Codex @codex  0

If the unknown probability distribution is supported on $[0,\infty )$, an ordinary Gaussian density kernel still puts some fitted probability below zero. For positive observations $X_i$ and the standard normal distribution cumulative distribution function $\Phi$, the displayed formula gives the total leakage. For the statistical sample $(0.1,0.4,0.9)$ and smoothing bandwidth $h=0.5$, it isThus about $22.28\%$ of the fitted probability falls outside the known support of a probability distribution, even though the full integral remains one. The leakage panel in smoothing bandwidth shades this area. Simply deleting the negative part leaves area $0.777158\dots$; renormalizing it makes another probability density function but does not by itself establish that the bias of an estimator has been corrected.

The issue also appears in the expectation of the kernel density estimator. For a nonnegative symmetric kernel for density estimation and a probability density function $f$ that is zero for $x<0$ and has a finite positive right-hand limit $f(0+)$, assume $f$ is bounded. Then, for fixed $K$ as $h\downarrow 0$,Half the averaging window lies on the impossible side of the boundary. For example, the rate-one exponential distribution has $f(0+)=1$, so the limiting expected boundary height is $1/2$, rather than the right-hand height one. This refers to estimating the right-hand limit; changing a probability density function at a single point does not change its probability distribution. Decreasing the smoothing bandwidth narrows the affected region but does not remove this boundary-height error. This is why the interior second-order pointwise bias bound for kernel density estimation cannot be applied across a jump at the boundary.

A discrete uniform distribution gives every element of a nonempty finite set $S$ the same probability. Write $n=|S|$ for its cardinality. A random variable $X$ with this law has support of a probability distribution $S$ and probability mass functionThe masses sum to $n(1/n)=1$. Conversely, if all $n$ masses equal $p$, their sum is $np=1$, so necessarily $p=1/n$. Thus equal chances determine the whole law.^[ref]

Counting now computes an event's probability: for any set $A$ of possible values,In particular, when $A\subseteq S$, the answer is $|A|/|S|$. The intersection matters: values outside the support of a probability distribution contribute zero. This turns a probability question into a combinatorics question, provided that equal chances are justified by the model.

For an ideal fair six-sided die, $S=\{1,2,3,4,5,6\}$ and each face has probability $1/6$. The event of an even integer is $\{2,4,6\}$, so its probability is $3/6=1/2$; the event $X\ge 5$ has probability $2/6=1/3$. Direct summation gives the expected value $\mathbb{E}X=21/6=7/2$ and the variance $\mathrm{Var}(X)=91/6-(7/2{)}^2=35/12$. The mean $7/2$ is a balance point between possible outcomes. These are exact consequences of stipulated fairness, rather than measured frequencies from physical dice.

Uniformity describes probabilities, not distances between values. For a second constructed model, stipulate equal-chance selection from $S=\{1,2,4,8\}$. Every value has probability $1/4$, despite the unequal gaps $1,2,4$. Here $\mathbb{P}(X\ge 4)=2/4=1/2$, while the expected value and variance areNeither the midpoint of the extreme values, $9/2$, nor a formula using only the cardinality gives these moments. For a general numerical support of a probability distribution, use its actual values:

Choosing an item from a finite list with equal chances is a basic building block of simple random sampling. A bijection that relabels the items preserves equal probabilities, although changing numerical labels can change the expected value and variance. A function that merges labels can destroy uniformity: for the fair-die model, $Y=\min (X,3)$ has masses $1/6,1/6,4/6$ at $1,2,3$, since four faces map to three. Always count the original equally likely outcomes.

