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Convolution combines two functions by adding their products over all ways of splitting an argument. For Lebesgue integrable functions $f,g$ on the real line, their convolution isThe arguments $y$ and $x-y$ always add to $x$. For $f,g$ in the L1 space, the integral exists almost everywhere and defines another element of the L1 space, with $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$ by Young's convolution inequality. Particular examples below have a value at every $x$.

To see what the integral counts, take the indicator function $q=1_{[0,1]}$. The product $q(y)q(x-y)$ equals one exactly when $y$ belongs to both intervals $[0,1]$ and $[x-1,x]$. Its integral is their overlap length, soAt $x=1/2$, the overlap is $[0,1/2]$ and the output is $1/2$; at $x=1$, the two intervals coincide and the output is one. The result is the shifted triangular function $\mathrm{tri}(x-1)$. As $x$ changes, $q(x-y)$ is a reflected and translated copy of $q(y)$: the moving overlap makes the entire graph visible without a difficult integration.

The same calculation gives the probability distribution of a sum of two independent random variables. If $X,Y$ are independent random variables with the continuous uniform distribution on $[0,1]$, each has probability density function $q$. Their sum has probability density function $q\ast q$ by convolution of probability densities. The probability density function at one is larger than near zero or two because there are more combinations of the two inputs giving that sum. Its total integral is one; its expected value is one and its variance is $1/6$. These follow equally from integrating the triangular function or from linearity of expectation and variance additivity for independent random variables.

Convolution also gives a useful averaging operation. For $\epsilon >0$, let $k_{\epsilon }(y)=1_{[-\epsilon ,\epsilon ]}(y)/(2\epsilon )$. Thenthe mean of $f$ in a neighborhood of $x$. A smaller $\epsilon$ examines a smaller neighborhood. For a continuous function $f$, this mean tends to $f(x)$ as $\epsilon \to 0$, an example of an approximate identity. Integrating a sharp input against a distributed convolution kernel can produce a smoother output; the triangular example turns two discontinuous inputs into a continuous output.

In signal processing, a linear time-invariant system combines shifted copies of its impulse response by the same rule. In discrete time, an input value at index $m$ contributes a scaled response $a_m{h}_{n-m}$ at index $n$, and discrete convolution sums these contributions. This is the superposition interpretation developed in Oppenheim, MIT Signals and Systems, Lecture 4, convolution sums and integrals. The convolution theorem then gives another description: transform both inputs with a Fourier transform and multiply their transforms. Combining shifts, combining independent quantities and multiplying frequency descriptions are three faces of the same operation.

Take $X$ with continuous uniform distribution on $(0,1)$ and set $Y=X^2$. This is an increasing invertible transformation on that interval. Its cumulative distribution function is $F_Y(y)=\sqrt{y}$, givingAlthough the probability density diverges near zero, it is integrable. The logarithmic stretch is integrable too:The inverse function $X=\sqrt{Y}$ recovers every input in the interval. The changed differential entropy records a changed coordinate volume, rather than a failure of recovery.

// Target: probability-and-statistics.bigb

For a continuous uniform distribution on an interval of length $L>0$, the probability density is $1/L$. Direct integration givesThus a uniform law on an interval of length $1/2$ has differential entropy $-\ln 2$, while length two gives $\ln 2$. Both laws have total probability one. The shorter interval has larger probability density, making $-\ln f(X)$ negative; this is neither a negative probability nor a negative discrete information entropy.

Partition $(0,1)$ into $n$ bins of width $\delta =1/n$, and let the discrete random variable $Q_n$ record the bin containing $X$. This is a quantization of $X$. Its probability mass function isand its discrete information entropy is $H_e(Q_n)=-{\sum }_j{p}_j\ln p_j$. Let ${\bar{f}}_j=p_j/\delta$ be the average probability density within bin $j$. Rearrangement gives the exact finite-bin identityFor a continuous probability density on $[0,1]$, uniform continuity makes the bin averages converge uniformly to the density. Since $t\ln t$ is continuous at zero with the stated convention, the right-hand side converges to $h_e(X)$. Thus the increasing resolution cost is $\ln n$; differential entropy is the remaining offset, rather than the entire discrete encoding cost.

