computes it in closed form for Gaussian vectors, the family in which it is affine.

The function m is the regression function of statistical learning, and the minimum-mean-square-error denoiser of image restoration.

Joint Gaussianity matters in item 2. Let X \sim \mathcal{N}(0, 1) and let \varepsilon = \pm 1 be an independent fair sign. Then Y = \varepsilon X is \mathcal{N}(0, 1) and uncorrelated with X, yet |Y| = |X|.

The pair is not a Gaussian vector: X + Y = 0 with probability 1/2, which no non-degenerate Gaussian variable does.

The conditional mean is affine in x, and the conditional covariance, the Schur complement of \Sigma_{XX}, does not depend on x. The Kalman filter, Gaussian-process regression and the Wiener denoiser are this theorem applied to different blocks.

The code samples the pair through a Cholesky factor and checks both numbers: once by averaging Y over a thin window around X = 3, and once by fitting the least-squares line of Y on X, which estimates the affine conditional mean.

import numpy as np
rng = np.random.default_rng(1)
mu = np.array([1.0, -1.0])
Sigma = np.array([[4.0, 2.4], [2.4, 4.0]])        # standard deviations 2, correlation 0.6
L = np.linalg.cholesky(Sigma)                       # Sigma = L L^T
Z = mu + rng.standard_normal((1_000_000, 2)) @ L.T  # rows ~ N(mu, Sigma)
x, y = Z[:, 0], Z[:, 1]
near = np.abs(x - 3.0) < 0.05                       # condition on X close to 3
m, se = y[near].mean(), y[near].std() / np.sqrt(near.sum())
print(f"E[Y | X=3] in a window: {m:.3f} +/- {2 * se:.3f}   (formula 0.200)")
slope, intercept = np.polyfit(x, y, 1)              # least-squares line of y on x
res = y - (slope * x + intercept)
print(f"slope {slope:.3f} (formula 0.600), residual variance {res.var():.3f} (formula 2.560)")
# E[Y | X=3] in a window: 0.219 +/- 0.029   (formula 0.200)
# slope 0.598 (formula 0.600), residual variance 2.554 (formula 2.560)

2. Modes of convergence and limit theorems

Most estimators in these notes are averages, or smooth functions of averages. This chapter says in which sense an average converges to its mean (laws of large numbers), at which scale it fluctuates around it (central limit theorem), and how both limits pass through smooth functions (Slutsky’s lemma, delta method).

Three counterexamples, with U uniform on (0, 1):

• X_n = n\,\mathbf{1}\{U < 1/n\} tends to 0 almost surely, but \mathbb{E}X_n = 1: no convergence in L^1.
• Independent X_n \sim \mathcal{B}(1/n) tend to 0 in every L^p, but not almost surely (Lemma 2.3 below, second part).
• If X \sim \mathcal{N}(0, 1), the constant sequence X_n = -X tends to X in law, but not in probability.

The Berry–Esseen theorem quantifies the speed. In dimension 1, the distance between the distribution function of the standardized mean and \Phi is at most C\rho/(\sigma^3\sqrt{n}), with \rho = \mathbb{E}|X - \mu|^3 and C a universal constant.

The code standardizes means of exponential variables, a skewed law, and measures the largest gap between their distribution function and \Phi (the Kolmogorov–Smirnov distance). The gap shrinks roughly like 1/\sqrt{n}, as Berry–Esseen predicts.

import numpy as np
from scipy import stats
rng = np.random.default_rng(2)
for n in [5, 30, 200]:
    xbar = rng.exponential(1.0, size=(20_000, n)).mean(axis=1)
    z = np.sqrt(n) * (xbar - 1.0)                 # mean 1 and variance 1 for Exp(1)
    ks = stats.kstest(z, "norm").statistic        # sup distance between cdfs
    inside = np.mean(np.abs(z) <= 1.96)
    print(f"n = {n:3d}   KS distance {ks:.3f}   P(|Z_n| <= 1.96) = {inside:.3f}")
# n =   5   KS distance 0.063   P(|Z_n| <= 1.96) = 0.955
# n =  30   KS distance 0.021   P(|Z_n| <= 1.96) = 0.952
# n = 200   KS distance 0.008   P(|Z_n| <= 1.96) = 0.951

3. Concentration inequalities

The central limit theorem describes fluctuations only in the limit. Learning theory, bandits and the analysis of algorithms need bounds valid for every n, with explicit constants. This chapter derives them from one idea: apply Markov’s inequality to an exponential.

Hoeffding’s inequality only sees the range of the variables. When the variance is much smaller than the squared range, as for rare events, Bernstein’s inequality is much sharper.

For small t, the bound is Gaussian with the true variance v; for large t, it decays exponentially. Union bounds over many variables cost only a logarithm:

The \log N is the price of choosing the best of N predictors on the same data, the first step of generalization bounds in statistical learning.

The code compares the bounds on \mathbb{P}(\bar X_n - p \ge 0.02) for n = 1000 Bernoulli variables with p = 0.05. The variance p(1-p) = 0.0475 is far below the squared range 1: Bernstein wins by a factor 18 over Hoeffding.

import numpy as np
from scipy import stats
n, p, t = 1000, 0.05, 0.02
exact = stats.binom.sf(np.ceil(n * (p + t)) - 1, n, p)   # P(mean - p >= t)
cheb = p * (1 - p) / (n * t**2)
hoeff = np.exp(-2 * n * t**2)
bern = np.exp(-n * t**2 / (2 * (p * (1 - p) + t / 3)))   # |X_i - p| <= 1
print(f"exact {exact:.1e}  Chebyshev {cheb:.1e}  Hoeffding {hoeff:.1e}  Bernstein {bern:.1e}")
# exact 3.5e-03  Chebyshev 1.2e-01  Hoeffding 4.5e-01  Bernstein 2.5e-02

4. Point estimation: likelihood, Fisher information, Cramér–Rao

A statistical model is a family of laws indexed by a parameter, and the data are drawn from one of them. This chapter measures the quality of an estimator of the parameter, builds the maximum-likelihood estimator, and proves that no unbiased estimator can beat the Fisher information, a bound that maximum likelihood attains asymptotically.

The usual cases follow by setting the derivative of \ell_n to zero:

• Bernoulli \mathcal{B}(p) and Poisson \mathcal{P}(\lambda): the MLE is the empirical mean \bar X_n;
• exponential \mathcal{E}(\lambda): \hat\lambda = 1/\bar X_n;
• Gaussian \mathcal{N}(\mu, \sigma^2): \hat\mu = \bar X_n, and \hat\sigma^2 of Example 4.2.