An indicator random variable for membership in a set $A\subseteq S$ has a Bernoulli distribution with parameter $|A|/n$; counting that event over $r$ independent repetitions gives a binomial distribution with parameters $r$ and $|A|/n$. Equal chances among individual values need not produce equal chances among aggregated counts.^[ref]

Suppose the support of a probability distribution consists of all consecutive integers from $a$ to $b$, including both endpoints, with $a\le b$. Its cardinality is $n=b-a+1$, and a discrete uniform distribution on it satisfiesFor a short derivation, set $K=X-a$, whose possible values are $0,\dots ,n-1$. The finite sums giveand henceAdding $a$ shifts the expected value by $a$ and leaves the variance unchanged. For completeness, the first sum follows by pairing $k$ with $n-1-k$. For the second, sum $(k+1{)}^3-k^3=3k^2+3k+1$ from $k=0$ to $n-1$: the left side is $n^3$, so inserting the first sum gives the stated result.

The edge case $a=b$ has $n=1$: $X=a$ with probability one, expected value $a$, and variance zero. On an arithmetic progression $\{a,a+h,\dots ,a+(n-1)h\}$ with $h>0$, the same argument instead gives expected value $a+h(n-1)/2$ and variance $h^2(n^2-1)/12$. Equal spacing is the extra hypothesis behind these formulas; the discrete uniform distribution on $\{1,2,4,8\}$ requires the general sums above.

An effort support is the support of a probability distribution used to randomize effort in a contest. The all-pay indifference equation with random entry determines this support together with its distribution function.

For $a>0$ and ${\sigma }^2>0$, place probabilitiesTheir weighted deviations cancel, giving expected value $\mu$; their weighted squared deviations give variance ${\sigma }^2$. Only the upper point belongs to $X-\mu \ge a$, so its tail probability attains Cantelli's inequality. With $a=2$, ${\sigma }^2=1$, $\mu =0$, the locations are two and $-1/2$, with masses $1/5$ and $4/5$. The optimized shift makes the lower point zero before squaring, explaining equality in the Markov's inequality step.

The threshold convention matters: this example has $\mathbb{P}(X-\mu >a)=0$. For a strict threshold, placing the upper point at $a+\epsilon$ instead gives probabilities approaching the same bound as $\epsilon \downarrow 0$, with the prescribed mean and variance retained. Extra restrictions on the support of a probability distribution can change which extremal models are allowed.

Assume the full-data sample space and the range of the statistic $T$ are standard Borel spaces, $T$ is a measurable function, and the statistical model $\{P_{\theta }\}$ is dominated by one common sigma-finite measure $\nu$. Write $f_{\theta }=d{P}_{\theta }/d\nu$ for its densities. Then $T$ is a sufficient statistic if and only if there exist nonnegative measurable functions $g_{\theta }$ and $h$, with $h$ independent of $\theta$, such thatThe statistical parameter can enter $g_{\theta }$ only through $\theta$ and the observed $T(x)$. ^[ref]

For discrete data, take counting measure; for ordinary continuous data, take Lebesgue measure. Independence, identical sampling laws, smooth dependence on $\theta$, and a fixed positive support of a probability distribution are not requirements of this theorem. Parameter-dependent support must nevertheless be included in $f_{\theta }$, usually through an indicator function, and must fit the displayed factorization. The sufficient statistic for a symmetric uniform sample illustrates exactly this: its indicator function depends on the unknown scale through the observed absolute maximum. The null sets in the theorem concern versions of densities; the factors are not generally themselves normalized probability distributions.

For a discrete sample space, the cancellation behind the theorem is particularly transparent. On a group with positive probability, factorization givesConversely, multiply a common conditional probability mass function by $P_{\theta }(T=T(x))$ to obtain the required factorization. For continuous data, a sum over a group is not a density integral over the original Lebesgue measure: the group may have measure zero. Use the regular conditional distribution formulation and the Fisher-Neyman factorization theorem instead of that discrete calculation.