For a continuous uniform distribution, every bin has probability $1/n$, and $H_e(Q_n)=\ln n$ exactly. For the continuous uniform distribution's squared transformation above, the exact bin probabilities areThe diagram computes these masses directly, including the bin next to the integrable singularity. It compares their corrected information entropy with the analytic differential entropy $\ln 2-1$.

Excess kurtosis subtracts $3$ from kurtosis, making the normal distribution baseline zero. The non-excess convention is sometimes called raw kurtosis; this still uses a fourth central moment, not the uncentered fourth moment $\mathbb{E}[X^4]$. Software may report either kurtosis or excess kurtosis, so the convention must be stated.^[ref]

For a standard normal distribution, the moment-generating function of a standard normal variable gives $\mathbb{E}[Z^4]=3$, hence excess kurtosis zero. For a continuous uniform distribution on $[-\sqrt{3},\sqrt{3}]$, the mean is zero, the variance is one, and direct integration givesThe rare-outlier three-point distribution can instead have kurtosis $100$ and excess kurtosis $97$ with mean zero and variance one. These are calculated properties of specified probability distributions, rather than a measured statistic.

For independent and identically distributed random variables $U_i$ with the continuous uniform distribution on $[0,1]$, let $M_n={\max }_{1\le i\le n}{U}_i$. Its cumulative distribution function is $x^n$ on $[0,1]$, so $M_n\stackrel{\mathrm{d}}{\to }1$. The endpoint error has a nonconstant limiting probability distribution: for fixed $z\ge 0$ and $n\ge z$,This survival function gives the displayed exponential distribution limit. More generally, for observations uniform on $[a,b]$, $n(b-M_n)/(b-a)$ has the same limit. The scaling identifies an error of order $1/n$, in contrast to the $1/\sqrt{n}$ fluctuations of a typical sample mean with nonzero finite variance.

For one observation from a continuous uniform distribution on $[0,\theta ]$ and prior density $\theta e^{-\theta }$, the likelihood cancels the factor $\theta$ and leaves $e^{-\theta }{1}_{\{\theta \ge x\}}$. Normalization gives posterior density $e^{-(\theta -x)}{1}_{\{\theta \ge x\}}$, for $x\ge 0$. Its mean is $x+1$ and variance one. The Bayes estimator under squared error loss is therefore $x+1$, unchanged by multiplying the loss by a positive constant.

The Kolmogorov axioms describe a consistent way to assign probabilities to events. Start with a sample space $\Omega$ and an event sigma-algebra $\mathcal{F}$: it contains $\Omega$ and is closed under complements and countable unions, hence also countable intersections. The probability measure is a function $\mathbb{P}:\mathcal{F}\to \mathbb{R}$ satisfyingThe triple $(\Omega ,\mathcal{F},\mathbb{P})$ is a probability space. The third axiom is countable additivity; the infinite series means the limit of a sequence of nonnegative partial sums. It applies to pairwise disjoint events, not to overlapping events and not to arbitrary uncountable unions.^[ref]

The sigma-algebra specifies which questions have probabilities, while the axioms specify how those probabilities fit together. For a finite sample space one can take the entire power set; a continuous model commonly uses a Borel sigma-algebra. Nonnegativity and normalization do not require equally likely outcomes or independence: those are additional modelling choices.

For an exact six-outcome model, take $\Omega =\{1,2,3,4,5,6\}$, $\mathcal{F}=\mathcal{P}(\Omega )$, and assign $\mathbb{P}(\{k\})=k/21$. The total is one because $1+2+3+4+5+6=21$. Define the probability of an event by summing the weights of its outcomes. The disjoint events $A=\{1,2\}$, $B=\{3,4\}$ and $C=\{5,6\}$ form a partition of a set withThis is a specified mathematical model, not measured die frequencies. The figure makes probability literal area: each region has the same height and a width proportional to its assigned mass. Regrouping disjoint events preserves their total area.