For a vector parameter, the same argument gives \mathrm{Cov}_\theta(T) \succeq I(\theta)^{-1}/n for every unbiased T of \theta. An unbiased estimator that attains the bound is called efficient.

The theorem gives the Wald interval \hat\theta_n \pm 1.96/\sqrt{n I(\hat\theta_n)}, and it says that maximum likelihood attains the Cramér–Rao bound in the limit.

The code checks it on the Cauchy location model, p_\theta(x) = 1/(\pi(1 + (x - \theta)^2)), whose Fisher information is 1/2. The sample mean is useless here: a mean of Cauchy variables is again Cauchy. The median is consistent, with asymptotic variance \pi^2/(4n); the MLE does better, with 2/n.

import numpy as np
from scipy import optimize
rng = np.random.default_rng(4)

def mle_cauchy(x):
    nll = lambda th: np.sum(np.log1p((x - th) ** 2))   # minus log-likelihood + const
    m = np.median(x)
    return optimize.minimize_scalar(nll, bounds=(m - 2, m + 2), method="bounded").x

n, reps = 200, 2000
X = rng.standard_cauchy((reps, n))                     # true location 0
mle = np.array([mle_cauchy(x) for x in X])
med = np.median(X, axis=1)
print(f"n Var(MLE)    = {n * mle.var():.2f}   (1/I = 2)")
print(f"n Var(median) = {n * med.var():.2f}   (pi^2/4 = {np.pi**2 / 4:.2f})")
# n Var(MLE)    = 2.08   (1/I = 2)
# n Var(median) = 2.59   (pi^2/4 = 2.47)

5. Confidence intervals and hypothesis tests

Self-study chapter. An estimate without its uncertainty cannot support a decision. This chapter turns the laws of ch. 2 and ch. 4 into intervals that contain the parameter with a prescribed probability, and into tests that control the probability of a false discovery, including when many tests are run at once.

When no exact pivot exists, the asymptotic normality of the MLE (Theorem 4.8) gives the Wald interval \hat\theta \pm z_{1-\alpha/2}/\sqrt{n I(\hat\theta)}. Its coverage can be poor at moderate n; the code at the end of the chapter measures it for a proportion.

A p-value is therefore a random variable, uniform when nothing happens. It is not the probability that H_0 is true, and a small p-value says nothing about the size of an effect (Wasserstein and Lazar, 2016).

Tests and intervals are two faces of one object. If, for every \theta_0, R(\theta_0) is a level-\alpha test of \theta = \theta_0, the set of \theta_0 that are not rejected is a confidence region of level 1 - \alpha.

Running many tests multiplies the chances of a false discovery. With m = 20 independent tests at level 0.05 and no true effect, the probability of at least one rejection is 1 - 0.95^{20} = 0.64.

The code computes the exact coverage of two 95 % intervals for a proportion with n = 50, summing over the 51 possible outcomes. The Wald interval \hat p \pm 1.96\sqrt{\hat p(1-\hat p)/n} undercovers, badly near 0. The Wilson interval inverts the score test, that is, solves |\hat p - p| \le z\sqrt{p(1-p)/n} for p:

\frac{\hat p + \frac{z^2}{2n}}{1 + \frac{z^2}{n}} \ \pm\ \frac{z}{1 + \frac{z^2}{n}} \sqrt{\frac{\hat p(1 - \hat p)}{n} + \frac{z^2}{4n^2}}.

import numpy as np
from scipy import stats
n, z = 50, stats.norm.ppf(0.975)
k = np.arange(n + 1)                                  # every possible number of successes
ph = k / n
half = z * np.sqrt(ph * (1 - ph) / n)
wald = (ph - half, ph + half)
c = (ph + z**2 / (2 * n)) / (1 + z**2 / n)
h = z * np.sqrt(ph * (1 - ph) / n + z**2 / (4 * n**2)) / (1 + z**2 / n)
wilson = (c - h, c + h)
for p in [0.05, 0.2, 0.5]:
    w = stats.binom.pmf(k, n, p)                      # exact coverage, no simulation
    cover = lambda I: np.sum(w * ((I[0] <= p) & (p <= I[1])))
    print(f"p = {p:.2f}   coverage: Wald {cover(wald):.3f}   Wilson {cover(wilson):.3f}")
# p = 0.05   coverage: Wald 0.920   Wilson 0.962
# p = 0.20   coverage: Wald 0.938   Wilson 0.951
# p = 0.50   coverage: Wald 0.935   Wilson 0.935

6. Linear regression

Self-study chapter. Linear regression explains a response by covariates. Least squares is the orthogonal projection of ch. 1 applied to data, and in the Gaussian model the Fisher–Cochran theorem of ch. 5 makes its uncertainty exact. Ridge regression then shows the bias–variance trade-off of ch. 4 at work.

Item 3 gives the 95 % interval and the t-test of \beta_j = 0 printed by every regression software:

\hat\beta_j \pm t_{n-p,\,0.975}\; \hat\sigma \sqrt{v_j}.

Comparing nested models uses the ratio of two independent \chi^2 variables, the F-test.

When p approaches n, or when columns of X are nearly collinear, (X^\top X)^{-1} has huge entries and least squares becomes unstable. Ridge regression adds a penalty.

Ridge trades a bias for a variance reduction, as in Example 4.2; ch. 7 reads it as a Bayesian posterior mean. The code checks Theorem 6.4 by simulation: the three t-intervals of a regression with n = 40 and p = 3 cover their coefficients 95 % of the time.

import numpy as np
from scipy import stats
rng = np.random.default_rng(6)
n, p, sigma = 40, 3, 1.5
X = np.column_stack([np.ones(n), rng.standard_normal((n, p - 1))])
beta = np.array([1.0, 2.0, 0.0])
G = np.linalg.inv(X.T @ X)
q = stats.t.ppf(0.975, n - p)
hits, reps = np.zeros(p), 5000
for _ in range(reps):
    y = X @ beta + sigma * rng.standard_normal(n)
    b = G @ X.T @ y                                   # least-squares estimate
    s2 = np.sum((y - X @ b) ** 2) / (n - p)           # unbiased noise variance
    hits += np.abs(b - beta) <= q * np.sqrt(s2 * np.diag(G))
print("coverage of the three 95% t-intervals:", np.round(hits / reps, 3))
# coverage of the three 95% t-intervals: [0.945 0.945 0.95 ]

7. Bayesian basics

Self-study chapter. The Bayesian approach treats the parameter as random, with a prior law that encodes what is known before the data. The data turn the prior into a posterior law, and every estimate, interval and decision is read off the posterior. Conditional expectation (ch. 1) and Gaussian conditioning do most of the work.