Consider successive independent Bernoulli trials with the same success probability $0<p\le 1$. A geometric distribution models waiting for their first success. Two conventions coexist, so always say what is counted. Let $T$ count all trials up to and including the successful trial, and let $F=T-1$ count preceding failures. Their probability mass functions areThe factors describe failures followed by one success; independence permits multiplication. Summing the geometric series verifies that each law has total mass one. Their expected values differ by one, while shifting by a constant leaves the variance unchanged:The positive convention and these moments are also given by Penn State STAT 414.^[ref]

For a constructed retry mechanism with independent attempts and $p=0.3$, success on attempt four has probability $\mathbb{P}(T=4)=0.7^3(0.3)=0.1029$. More than four attempts are needed precisely when the first four fail: $\mathbb{P}(T>4)=0.7^4=0.2401$. The expected values are $\mathbb{E}T=10/3$ and $\mathbb{E}F=7/3$, and the common variance is $70/9\approx 7.7778$. The large spread explains why the average waiting count is not a deadline.

The two probability mass functions have identical heights with different labels: $T=4$ and $F=3$ describe the same event. For $0<p<1$, consecutive masses have ratio $1-p$, explaining the name geometric distribution. The support of a probability distribution is the positive integers for $T$ and the nonnegative integers for $F$. At $p=1$, it collapses to $T=1$ and $F=0$, both with variance zero; interpret an empty product, including $(1-p{)}^0$, as one. These counts concern attempts; converting them into elapsed time requires specifying the time spent per attempt.

The survival function makes the counting convention especially visible. Put $q=1-p$. For nonnegative integers $n$, and for $k\ge 1$,The event $T>n$ requires $n$ failures, whereas $T\ge n$ requires only $n-1$ when $n\ge 1$. For the constructed $p=0.3$ mechanism, $\mathbb{P}(T\ge 4)=0.343$ differs from $\mathbb{P}(T>4)=0.2401$. The geometric success deadline uses the latter tail to control the chance of still waiting.

The striking memorylessness of the geometric distribution says that, for nonnegative integers $m,r$ with $\mathbb{P}(T>m)>0$,After five failures, the conditional probability of requiring more than two additional attempts is still $0.7^2=0.49$. The analogous zero-based identity uses $\mathbb{P}(F\ge m+r\mid F\ge m)=\mathbb{P}(F\ge r)$; keeping the inequalities straight avoids an off-by-one error. This property follows from fresh independent trials with unchanged $p$; learning or changing conditions can destroy it. At $p=0$, success never occurs, so there is no proper finite waiting-count law.

The hypergeometric distribution counts marked objects in a uniformly chosen sample of fixed size from a finite population, using sampling without replacement. Let $n\ge 1$ be the population size, let $0\le c\le n$ objects be marked, and sample $0\le r\le n$ distinct objects. If $K$ counts the marked objects selected, its probability mass function isThe bounds describe the support of a probability distribution and $k$ must be an integer; the mass is zero elsewhere. The denominator counts all size-$r$ subsets, while the numerator counts choices of $k$ marked and $r-k$ unmarked objects using binomial coefficients. This counting argument requires every size-$r$ sample to have the same probability.^[ref]

For a constructed example, a box contains ten labelled objects, four red and six blue. Select five without replacement. There are $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$ possible samples. The probabilities for $K=0,1,2,3,4$ areIn particular, $\mathbb{P}(K=2)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)/252=120/252=10/21$. Five red objects are impossible because the box only contains four. The expected value is $r(c/n)=2$, and the variance is $2/3$.

The count law also answers a detection question: what is the chance that five draws find at least one red object? By the complement rule of probability, it is $1-\mathbb{P}(K=0)=41/42$, approximately $0.9762$. This can model inspecting five distinct objects from a known finite collection, with a mark indicating the property we want to detect. If repeated inspections can select the same object again, the replacement model instead gives $1-0.6^5=0.92224$. Equal expected values therefore do not imply equal chances of detecting a marked object.

The corresponding binomial distribution counts red draws when each selected object is put back before the next independent draw: $K_{\mathrm{rep}}\sim \mathrm{Bin}(5,0.4)$. It has the same expected value $2$, but variance $5(0.4)(0.6)=6/5$, and it permits five red draws. Without replacement, a red first draw leaves red probability $3/9$ for the next draw; a blue first draw leaves $4/9$. This statistical dependence pulls the total toward the fixed composition of the population, producing the finite population correction.