Several familiar rules are consequences, not further axioms. Finite additivity of a set function follows by appending empty events to the sequence in countable additivity; $\mathbb{P}(\varnothing )=0$ follows already by writing $\Omega =\Omega \cup \varnothing$ as a disjoint union. The complement rule of probability gives $\mathbb{P}(A^c)=1-\mathbb{P}(A)$ and therefore $0\le \mathbb{P}(A)\le 1$. Splitting larger events into a smaller event and a set difference shows $A\subseteq B\Rightarrow \mathbb{P}(A)\le \mathbb{P}(B)$.

For overlapping events, the inclusion-exclusion principle corrects double counting:In the six-outcome model, $D=\{2,3\}$ and $E=\{3,4\}$ give $5/21+7/21-3/21=9/21$, agreeing with the weight of $\{2,3,4\}$. This decomposition is useful because a complicated event can be assembled from simpler questions with known probabilities.

On a finite sample space, a family of pairwise disjoint events has only finitely many nonempty members, so finite additivity of a set function already handles every countable disjoint union. On an infinite sample space, the countable axiom additionally controls what is left beyond every finite stage. In the fair-coin waiting model of continuity of probability, the finite-stage calculations $\mathbb{P}(\text{first head at toss }j)=2^{-j}$ sum to $1-2^{-n}$ through toss $n$. Countable additivity makes the probability of a head at some finite toss equal to ${\sum }_{j\ge 1}{2}^{-j}=1$; finite-stage sums alone do not justify that passage. For the continuous uniform distribution on $[0,1]$, each singleton instead has probability zero, while their uncountable union has probability one. Countable, rather than unrestricted, additivity accommodates both models.

Historically, Kolmogorov's 1933 book Grundbegriffe der Wahrscheinlichkeitsrechnung established the influential axiomatic framework connecting probability theory with measure theory. The three conditions above are the modern presentation on a sigma-algebra. Shafer and Vovk's historical study explains that the original presentation began with finite additivity on an algebra of sets and added continuity for decreasing events with empty intersection, which supplies the countable extension.^[ref]

The ideal measurement calibration is a straight line through the equal-pressure zero. For the stipulated water mass density and gravitational acceleration above, its slope is $9.81\mathrm{Pa}$ per vertical millimetre. A lower liquid mass density gives a larger length change for the same pressure, improving the height-reading contribution to measurement resolution but requiring more vertical space for a given pressure range. The pressure scale depends on the actual mass density, which depends on temperature; $1000\mathrm{kg}{\mathrm{m}}^{-3}$ here is a chosen model value, not an exact property of all water samples.

Suppose each liquid meniscus height is rounded to the nearest scale division $q=1\mathrm{mm}$. Each height measurement error lies within $q/2$, so the error in their difference lies within $q$. The resulting pressure error is bounded by $\rho gq=9.81\mathrm{Pa}$. To quantify a standard uncertainty, make the additional model assumption that the two rounding errors are independent random variables, each having a continuous uniform distribution on $[-q/2,q/2]$. Each then has variance $q^2/12$; adding the two variances givesHere $u$ denotes standard uncertainty. The $9.81\mathrm{Pa}$ worst-case bound and $4.0049\mathrm{Pa}$ standard uncertainty answer different questions. At $588.6\mathrm{Pa}$ they are respectively $1.67\%$ and $0.680\%$ of the reading. Shared reading bias or a different reading method can violate the assumed independence.

If the mass density, gravitational acceleration and height difference also have small independent standard uncertainties, a first-order estimate follows from the partial derivatives of $\Delta p=\rho g\Delta h$:This displays why a finer scale does not remove uncertainty in mass density or gravitational acceleration. A pressure transducer compared against the column needs its own differential-pressure calibration to establish an electrical response.