The code computes the posterior of Example 7.3. The 95 % credible interval almost coincides with the Wald interval [0.894, 0.930] of Example 2.8, as Theorem 7.7 predicts. The posterior probability that the accuracy exceeds 0.90 is 0.89: suggestive, not conclusive.

from scipy import stats
k, n = 912, 1000
post = stats.beta(1 + k, 1 + n - k)                   # uniform prior on p
lo, hi = post.ppf([0.025, 0.975])
print(f"posterior mean {post.mean():.4f}, 95% credible interval [{lo:.4f}, {hi:.4f}]")
print(f"P(p > 0.90 | data) = {post.sf(0.90):.3f}")
# posterior mean 0.9112, 95% credible interval [0.8928, 0.9280]
# P(p > 0.90 | data) = 0.891

Outside conjugate models the posterior has no closed form; its normalizing integral is the obstacle. Markov chain Monte Carlo (ch. 8) samples it without computing that integral, and variational inference approximates it by optimization.

8. Martingales and Markov chains

Self-study chapter. A gambler’s fortune, a stream of observations analysed on the fly, the trajectory of a sampler: these sequences are not independent. Martingales and Markov chains are the two structures that keep enough of independence to prove concentration and limit theorems. This chapter gives the discrete-time results that the MVA courses on stochastic calculus, sequential learning and sampling build on.

Martingales

Four martingales that recur in the MVA courses, each checked with Proposition 1.2:

• S_n = \sum_{i \le n} X_i for independent centred X_i, and S_n^2 - n\sigma^2 when they also share the variance \sigma^2;
• the likelihood ratio L_n = \prod_{i \le n} p_1(X_i)/p_0(X_i) for i.i.d. X_i with density p_0 and another density p_1 that vanishes where p_0 does, since \mathbb{E}[p_1(X)/p_0(X)] = \int p_1 = 1;
• the Doob martingale M_k = \mathbb{E}[f(X_1, \dots, X_n) \mid X_1, \dots, X_k], for k = 0, \dots, n;
• (q/p)^{S_n} for a walk with steps +1 and -1 of probabilities p and q = 1 - p.

Applied to the likelihood ratio L_n under the null, which has mean 1, Ville’s inequality says that the probability that L_n ever exceeds 1/\alpha is at most \alpha. This gives tests that stay valid however often one looks at the data, the starting point of the anytime-valid inference of ch. 9.

Two further results are used without proof in the MVA courses. Doob’s inequality bounds the maximum of a nonnegative submartingale:

\mathbb{P}\Big(\max_{k \le n} M_k \ge c\Big) \le \frac{\mathbb{E}[M_n]}{c}.

The martingale convergence theorem says that a martingale bounded in L^1 converges almost surely (Williams, 1991, ch. 11).

Markov chains

A Markov chain can thus be designed to have a prescribed invariant law \pi, and its time averages then estimate expectations under \pi. This is Markov chain Monte Carlo.

The code samples the posterior \mathrm{Beta}(913, 89) of ch. 7 with a random-walk Metropolis–Hastings chain, using only the unnormalized log-posterior. Time averages along the chain recover the exact posterior mean and tail probability (Theorem 8.10, item 2).

import numpy as np
from scipy import stats
rng = np.random.default_rng(8)
k, n = 912, 1000

def log_target(p):                                    # log posterior, up to a constant
    return k * np.log(p) + (n - k) * np.log1p(-p) if 0 < p < 1 else -np.inf

p, chain, accepted = 0.5, np.empty(50_000), 0
for t in range(chain.size):
    proposal = p + 0.02 * rng.standard_normal()       # symmetric random walk
    if np.log(rng.uniform()) < log_target(proposal) - log_target(p):
        p, accepted = proposal, accepted + 1
    chain[t] = p
kept = chain[5_000:]                                  # discard the burn-in
exact = stats.beta(k + 1, n - k + 1)
print(f"acceptance rate {accepted / chain.size:.2f}")
print(f"mean: MCMC {kept.mean():.4f}, exact {exact.mean():.4f}")
print(f"P(p > 0.9): MCMC {np.mean(kept > 0.9):.3f}, exact {exact.sf(0.9):.3f}")
# acceptance rate 0.47
# mean: MCMC 0.9111, exact 0.9112
# P(p > 0.9): MCMC 0.891, exact 0.891

9. Research directions and open problems

Self-study chapter. The tools of this page are classical, but each one has a research frontier where its hypotheses fail: heavy tails, models that are only approximately right, data analysed while they arrive, more parameters than observations. This chapter points to a few active lines, with papers to start from.

Mean estimation under heavy tails

Hoeffding’s and Bernstein’s inequalities need bounded or light-tailed variables. With only a finite variance, the empirical mean has polynomial tails: Chebyshev’s bound cannot be improved in general. Catoni (2012) built M-estimators of the mean with sub-Gaussian deviations under a finite variance only.

The median-of-means estimator, which dates back to the 1980s, achieves the same with a three-line proof (Exercise 16). Lugosi and Mendelson (2019) trace its history and survey the field, including the multivariate case.

Open questions concern the optimal constants and computation in high dimension. Hopkins (2020) gave the first polynomial-time multivariate estimator with sub-Gaussian rates, and Lee and Valiant (2021) obtained the optimal constant in dimension one.

Distribution-free prediction: conformal inference

Confidence intervals for a parameter assume a model. Conformal prediction (Vovk, Gammerman and Shafer, 2005) builds prediction sets with finite-sample coverage for any predictor and any distribution, under exchangeability alone; Exercise 17 proves the key step. Lei, G’Sell, Rinaldo, Tibshirani and Wasserman (2018) developed it for regression, and Angelopoulos and Bates (2023) give an introduction.

The guarantee is marginal, averaged over the inputs. Barber, Candès, Ramdas and Tibshirani (2021) show that coverage conditional on the input cannot be guaranteed for every distribution, except by uninformative sets. Extensions to covariate shift (Tibshirani, Barber, Candès and Ramdas, 2019) and to non-exchangeable data (Barber, Candès, Ramdas and Tibshirani, 2023) are active.

Anytime-valid inference

A p-value checked after every new batch of data, until it falls below 0.05, gives far more than 5 % false positives: the stopping time is chosen by the data. Ville’s inequality (Proposition 8.5) suggests the fix: use nonnegative supermartingales, which stay valid at any stopping time.

Confidence sequences (Howard, Ramdas, McAuliffe and Sekhon, 2021) and e-values (Ramdas, Grünwald, Vovk and Shafer, 2023) develop this into a theory of sequential inference. A/B testing on the web was one motivation (Johari, Koomen, Pekelis and Walsh, 2017). Open problems include optimal power under composite alternatives and inference under adaptive data collection.