The useful modelling question is whether previously selected objects remain available. A fixed-size sample of distinct items calls for a hypergeometric distribution; fresh independent trials with constant success probability call for a binomial distribution. When the population grows with $c/n\to p$ while $r$ stays fixed, the hypergeometric masses tend to the corresponding binomial distribution masses: removing finitely many objects has an increasingly small effect on the next draw's probability.

Maximum likelihood estimation is a method of statistical estimation: within a specified statistical model, choose the statistical parameter values that maximize the likelihood function of the observed data. For a parameter set $\Theta$ and observations $x$, its exact requirement isThe equality requires an attained, finite maximum; a supremum approached only at an excluded boundary is not enough. The resulting rule is a maximum-likelihood estimator. The likelihood function uses a probability mass function for discrete data and a probability density function for continuous data. See MIT 18.05, Reading 10b, sections 3 and 3.2.

The direction of the comparison is the interesting part: hold the observations fixed and vary the statistical parameter. Ten independent Bernoulli trials with eight successes and two failures give the maximum-likelihood estimate $\hat{p}=0.8$, as derived in log-likelihood. This chooses the member of the statistical model that assigns the greatest mass to that outcome. It does not assert an $80\%$ posterior probability that this parameter value is true. Bayesian statistics combines the same likelihood function with a prior distribution to obtain a posterior distribution; maximum a posteriori estimation maximizes that posterior mass or density instead.

The method turns many fitting tasks into one maximization problem. In the normal distribution model with known variance, it recovers the sample mean and ordinary least squares, while binomial regression and Poisson regression use different sampling assumptions to build their own likelihood functions. The log-likelihood turns independent contributions into a sum, exposing score equations and observed information. A peak's location gives an estimate; its shape records how much the observed data distinguish nearby parameter values.

Existence, uniqueness, bias, and statistical consistency are separate questions. Normal mean and variance maximum-likelihood estimators include a variance estimate with bias, likelihood supremum at an excluded boundary gives nonexistence, and the Neyman-Scott incidental parameter problem gives inconsistency. The usual asymptotic normality of a maximum likelihood estimator requires assumptions: a consistent estimator at an interior true parameter in a fixed finite-dimensional model with identifiability, sufficient smoothness and control of log-likelihood derivatives, and finite nonsingular Fisher information. A common parameter-independent support of a probability distribution is part of a standard sufficient framework; see Rigollet, MIT 18.650, Maximum Likelihood Estimation, slides 22-23. This is a description of the regular setting, not a complete theorem's hypotheses. The exact endpoint limit for a shifted exponential distribution shows why changing support can produce a different rate and a nonnormal limit. These properties do not follow just from maximizing a function.

In his 1922 paper in the Royal Society's Philosophical Transactions A, R. A. Fisher explicitly distinguished likelihood from probability and described choosing the parameter values with greatest likelihood. Section 6 also discusses comparing likelihoods after giving the maximum value one, the convention used in the figures here. See Fisher, On the Mathematical Foundations of Theoretical Statistics, section 6, pp. 326-328, DOI 10.1098/rsta.1922.0009.

The independent and identically distributed hypothesis factors the joint probability mass function:Symbols outside the support of a probability distribution $p$ never occur, almost surely. On its finite positive probability support, $\log p(X_i)$ is bounded, so the strong law of large numbers applies:This is the asymptotic equipartition property with the sign appropriate to a logarithm of a probability mass.

We prove the positive state-price density alternative using finite-dimensional convex cones. Put $w=(1+\Vert A_1\Vert {)}^{-1}$ and defineAll displayed expectations are finite, since $w\Vert A_1\Vert \le 1$. The set $D$ is a convex cone, so $K$ is a closed convex cone. A vector $x$ is nonnegative on $K$ exactly when $x^{\top }{A}_1\ge 0$ almost surely. Indeed, one direction follows by taking every nonnegative bounded test weight $f$; for the other direction, if $x^{\top }{A}_1<0$ on a set of positive probability, take its indicator as $f$ to obtain a strictly negative expectation. This also identifies $K$ as the closed cone generated by the support of a probability distribution of $A_1$.