A normal Q-Q plot compares order statistics with standard normal quantiles $z_i={\Phi }^{-1}(p_i)$. For a normal distribution $N(\mu ,{\sigma }^2)$ with $\sigma >0$, the population quantile function is $\mu +\sigma {\Phi }^{-1}(p)$, so the diagnostic sample relation isThe appropriate reference is a line with an intercept and positive slope, not necessarily the identity line. A line can be fitted through two central sample quantiles; R's qqline uses the quantile levels $p=0.25$ and $p=0.75$ by default. Neither that line nor its fitting rule is part of the definition of a normal Q-Q plot. ^[ref]

To isolate curvature, retain the same 19 plotting positions and construct two more ordered sequences:The $b_i$ are continuous uniform distribution quantiles on $[-1,1]$, while the $c_i$ apply the hyperbolic sine to the reference quantiles. Both examples are symmetric about zero. At ranks $1,10,19$, $b_i$ is approximately $-0.947368,0,0.947368$ and $c_i$ is approximately $-1.812618,0,1.812618$. Again, these are constructed values rather than random draws or real data.

Read these shapes relative to their central slope. The uniform distribution example has $b(z)=2\Phi (z)-1$, with derivative $b^{\prime }(0)=\sqrt{2/\pi }\approx 0.797885$; at the upper endpoint the tangent predicts $1.546246$, while $b_{19}=0.947368$. Its lower end lies above the tangent and its upper end below it: the outer quantiles are compressed. The hyperbolic sine example has $c^{\prime }(0)=0.7$; its tangent predicts $1.356552$ at the upper endpoint, while $c_{19}=1.812618$. Its lower end lies below the tangent and its upper end above it: the outer quantiles are stretched. Unequal departures at the two ends can suggest skewness. The direction of these descriptions assumes that reference quantiles are horizontal and observed values vertical. The graphs illustrate finite-range shape differences; a sample pattern cannot establish an asymptotic tail probability law or membership in a heavy-tailed distribution class.

A nearly straight normal Q-Q plot is a diagnostic, not a proof of a normal distribution or a formal hypothesis test. Random statistical samples fluctuate, particularly at extreme order statistics, and apparent curvature depends on sample size, plotting positions and the reference line. A calibrated hypothesis test or uncertainty band requires its own assumptions and must account for parameters estimated from the same observations. Sorting also discards the original observation order: rearranging a time series leaves its Q-Q plot unchanged, so the plot cannot establish independence.

For a normal linear model, R's diagnostic normal Q-Q plot uses standardized regression residuals, also called internally studentized residuals. ^[ref] Even with independent errors, the regression residuals have covariance matrix ${\sigma }^2(I-H)$, where $H$ is the hat matrix; fitting parameters generally creates dependencies. Dividing by a shared estimated standard deviation creates further dependence, and the distribution of an internally studentized Gaussian residual is not exactly the standard normal distribution. The normal Q-Q plot therefore helps assess error shape alongside other diagnostics rather than certifying independent normal random variables.

Consider a noninteracting electron in a finite tight-binding model with open ends:The vectors $|n⟩$ form an orthonormal basis of site states. The hopping energy is $t$; it is not the time coordinate. Draw each ${ϵ}_n$ independently from the continuous uniform distribution on $[-W/2,W/2]$ and then hold it fixed. The disorder strength $W/t$ is dimensionless. Open ends mean no hopping bond connects sites $N$ and $1$; equivalently set ${\psi }_0={\psi }_{N+1}=0$. There are no electron interactions, phonons or wave absorption in this model.

The eigenvalue equation in the interior isAs a simple local calculation, if a state's peak is at $n_0$, a neighbouring site with $|E-{ϵ}_n|\gg t$ receives a small leading contribution ${\psi }_n\simeq -t{\psi }_{n_0}/(E-{ϵ}_n)$. Several successive large mismatches multiply small factors. This explains why large disorder can restrict spreading, but is not a proof of Anderson localization: near-resonant denominators and alternative interference paths prevent treating this expansion as uniformly valid.