Regression with more parameters than observations

When p \ge n, least squares interpolates the data and Proposition 6.5 no longer applies. Belkin, Hsu, Ma and Mandal (2019) documented a second descent of the test error beyond the interpolation threshold. Bartlett, Long, Lugosi and Tsigler (2020) characterized when interpolation is benign in linear regression, and Hastie, Montanari, Rosset and Tibshirani (2022) computed the asymptotic risk of minimum-norm least squares with random matrix theory.

How far these results extend to neural networks is open; the Large Random Matrices and Theoretical Foundations of Deep Learning courses meet these questions.

Inference after selection and with machine-learning predictions

Tests computed on a model chosen from the same data are not valid at their nominal level. Valid post-selection inference (Berk, Brown, Buja, Zhang and Zhao, 2013) and knockoffs, which control the false discovery rate of variable selection (Barber and Candès, 2015), are two answers.

Prediction-powered inference (Angelopoulos, Bates, Fannjiang, Jordan and Zrnic, 2023) combines a few labelled data with many machine-learning predictions while keeping valid confidence intervals.

Bayesian methods when the model is wrong, and sampling in high dimension

Under misspecification, the posterior concentrates around the Kullback–Leibler projection, but its spread is not the sampling spread of the estimator (Kleijn and van der Vaart, 2012): credible intervals can be too narrow or too wide. On the computational side, non-asymptotic bounds for Langevin-based samplers in high dimension (Durmus and Moulines, 2019) quantify how many steps an MCMC method needs; the Stochastic Calculus in Machine Learning course cites this paper.

Exercises

1. ★ Changes of variables. Let U be uniform on (0, 1) and X \sim \mathcal{N}(0, 1). (a) Find the density of U^2. (b) Show that -2\log U is exponential with mean 2. (c) Find the density of X^2, its mean and its variance.

   Solution: (a) \mathbb{P}(U^2 \le y) = \sqrt{y}, so the density is 1/(2\sqrt{y}) on (0, 1).

   2. For y > 0, -2\log U > y exactly when U < e^{-y/2}, an event of probability e^{-y/2}: the exponential law with mean 2.

   3. \mathbb{P}(X^2 \le y) = 2\Phi(\sqrt{y}) - 1; differentiating gives the density e^{-y/2}/\sqrt{2\pi y} for y > 0, the \chi^2_1 law of ch. 5. Its mean is \mathbb{E}X^2 = 1 and its variance \mathbb{E}X^4 - 1 = 3 - 1 = 2.

2. ★ Conditioning on a sum of dice. Throw two fair dice D_1, D_2 and let S = D_1 + D_2. Compute \mathbb{E}[D_1 \mid S], then \mathrm{Var}(D_1 \mid S = 4), and check the tower property.

   Solution: by symmetry \mathbb{E}[D_1 \mid S] = \mathbb{E}[D_2 \mid S], and the two sum to \mathbb{E}[S \mid S] = S; hence \mathbb{E}[D_1 \mid S] = S/2.

   Given S = 4, the outcomes (1,3), (2,2), (3,1) are equally likely: D_1 is uniform on \{1, 2, 3\}, with variance 2/3. Tower property: \mathbb{E}[S/2] = 7/2 = \mathbb{E}D_1.

3. ★ Conditioning a Gaussian triple. Let (X_1, X_2, Y) be centred Gaussian with unit variances, \mathrm{Cov}(X_1, X_2) = 0.5 and \mathrm{Cov}(X_i, Y) = 0.6. Compute \mathbb{E}[Y \mid X_1, X_2] and \mathrm{Var}(Y \mid X_1, X_2), and compare with conditioning on X_1 alone.

   Solution: apply Theorem 1.8 with

   \Sigma_{XX}^{-1} = \frac{1}{0.75}\begin{pmatrix} 1 & -0.5 \\ -0.5 & 1 \end{pmatrix}, \qquad \Sigma_{YX}\Sigma_{XX}^{-1} = (0.4,\ 0.4).

   Hence

   \mathbb{E}[Y \mid X_1, X_2] = 0.4\,(X_1 + X_2), \qquad \mathrm{Var}(Y \mid X_1, X_2) = 1 - 0.4 \times 0.6 \times 2 = 0.52.

   On X_1 alone: mean 0.6 X_1 and variance 0.64. The second observation is correlated with the first, so it reduces the variance less than an independent one would.

4. ★ Chebyshev, Hoeffding or the central limit theorem? A poll estimates a proportion within \pm 0.02 with probability 0.95. How many people are needed according to Chebyshev (worst variance 1/4), Hoeffding, and the central limit theorem?

   Solution: Chebyshev: n \ge 1/(4 \times 0.05 \times 0.02^2) = 12\,500. Hoeffding: n \ge \log(40)/(2 \times 0.02^2) = 4\,612. Central limit theorem: n \ge (1.96/0.02)^2/4 = 2\,401. The last number is the one pollsters use, but it is only asymptotic; Hoeffding’s is a guarantee.

5. ★ (Python) The delta method at work. For n = 50 exponential observations with \lambda = 1, simulate the estimator 1/\bar X_n 10,000 times; ch. 4 shows that it is the maximum-likelihood estimator (MLE). Compare its mean with n/(n-1), its standard deviation with the delta-method value \lambda/\sqrt{n} (Theorem 2.7 with g(x) = 1/x), and the coverage of the asymptotic interval with 95 %.

   Hints and expected result:

   import numpy as np
   rng = np.random.default_rng(5)
   n, lam = 50, 1.0
   est = 1 / rng.exponential(1 / lam, size=(10_000, n)).mean(axis=1)   # MLE 1 / mean
   print(f"mean of the MLE {est.mean():.4f}   (exact n/(n-1) = {n / (n - 1):.4f})")
   print(f"sd of the MLE   {est.std():.4f}   (delta method {lam / np.sqrt(n):.4f})")
   z = (est - lam) / (lam / np.sqrt(n))
   print(f"P(|z| <= 1.96) = {np.mean(np.abs(z) <= 1.96):.3f}")
   # mean of the MLE 1.0200   (exact n/(n-1) = 1.0204)
   # sd of the MLE   0.1475   (delta method 0.1414)
   # P(|z| <= 1.96) = 0.939

   The bias and the skewness of 1/\bar X_n are still visible at n = 50: the coverage is 93.9 %, not 95 %.