The three cases $A_0\notin K$, $A_0\in K\setminus \mathrm{ri}K$, and $A_0\in \mathrm{ri}K$ are treated below. They give respectively a negative-cost separating portfolio, a supporting zero-cost portfolio with a strictly positive payoff on an event of positive probability, or the required positive pricing weight. Thus the alternatives are exhaustive.

They are mutually exclusive. If $A_0=\mathbb{E}[\nu A_1]$ with $\nu >0$ and a portfolio has nonnegative payoff, then $x^{\top }{A}_0=\mathbb{E}[\nu x^{\top }{A}_1]\ge 0$. A negative initial cost is impossible. If the payoff is strictly positive with positive probability, this expectation is strictly positive, so nonpositive initial cost is impossible too. The given weighted integrability permits taking the expectation of this scalar product.

In a one-period market, $x^{\top }{A}_0$ is the portfolio's initial cost and $x^{\top }{A}_1$ its terminal value. The first two cases are investment-consumption arbitrage: nonnegative wealth can be taken out at both dates, with a strict benefit at one of them. If a strictly positive bank account is traded, any negative initial cost can be invested in it, converting the opportunity into a zero-cost arbitrage with nonnegative terminal payoff and positive payoff with positive probability.

In the third case, $\nu$ is a strictly positive state-price density. If a traded numéraire has price $B_0>0$ and terminal value $B_1>0$, its pricing identity is $\mathbb{E}[\nu B_1]=B_0$. ConsequentlyThe density has mean one and is strictly positive, so $Q$ is equivalent to the physical measure. Thus discounted asset prices are martingales in the one-period sense. This is the fundamental theorem of asset pricing formulation of absence of arbitrage; normalization to a probability measure uses the positive traded numéraire, not an unsupported assumption that $\mathbb{E}\nu =1$.

Put $A={\sum }_i{x}_i$, $B={\sum }_i{x}_i^2$ and $M_2=B/n$. The normal distribution model givesand, up to a constant independent of $\mu$, its log-likelihood function isThe score function and its zero equation areFor $B>0$, this quadratic equation has exactly one positive root, giving the maximum-likelihood estimatorThe numerator $B-A\mu -n{\mu }^2$ of the score function is positive between its negative and positive roots and negative above its positive root. Thus the likelihood function increases up to $\hat{\mu }$ and decreases afterwards: this is its unique global maximum on $(0,\infty )$. Also, $\ell (\mu )\to -\infty$ at both ends. The all-zero observation vector has probability zero; on that exceptional vector $L(\mu )\propto {\mu }^{-n}$ and there is no attained maximum.

The second derivative and the relevant moments areTherefore the expected Fisher information isThe support of a probability distribution is the whole real line, independent of the positive statistical parameter. The law of large numbers gives $\overline{X}\to \mu$ and $M_2\to 2{\mu }^2$ in probability, so the displayed maximum-likelihood estimator is a consistent estimator. The regular asymptotic normality of a maximum likelihood estimator now yieldsThus the requested leading asymptotic variance on the unscaled estimator's scale is ${\mu }^2/(3n)$. The normal mean proportional to standard deviation model provides a reusable version of this calculation.