For the first original calculation, take $N=80$, $t=1$, $W/t=6$ and the fixed random seed $20261015$. Diagonalizing the resulting real symmetric matrix with numpy.linalg.eigh and selecting the eigenvalue closest to zero gives $E/t\simeq -0.05962$, a probability peak at site $29$, and $P\simeq 6.0031$. The selected eigenvector has total probability one. This is one chosen realization, not a disorder average or a measured specimen.

The clean comparison needs care. With $W=0$, the open chain has standing wavesFor even $N$, the identity ${\sin }^{4}x=(3-4\cos 2x+\cos 4x)/8$ and finite trigonometric sums give ${\sum }_n|{\psi }_n^{(m)}{|}^4=3/[2(N+1)]$, hence $P=2(N+1)/3$. Thus the clean $80$-site comparison has $P=54$, although it extends across the chain. Extended does not mean uniform probability. For odd $N$, the special central mode $m=(N+1)/2$ has zeros on alternating sites and instead $P=(N+1)/2$; the even-size assumption avoids that exception.

The second original calculation compares $N=40$ and $80$ over $W/t=0,0.5,1,2,3,4,6,8$. For each size, draw $24$ independent site patterns using seed $20261016$, with the size included in the random-seed construction. Scale each fixed pattern across disorder strengths. At every point, average $P$ over the four energy eigenstates closest to $E=0$. The solid line is the median across these sample averages; shading spans their $10$th–$90$th percentiles. The selected states can change as $W/t$ varies: lines connect one sample's selections rather than tracking a single mode.

At $W/t=0$ the two clean values are $27\frac{1}{3}$ and $54$. At $W/t=8$, the computed median sample averages are approximately $3.17$ and $3.01$. In the latter regime, doubling $N$ hardly changes the typical occupied site count, so $P/N$ falls. At weak disorder the larger chain permits more spatial extent. Neither a single curve nor the overlap of finite-size bands proves the infinite-system statement under mobility edge; a long localization length can hide confinement in either size.

For an observed $x\ge 0$, the continuous uniform distribution likelihood is $p(x\mid \theta )={\theta }^{-1}{1}_{\{\theta \ge x\}}$, with $\theta >0$. Multiplying by the stated prior cancels the factor $\theta$, giving $e^{-\theta }{1}_{\{\theta \ge x\}}$. Its integral over the permitted parameter values is $e^{-x}$. Hence the posterior distribution isThis is the exponential posterior for a uniform endpoint. Its posterior mean is $x+1$ and its variance is one. The posterior expected squared-error loss decomposes asSince $c>0$, its unique minimum occurs at the Bayes estimate $\boxed{{\hat{\theta }}_B=x+1}$. The positive loss multiplier changes the risk, not the optimizing action.

The angular interval has width $\pi$, so its continuous uniform distribution has probability density functionIntegrating this probability density function gives the cumulative distribution functionValues of the probability density function at the two individual endpoints may be assigned arbitrarily without changing any probability.

The new angular interval has width $\pi /2$, so its probability density function is $2/\pi$ there. Its endpoints stay away from the poles of the tangent function, making $Y$ bounded and its expected value finite. Integrate over the continuous uniform distribution of angles:The two cosines are $\sqrt{3}/2$ and $1/2$. HenceIt has units of length and is positive because the permitted positive angles extend farther than the negative ones.

The sample maximum has cumulative distribution functionThe product $x^n$ follows from independence of random variables and the continuous uniform distribution of each observation. For every $x<1$, this cumulative distribution function tends to zero; for $x>1$ it equals one. These are exactly the values of the cumulative distribution function of the constant random variable $1$ at its continuity points. Therefore, by the definition of convergence in distribution,At $x=1$ the limiting cumulative distribution function is discontinuous, so the definition does not require checking that point, although $F_{M_n}(1)=1$ also holds.