6. ★ A Hoeffding interval for a test error. A classifier makes 150 errors on 2,000 fresh test examples. Give the 99 % Hoeffding interval for its error rate and compare it with the interval from the central limit theorem.

   Solution: \hat p = 0.075. Hoeffding’s half-width is \sqrt{\log(200)/4000} = 0.036, giving [0.039, 0.111]. The central limit theorem gives the half-width 2.576\sqrt{0.075 \times 0.925/2000} = 0.015, giving [0.060, 0.090]. Hoeffding ignores that the variance p(1-p) is small; Bernstein-type bounds close most of the gap.

7. ★ Maximum likelihood for the exponential law. Derive the MLE of \lambda for i.i.d. \mathcal{E}(\lambda) data, its Fisher information and the asymptotic 95 % interval, then apply it with n = 40 and \bar X = 2.5.

   Solution: \ell_n(\lambda) = n\log\lambda - \lambda\sum_i X_i gives \hat\lambda = 1/\bar X = 0.4. The second derivative of \log p_\lambda is -1/\lambda^2, so I(\lambda) = 1/\lambda^2.

   Theorem 4.8 gives the interval below, that is [0.276, 0.524]:

   \hat\lambda \pm \frac{1.96\, \hat\lambda}{\sqrt{n}} = 0.4 \pm 0.124.

8. ★★ Sub-Gaussian maxima. (a) Show that a fair sign \varepsilon satisfies \cosh\lambda \le e^{\lambda^2/2}, the moment generating function bound of Definition 3.2. (b) For N standard Gaussian variables, show that the expected maximum is at most \sqrt{2\log N}, and evaluate it for N = 1000. (c) Why is independence not needed?

   Solution: (a) Compare the series term by term: \lambda^{2k}/(2k)! \le \lambda^{2k}/(2^k k!), because (2k)! \ge 2^k k!. (b) Proposition 3.8 with \sigma = 1: the bound is \sqrt{2\log 1000} = 3.72, while the true expected maximum of 1000 independent standard Gaussians is about 3.24. (c) The proof bounds a maximum by a sum inside Jensen’s inequality, which needs no independence.

9. ★★ Efficiency of variance estimators. For \mathcal{N}(\mu, v) data with \mu known, compute the Fisher information about v = \sigma^2 and show that \frac1n\sum_i (X_i - \mu)^2 is efficient. With \mu unknown, compare \mathrm{Var}(S^2) with the bound.

   Solution: \log p_v(x) = -\frac12\log(2\pi v) - (x - \mu)^2/(2v), whose score is -\frac{1}{2v} + \frac{(x - \mu)^2}{2v^2}. Its variance is \mathrm{Var}((X - \mu)^2)/(4v^4) = 2v^2/(4v^4), so I(v) = 1/(2v^2) and the bound is 2v^2/n.

   The estimator with known \mu is unbiased with variance 2v^2/n: efficient. With \mu unknown, \mathrm{Var}(S^2) = 2v^2/(n-1) exceeds the bound, by a ratio that tends to 1.

10. ★★ Power and sample size of a z-test. Data X_i \sim \mathcal{N}(\mu, 1); test \mu = 0 against \mu > 0 at level 5 %.

    1. Compute the power at \mu = 0.3 for n = 50. (b) Find the smallest n giving power 0.9 at \mu = 0.3.

    Solution: (a) Example 5.9 gives the power \Phi(0.3\sqrt{50} - 1.645) = \Phi(0.476) = 0.683. (b) One needs 0.3\sqrt{n} \ge 1.645 + 1.282 = 2.927, that is n \ge 95.2, so n = 96.

11. ★★ (Python) Ridge against least squares. Simulate a regression with n = 50 observations, p = 40 strongly correlated covariates and \beta \sim \mathcal{N}(0, I_p). Estimate the risk \mathbb{E}\|\hat\beta_\lambda - \beta\|^2 over a grid of \lambda, and relate the result to Proposition 6.7.

    Hints and expected result:

    import numpy as np
    rng = np.random.default_rng(11)
    n, p, sigma = 50, 40, 1.0
    C = 0.9 ** np.abs(np.subtract.outer(np.arange(p), np.arange(p)))   # correlated covariates
    X = rng.multivariate_normal(np.zeros(p), C, size=n)
    beta = rng.standard_normal(p)
    lams = [0.0, 0.3, 1.0, 3.0, 10.0, 30.0]
    err = np.zeros(len(lams))
    for _ in range(500):
        y = X @ beta + sigma * rng.standard_normal(n)
        for j, lam in enumerate(lams):
            b = np.linalg.solve(X.T @ X + lam * np.eye(p), X.T @ y)       # ridge; OLS at 0
            err[j] += np.sum((b - beta) ** 2) / 500
    for lam, e in zip(lams, err):
        print(f"lambda = {lam:5.1f}   E||b - beta||^2 = {e:6.2f}")
    # lambda =   0.0   E||b - beta||^2 =  47.58
    # lambda =   0.3   E||b - beta||^2 =  17.58
    # lambda =   1.0   E||b - beta||^2 =  16.57
    # lambda =   3.0   E||b - beta||^2 =  19.17
    # lambda =  10.0   E||b - beta||^2 =  24.94
    # lambda =  30.0   E||b - beta||^2 =  31.04

    The risk is U-shaped: the small eigenvalues d_j of a correlated design make the variance term \sigma^2/d_j of least squares explode, and a moderate \lambda removes most of it at the price of a small bias.

12. ★★ Laplace’s rule of succession. Take a uniform prior on p and observe k successes in n Bernoulli trials.

    1. Show that the posterior probability that trial n + 1 succeeds is (k + 1)/(n + 2). (b) For k = 0 and n = 10, find the 95 % one-sided upper credible bound for p.

    Solution: (a) The predictive probability is the posterior mean of p, here (1 + k)/(2 + n) by Example 7.3.

    2. The posterior is \mathrm{Beta}(1, 11), with \mathbb{P}(p \le u) = 1 - (1 - u)^{11}. Setting this to 0.95 gives u = 1 - 0.05^{1/11} = 0.238, close to the “rule of three” 3/n = 0.3. The MLE, 0, is useless here.

13. ★★ Bonferroni or Benjamini–Hochberg? Ten hypotheses give the sorted p-values 0.001, 0.006, 0.009, 0.020, 0.041, 0.10, 0.27, 0.48, 0.66, 0.91. Which are rejected at \alpha = 0.05 by Bonferroni, and by Benjamini–Hochberg? With 20 independent true nulls tested at 0.05 each, what is the family-wise error rate?

    Solution: Bonferroni’s threshold is 0.005: one rejection. Benjamini–Hochberg compares p_{(i)} with 0.005\, i; the largest i that passes is i = 4, since 0.020 \le 0.020, so four rejections.