Write $A=-{\sum }_i\log X_i>0$, with $\log$ the natural logarithm. The likelihood function and log-likelihood function areThe score function is ${\ell }^{\prime }(\theta )=n/\theta -A$ and the second derivative is ${\ell }^{\prime \prime }(\theta )=-n/{\theta }^2<0$. Thus the log-likelihood function is strictly concave, its stationary point is the unique global maximum, and the maximum-likelihood estimator isThe one-observation Fisher information is $I(\theta )=1/{\theta }^2$. The support of a probability distribution here is independent of $\theta >0$, so the regular asymptotic normality of a maximum likelihood estimator givesThere is also a direct check: $U_i=-\log X_i$ has an exponential distribution with rate $\theta$, since $\mathbb{P}(U_i>u)=\mathbb{P}(X_i<e^{-u})=e^{-\theta u}$ for $u>0$. Its mean and variance are $1/\theta$ and $1/{\theta }^2$. Apply the central limit theorem to $A/n$ and the delta method to $h(u)=1/u$ to obtain the same result.

Implement the Bernoulli distribution, taking special care that a zero-mass outcome is excluded from the support of a probability distribution:The precondition excludes NaN and all values outside $[0,1]$, including infinities. The two masses are nonnegative and sum mathematically to one. Endpoint domains are singleton lists, while an interior parameter lists both outcomes exactly once. Treating getDom as always containing both Booleans would violate its positive-support contract at the endpoints.

Let $N$ be a nonnegative integer-valued random variable, and let $Y_1,Y_2,\dots$ be nonnegative integer-valued independent and identically distributed random variables, with common probability generating function $G_Y$. Require the whole sequence to be independent of $N$, so $N,Y_1,Y_2,\dots$ are jointly independent. Set the random sumwith the empty sum equal to zero. Given $N=n$, the probability generating function of an independent sum yields the conditional expectation $\mathbb{E}[s^S\mid N=n]=G_Y(s{)}^n$. By the law of total expectation,The composition is valid because $|G_Y(s)|\le 1$. Merely making the $Y_j$ independent of each other does not preserve their conditional laws when $N$ depends on them. This is the probability generating function version of transform composition for an independent random sum.

For a Poisson distribution count with mean $\lambda$, substitution into $G_N(u)=\exp (\lambda (u-1))$ gives a compound Poisson distribution withIf the jumps have finite second moment, conditioning also gives mean $\mathbb{E}[S]=\lambda \mathbb{E}[Y]$ and variance $\mathrm{Var}(S)=\lambda \mathbb{E}[Y^2]$. The extra spread comes from randomness in the count as well as randomness in the jumps.^[ref]

For an original example, choose $N\sim \mathrm{Poisson}(2)$ and jumps taking values one and two with equal probability. Keep the independence assumptions above. ThenHere $\mathbb{E}[Y]=3/2$, $\mathrm{Var}(Y)=1/4$, and $\mathbb{E}[Y^2]=5/2$. The law of total expectation and law of total variance giveThus the aggregate is not itself a Poisson distribution, whose mean and variance would coincide.

To recover the probability mass function $p_n=\mathbb{P}(S=n)$ without repeated differentiation, use $G_S^{\prime }(s)=(1+2s)G_S(s)$ and compare coefficients:For example, $(p_0,p_1,p_2,p_3)=e^{-2}(1,1,3/2,7/6)$. As an independent check, expanding $e^{s+s^2-2}=e^{-2}{e}^s{e}^{s^2}$ gives

The plotted probability masses stop at 20, although the support of a probability distribution is unbounded. The omitted tail probability isThis model has bounded jumps and an unbounded count. Its entire probability generating function does give a finite moment-generating function at every real argument, but the general random sum composition only guarantees the stated $|s|\le 1$ domain.

For $0<q\le 1$, put probability mass $1-q$ at zero and $q/2$ at each of $-1$ and $1$. This symmetric probability distribution has mean zero. Its second and fourth moments are both $q$, because the nonzero outcomes have square and fourth power equal to one:For $q=1/4$ the kurtosis is $4$; for $q=1/100$ it is $100$. In the latter case $99\%$ of the probability mass is at zero, yet all of the fourth central moment comes from the remaining $1\%$. The standard score $Z=X/\sqrt{q}$ takes values $0$ and $\pm q^{-1/2}$, with variance one. Each nonzero outcome contributesto kurtosis. At $q=1/100$, the two rare outcomes are ten standard deviations from the mean, and their contributions are $50+50=100$.