Set $\alpha =\arctan c\in (0,\pi /2)$. In polar coordinates, the event is $\sin \Theta >c|\cos \Theta |$, the angular sectorBecause $\Theta$ has the continuous uniform distribution on an interval of length $2\pi$, the probability is the sector's angular length divided by $2\pi$:The symmetry of the standard normal distribution in $Y$ makes the lower sector $Y<-c|X|$ equally likely. Their disjoint union is $|Y/X|>c$, apart from $X=0$, an event of probability zero. Therefore the requested cumulative distribution function isThis is the unit-scale half-Cauchy distribution. At $z=0$ there is no atom, and as $z\to \infty$ the cumulative distribution function tends to one.

The left panel uses $c=1$, giving a sector of angle $\pi /2$ and probability $1/4$. The right panel displays the domain $u<v<2u$ for the next change of variables for probability densities; a vertical section through it determines the conditional range.

The joint probability density of the two independent standard normal random variables is $(2\pi {)}^{-1}{e}^{-(x^2+y^2)/2}$. The polar coordinates map has absolute Jacobian determinant $r$, so the change of variables for probability densities givesAt the origin the angle may be assigned arbitrarily: that point has probability zero. The joint probability density factorizes as a Rayleigh distribution density $r{e}^{-r^2/2}$ and a continuous uniform distribution density $1/(2\pi )$. Hence $R$ and $\Theta$ are independent.

For $u>0$, substitute $s=r^2/2$ to findTherefore $R^2\sim \mathrm{Exp}(1/2)$, equivalently ${\chi }_2^2$, and applying a measurable function to $R$ preserves its independence from $\Theta$. Explicitly, with $S=R^2$,Thus $\Theta \sim \mathrm{Unif}[0,2\pi )$ and $S\sim \mathrm{Exp}(1/2)$ independently. This polar decomposition of a planar standard normal vector separates the uniformly distributed direction from its random squared distance.

The adapted process is bounded by $1$, so every $X_n$ is an integrable random variable, indeed a square-integrable random variable. Independence of $U_{n+1}$ from ${\mathcal{F}}_n$ and its continuous uniform distribution giveThis calculation with a random threshold is justified by conditioning on the ${\mathcal{F}}_n$-measurable value of $X_n$: for a fixed $t\in [0,1]$, the Lebesgue integral ${\int }_0^1{1}_{\{u\le t\}}du$ equals $t$. More formally, independence integrates the bounded Borel measurable function $(t,u)\mapsto 1_{\{u\le t\}}$ over its second coordinate. Thereforewhich verifies the martingale property for this filtration.

Since $B_{n+1}^2=B_{n+1}$, the conditional expectation of the squared martingale difference isThe law of total expectation yields $\mathbb{E}[(X_{n+1}-X_n{)}^2]=\frac{1}{4}\mathbb{E}[X_n(1-X_n)]$. In particular, $\mathbb{E}{X}_n=x$ for every $n$.

Let $M=\max \{b,y_1,\dots ,y_n\}$. The continuous uniform distribution likelihood contributes ${\varphi }^{-n}{1}_{\{\varphi \ge {\max }_i{y}_i\}}$, while the Pareto distribution prior contributes ${\varphi }^{-a-1}{1}_{\{\varphi \ge b\}}$. Their product proves Uniform-Pareto conjugacy:Thus the posterior distribution is Pareto with shape $a+n$ and lower bound $M$. Its density is strictly decreasing on its support, so the highest posterior density interval begins at $M$. Since the upper-tail probability is $(M/u{)}^{a+n}$, the required upper endpoint satisfies $(M/u{)}^{a+n}=0.05$. The Pareto highest density interval is therefore

Assume $a<b$. An increasing affine map sends the two endpoints to zero and one when $ca+d=0$ and $cb+d=1$. Solving givesIndeed, for $0\le t\le 1$, the cumulative distribution function of the transformed random variable iswhich is the cumulative distribution function of the continuous uniform distribution on $[0,1]$.

A decreasing affine map works too, by swapping the endpoints:These are the two possible choices, since the image interval must have endpoints zero and one. The case $a=b$ cannot be transformed into a nondegenerate continuous uniform distribution by deterministic constants.