    Without correction, the family-wise error rate is 1 - 0.95^{20} = 0.64; with Bonferroni, it is 1 - 0.9975^{20} = 0.049.

14. ★★ Ruin with a small edge. A walk takes steps +1 with probability p = 0.49 and -1 with probability q = 0.51. It starts at x = 10 and stops at 0 or at N = 20.

    Show that (q/p)^{S_n} is a martingale and compute the probability of reaching N first.

    Solution: the next step multiplies (q/p)^{S_n} by q/p or by p/q, so

    \mathbb{E}\big[(q/p)^{S_{n+1}} \mid \mathcal{F}_n\big] = (q/p)^{S_n}\Big(p \cdot \frac{q}{p} + q \cdot \frac{p}{q}\Big) = (q/p)^{S_n}.

    The stopped martingale is bounded, so Theorem 8.3 gives, with r = q/p,

    \mathbb{P}(\text{reach } N \text{ first}) = \frac{1 - r^{x}}{1 - r^{N}} = 0.401.

    A 1 % edge per step moves the fair answer 0.5 to 0.40.

15. ★★ (Python) A Markov chain and its invariant law. For the three-state chain below, compute the invariant law, check that the rows of P^{20} approach it, compare with the occupation frequencies of a simulated path, and test detailed balance.

    Hints and expected result:

    import numpy as np
    P = np.array([[0.7, 0.2, 0.1],
                  [0.3, 0.4, 0.3],
                  [0.2, 0.3, 0.5]])
    w, V = np.linalg.eig(P.T)                         # pi P = pi: left eigenvector for 1
    pi = np.real(V[:, np.argmin(np.abs(w - 1))]); pi /= pi.sum()
    print("invariant law", np.round(pi, 4))
    print("row of P^20  ", np.round(np.linalg.matrix_power(P, 20)[0], 4))
    rng = np.random.default_rng(15)
    x, counts = 0, np.zeros(3)
    for _ in range(100_000):
        x = rng.choice(3, p=P[x]); counts[x] += 1
    print("frequencies  ", np.round(counts / counts.sum(), 4))
    F = pi[:, None] * P                               # flows pi(x) P(x, y)
    print("reversible:", np.allclose(F, F.T))
    # invariant law [0.4565 0.2826 0.2609]
    # row of P^20   [0.4565 0.2826 0.2609]
    # frequencies   [0.4598 0.2803 0.2599]
    # reversible: False

    The chain is irreducible and aperiodic (self-loops), so Theorem 8.10 applies; it is not reversible, which shows that detailed balance is sufficient but not necessary for invariance.

16. ★★★ Median of means. Split n = km i.i.d. observations with mean \mu and variance \sigma^2 into k blocks of size m.

    Let \hat\mu be the median of the k block means. Show that

    \mathbb{P}\Big(|\hat\mu - \mu| > 2\sigma/\sqrt{m}\Big) \le e^{-k/8},

    and deduce a deviation of order \sigma\sqrt{\log(1/\delta)/n} with probability 1 - \delta, under a finite variance only. Then compare the empirical mean and the median of means on Student data with 3 degrees of freedom.

    Hints and expected result: by Chebyshev, each block mean is off by more than 2\sigma/\sqrt{m} with probability at most 1/4. The median is that far off only if at least k/2 blocks are, and Hoeffding’s inequality on the number of bad blocks gives e^{-2k(1/4)^2}. Take k = \lceil 8\log(1/\delta) \rceil.

    In simulations with n = 960, k = 24 blocks of m = 40 (that is, \delta = 0.05) and 10^5 repetitions, the empirical mean is better in the bulk. Its standard deviation is 0.056 against 0.063, and its 99 % quantile of |\hat\mu - \mu| is 0.146 against 0.163.

    The median of means wins only far in the tail: its 99.99 % quantile is about 0.25, against more than 0.30 for the mean. Its guarantee is about rare large deviations, not typical errors. Lugosi and Mendelson (2019) discuss the constants (ch. 9).

17. ★★★ Why conformal prediction works. Let S_1, \dots, S_{n+1} be exchangeable real scores, almost surely distinct. Show that

    \mathbb{P}\Big(S_{n+1} \le S_{(\lceil (n+1)(1-\alpha) \rceil)}\Big) \ge 1 - \alpha,

    where S_{(j)} is the j-th smallest of S_1, \dots, S_n and \alpha \ge 1/(n+1), so that the index is at most n. Turn this into a prediction interval for regression and discuss what the guarantee does not say.

    Hints and expected result: by exchangeability, the rank of S_{n+1} among the n + 1 scores is uniform on \{1, \dots, n + 1\}. The event holds when this rank is at most j = \lceil (n+1)(1-\alpha) \rceil, which has probability j/(n+1) \ge 1 - \alpha.

    With residual scores |y - \hat f(x)| computed on a calibration set independent of the fit of \hat f, the interval \hat f(x) \pm S_{(j)} covers a new response with probability at least 1 - \alpha, for any \hat f and any distribution. The coverage is marginal, not conditional on x (ch. 9).

Projects and assessment

No assessment is published for the refresher courses (see the fiche): they are optional, without registration, and mandatory only for the I&I track. This section offers two tools instead, neither of them official. The diagnostic self-check, in the format of a short written exam, tells you which chapters to work on before the semester starts. The two small practice projects apply the page to realistic questions, at the scale of an afternoon or two.

Diagnostic self-check

Format: 60 minutes, no documents, 13 short questions, 20 points; each question names the chapter that answers it. Grade yourself with the expected answers. Below 12/20, follow the four steps of the session plan in order; between 12 and 16, work on the chapters of the missed questions; above 16, the recap boxes are enough.

1. Question (1.5 points) — ch. 0. Let U be uniform on (0, 1). What is the law of -\log(U)/3? Give its mean and its variance.

   Expected: by Example 0.4, the exponential law with parameter 3, with mean 1/3 and variance 1/9.

2. Question (1.5 points) — ch. 1. Throw a fair die, get X, then toss X fair coins and let Y be the number of heads. Compute the mean and the variance of Y.

   Expected: \mathbb{E}[Y \mid X] = X/2 and \mathrm{Var}(Y \mid X) = X/4. Hence \mathbb{E}Y = 1.75, and by the law of total variance (Proposition 1.4), \mathrm{Var}(Y) = 3.5/4 + (35/12)/4 = 1.604.

3. Question (2 points) — ch. 1. Let (X, Y) be a centred Gaussian vector with variances 1 and 2, and covariance 1. Give the law of Y given X = 2. Are Y - X and X independent?

   Expected: Theorem 1.8 gives \mathcal{N}(2, 1). \mathrm{Cov}(Y - X, X) = 1 - 1 = 0, and (Y - X, X) is a Gaussian vector, so they are independent (Proposition 1.7).

4. Question (1.5 points) — ch. 2. Give a sequence that converges in probability but not almost surely, and one that converges in law but not in probability.

   Expected: independent X_n \sim \mathcal{B}(1/n), by the second Borel–Cantelli lemma; X_n = -X with X \sim \mathcal{N}(0, 1), which has the law of X but stays at distance 2|X| from it.

5. Question (2 points) — ch. 2. Let X_i be i.i.d. with mean \mu \ne 0 and variance \sigma^2. What is the asymptotic law of \sqrt{n}\,(\bar X_n^2 - \mu^2)?

   Expected: the delta method (Theorem 2.7) with g(x) = x^2 and g'(\mu) = 2\mu gives \mathcal{N}(0, 4\mu^2\sigma^2).

6. Question (1.5 points) — ch. 3. How many i.i.d. observations with values in [0, 1] does Hoeffding’s inequality require to estimate their mean within 0.05 with probability 0.95?

   Expected: 2e^{-2n(0.05)^2} \le 0.05, that is n \ge \log(40)/0.005 = 737.8, so n = 738.

7. Question (2 points) — ch. 3. Prove that \mathbb{P}(X \ge t) \le e^{-t^2/(2\sigma^2)} for X \sim \mathcal{N}(0, \sigma^2) and t > 0.

   Expected: \mathbb{E}e^{\lambda X} = e^{\lambda^2\sigma^2/2}. Markov’s inequality on e^{\lambda X} gives the bound e^{-\lambda t + \lambda^2\sigma^2/2} for every \lambda > 0, and \lambda = t/\sigma^2 minimizes it (Propositions 3.1 and 3.3).

8. Question (2 points) — ch. 4. For i.i.d. Poisson \mathcal{P}(\lambda) data, give the MLE, the Fisher information, and say whether the MLE is efficient.

   Expected: \hat\lambda = \bar X_n and I(\lambda) = 1/\lambda. The MLE is unbiased with variance \lambda/n, the Cramér–Rao bound: it is efficient (Example 4.7).

9. Question (1.5 points) — ch. 4. True or false: an unbiased estimator always has a smaller quadratic risk than a biased one.

   Expected: false. The risk is the squared bias plus the variance; for Gaussian data, the biased variance estimator with factor 1/n has a smaller risk than S^2 (Example 4.2).

10. Question (1.5 points) — ch. 5. A test returns p = 0.03. Which statements are correct? (a) The null hypothesis has probability 0.03. (b) If the null hypothesis is true, a p-value at most 0.03 occurs with probability 0.03. (c) The effect is large.

    Expected: only (b), for a continuous test statistic (Proposition 5.7). A p-value is not a probability of the hypothesis, and it does not measure the size of an effect.

11. Question (1 point) — ch. 6. Give the covariance of the least-squares estimator, and say what ridge regression changes.

    Expected: \sigma^2 (X^\top X)^{-1}. Ridge adds a bias towards 0 and reduces the variance; it is the posterior mean under a Gaussian prior (Propositions 6.7 and 7.5).

12. Question (1 point) — ch. 7. With a prior \mathrm{Beta}(2, 2) and 7 successes in 10 trials, give the posterior and the posterior mean.

    Expected: \mathrm{Beta}(9, 5), with mean 9/14 = 0.643 (Example 7.3).

13. Question (1 point) — ch. 8. A two-state chain jumps from state 1 to state 2 with probability a, and from 2 to 1 with probability b, with a + b > 0. Find its invariant law.

    Expected: detailed balance \pi_1 a = \pi_2 b gives \pi = \big(b/(a+b),\ a/(a+b)\big) (Definition 8.8).

Practice projects

Two small projects, one solo and one for a pair, each at the scale of an afternoon or two.

1. [Solo · ≈ 6 h] Interval estimation for a binomial proportion: a paper study. How well do the textbook intervals for a proportion cover at moderate sample sizes? Reproduce the coverage figures of Brown, Cai and DasGupta (2001) for the Wald, Wilson, Agresti–Coull and Jeffreys intervals.
   • Method: exact coverage, by summing binomial probabilities as in the code of ch. 5, over 1,000 values of p for n = 30 and n = 100. Add the Hoeffding interval (ch. 3) and the credible interval under a uniform prior (ch. 7).
   • Success criterion: coverage curves that show the oscillations of the Wald interval and its collapse near 0 and 1, and a table of mean coverage and mean length for each interval.
   • Deliverables: one script, the figures, and a 2-page note ending with a recommendation backed by numbers. Chapters 2, 3, 5, 7; evaluated with the rubric below.
2. [Group of 2 · ≈ 6 h each] Peeking at an A/B test. Visitors are split at random between two versions of a page, with the same 10 % conversion rate. An analyst runs a two-sided z-test of equal rates after every 100 visitors and stops at the first p-value below 0.05, up to 10,000 visitors. How often does an experiment with no true effect end as a “success”? Johari, Koomen, Pekelis and Walsh (2017) studied this problem on a commercial A/B testing platform.
   • Work split: member 1 simulates the false-positive rate as a function of the number of looks, and explains it with optional stopping (ch. 8). Member 2 builds two corrections: Bonferroni over the looks (ch. 5), and a test that rejects when a likelihood-ratio martingale exceeds 1/\alpha (Proposition 8.5).
   • Success criterion: a false-positive rate near 37 % with 100 looks and no correction; at most 5 % for both corrections; the power of each correction at a stated true effect.
   • Deliverables: shared code, one figure per member, a 3-page joint note and a 10-minute talk. Chapters 3, 5, 8; evaluated with the rubric below.

Self-assessment grid

References

Official references

The refresher courses publish no programme and no references (master-mva.com planning, 2026–2027). The works below are standard textbooks and the papers cited on this page.

Textbooks

• P. Billingsley, Probability and Measure, 3rd edition, Wiley, 1995.
• R. Durrett, Probability: Theory and Examples, 5th edition, Cambridge University Press, 2019.
• J.-F. Le Gall, Measure Theory, Probability, and Stochastic Processes, Graduate Texts in Mathematics 295, Springer, 2022.
• D. Williams, Probability with Martingales, Cambridge University Press, 1991.
• J. Jacod and P. Protter, Probability Essentials, 2nd edition, Springer, 2004.
• A. W. van der Vaart, Asymptotic Statistics, Cambridge University Press, 1998.
• L. Wasserman, All of Statistics: A Concise Course in Statistical Inference, Springer, 2004.
• G. Casella and R. L. Berger, Statistical Inference, 2nd edition, Duxbury, 2002.
• E. L. Lehmann and J. P. Romano, Testing Statistical Hypotheses, 3rd edition, Springer, 2005.
• S. Boucheron, G. Lugosi and P. Massart, Concentration Inequalities: A Nonasymptotic Theory of Independence, Oxford University Press, 2013.
• R. Vershynin, High-Dimensional Probability: An Introduction with Applications in Data Science, Cambridge University Press, 2018.
• D. A. Levin and Y. Peres, Markov Chains and Mixing Times, 2nd edition, American Mathematical Society, 2017.
• C. P. Robert and G. Casella, Monte Carlo Statistical Methods, 2nd edition, Springer, 2004.
• A. Gelman, J. B. Carlin, H. S. Stern, D. B. Dunson, A. Vehtari and D. B. Rubin, Bayesian Data Analysis, 3rd edition, CRC Press, 2013.
• T. Hastie, R. Tibshirani and J. Friedman, The Elements of Statistical Learning, 2nd edition, Springer, 2009.

Classical papers (ch. 3–8)

• H. Chernoff, “A measure of asymptotic efficiency for tests of a hypothesis based on the sum of observations”, Annals of Mathematical Statistics, 1952.
• W. Hoeffding, “Probability inequalities for sums of bounded random variables”, Journal of the American Statistical Association, 1963.
• J. Neyman and E. S. Pearson, “On the problem of the most efficient tests of statistical hypotheses”, Philosophical Transactions of the Royal Society of London, Series A, 1933.
• S. S. Wilks, “The large-sample distribution of the likelihood ratio for testing composite hypotheses”, Annals of Mathematical Statistics, 1938.
• E. B. Wilson, “Probable inference, the law of succession, and statistical inference”, Journal of the American Statistical Association, 1927.
• Y. Benjamini and Y. Hochberg, “Controlling the false discovery rate: a practical and powerful approach to multiple testing”, Journal of the Royal Statistical Society, Series B, 1995.
• A. E. Hoerl and R. W. Kennard, “Ridge regression: biased estimation for nonorthogonal problems”, Technometrics, 1970.
• N. Metropolis, A. W. Rosenbluth, M. N. Rosenbluth, A. H. Teller and E. Teller, “Equation of state calculations by fast computing machines”, Journal of Chemical Physics, 1953.
• W. K. Hastings, “Monte Carlo sampling methods using Markov chains and their applications”, Biometrika, 1970.
• J. Ville, Étude critique de la notion de collectif, Gauthier-Villars, 1939.
• L. D. Brown, T. T. Cai and A. DasGupta, “Interval estimation for a binomial proportion”, Statistical Science, 2001.
• R. L. Wasserstein and N. A. Lazar, “The ASA’s statement on p-values: context, process, and purpose”, The American Statistician, 2016.
• R. Johari, P. Koomen, L. Pekelis and D. Walsh, “Peeking at A/B tests: why it matters, and what to do about it”, Proceedings of the 23rd ACM SIGKDD International Conference on Knowledge Discovery and Data Mining (KDD), 2017.

Research directions (ch. 9)

• O. Catoni, “Challenging the empirical mean and empirical variance: a deviation study”, Annales de l’Institut Henri Poincaré, Probabilités et Statistiques, 2012.
• G. Lugosi and S. Mendelson, “Mean estimation and regression under heavy-tailed distributions: a survey”, Foundations of Computational Mathematics, 2019.
• S. B. Hopkins, “Mean estimation with sub-Gaussian rates in polynomial time”, Annals of Statistics, 2020.
• J. C. H. Lee and P. Valiant, “Optimal sub-Gaussian mean estimation in ℝ”, IEEE 62nd Annual Symposium on Foundations of Computer Science (FOCS), 2021.
• V. Vovk, A. Gammerman and G. Shafer, Algorithmic Learning in a Random World, Springer, 2005.
• J. Lei, M. G’Sell, A. Rinaldo, R. J. Tibshirani and L. Wasserman, “Distribution-free predictive inference for regression”, Journal of the American Statistical Association, 2018.
• A. N. Angelopoulos and S. Bates, “Conformal prediction: a gentle introduction”, Foundations and Trends in Machine Learning, 2023.
• R. F. Barber, E. J. Candès, A. Ramdas and R. J. Tibshirani, “The limits of distribution-free conditional predictive inference”, Information and Inference: A Journal of the IMA, 2021.
• R. J. Tibshirani, R. F. Barber, E. J. Candès and A. Ramdas, “Conformal prediction under covariate shift”, Advances in Neural Information Processing Systems (NeurIPS), 2019.
• R. F. Barber, E. J. Candès, A. Ramdas and R. J. Tibshirani, “Conformal prediction beyond exchangeability”, Annals of Statistics, 2023.
• S. R. Howard, A. Ramdas, J. McAuliffe and J. Sekhon, “Time-uniform, nonparametric, nonasymptotic confidence sequences”, Annals of Statistics, 2021.
• A. Ramdas, P. Grünwald, V. Vovk and G. Shafer, “Game-theoretic statistics and safe anytime-valid inference”, Statistical Science, 2023.
• M. Belkin, D. Hsu, S. Ma and S. Mandal, “Reconciling modern machine-learning practice and the classical bias–variance trade-off”, Proceedings of the National Academy of Sciences, 2019.
• P. L. Bartlett, P. M. Long, G. Lugosi and A. Tsigler, “Benign overfitting in linear regression”, Proceedings of the National Academy of Sciences, 2020.
• T. Hastie, A. Montanari, S. Rosset and R. J. Tibshirani, “Surprises in high-dimensional ridgeless least squares interpolation”, Annals of Statistics, 2022.
• R. Berk, L. Brown, A. Buja, K. Zhang and L. Zhao, “Valid post-selection inference”, Annals of Statistics, 2013.
• R. F. Barber and E. J. Candès, “Controlling the false discovery rate via knockoffs”, Annals of Statistics, 2015.
• A. N. Angelopoulos, S. Bates, C. Fannjiang, M. I. Jordan and T. Zrnic, “Prediction-powered inference”, Science, 2023.
• B. J. K. Kleijn and A. W. van der Vaart, “The Bernstein–von Mises theorem under misspecification”, Electronic Journal of Statistics, 2012.
• A. Durmus and É. Moulines, “High-dimensional Bayesian inference via the unadjusted Langevin algorithm”, Bernoulli, 2019.