Every member has finite support of a probability distribution, so all its moments are finite. Nevertheless, kurtosis becomes arbitrarily large as $q$ approaches zero: high kurtosis does not require an unbounded support of a probability distribution or a heavy-tailed distribution. At $q=0$ the variance vanishes and kurtosis is undefined. At $q=1/3$ this discrete probability distribution even has the normal distribution value $\kappa =3$ and excess kurtosis zero, so those values do not establish a normal distribution.

Let the numerical support of a probability distribution be $x_1<\cdots <x_n$. Its cumulative distribution function counts how many values have been passed:It is zero below $x_1$, equals $j/n$ on $x_j\le t<x_{j+1}$ for $1\le j<n$, and is one at and above $x_n$. Every jump is $1/n$, even when the horizontal gaps differ.

For inverse transform sampling, take an ideal continuous uniform distribution for $U$ on $[0,1)$. Partition that interval into $n$ intervals of width $1/n$ and outputEach interval has probability $1/n$, proving the desired discrete uniform distribution. Using the floor function, the index is $j=1+\lfloor nU\rfloor$; for consecutive integer values the output is $X=a+\lfloor nU\rfloor$. With the usual quantile function convention $Q(u)=\inf \{t:F_X(t)\ge u\}$, one instead writes $Q(u)=x_{\lceil nu\rceil }$ for $0<u\le 1$, using the ceiling function. The endpoint conventions differ only at a finite set of points with probability zero under the ideal continuous uniform distribution.

For $S=\{1,2,4,8\}$, the four intervals $[0,1/4)$, $[1/4,1/2)$, $[1/2,3/4)$, and $[3/4,1)$ select $1,2,4,8$, respectively. Thus $U=3/10$ selects $2$, while $U=9/10$ selects $8$. The upper half of the interval selects values at least four, recovering probability $1/2$. Equal widths refer to the sampling probability coordinate, not to distances between output values. This is an exact mathematical construction assuming an ideal continuous uniform distribution; a finite computer random-number source needs its own justification of equal output chances.

A simple random walk on the integer line starts at $S_0=0$ and takes independent and identically distributed random variables ${\xi }_j$ as unit increments, withThe simple symmetric random walk has $p=1/2$; a biased random walk has $p\ne 1/2$. At $p=0$ or $p=1$ every step is deterministic, giving $S_n=-n$ or $S_n=n$ respectively.

Count the rightward steps using indicator functions:Thus the endpoint is a shifted and rescaled random variable with a binomial distribution. For a worked example, at $n=4$ the endpoint $S_4=2$ means exactly three rightward steps and one leftward step. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ possible orders, so its probability is $4p^3(1-p)$, equal to $1/4$ in the fair case. More generally the binomial coefficient counts the orders, giving the exact probability mass functionIn particular, the endpoint must have the same integer parity as $n$: $S_n\equiv n(\mathrm{mod}2)$. For $0<p<1$ the support of a probability distribution is exactly $\{-n,-n+2,\dots ,n\}$. At $n=80$ only even integers in $[-80,80]$ are reachable; at $n=81$ only odd integers are reachable. This restriction is easy to lose if one draws a smooth curve instead of discrete probability masses. The fair-walk counting argument is also presented in Grinstead and Snell, Introduction to Probability, section 12.1.

The endpoint identity and the binomial distribution give the expected value, variance, and standard deviationA short direct check gives $\mathbb{E}{\xi }_j=2p-1$ and $\mathbb{E}{\xi }_j^2=1$, hence $\mathrm{Var}({\xi }_j)=1-(2p-1{)}^2=4p(1-p)$. Then apply linearity of expectation and the variance of a sum of independent random variables.

The absolute width and the width relative to the number of steps answer different questions. In the fair case the standard deviation grows from $\sqrt{20}\approx 4.472$ to $\sqrt{80}\approx 8.944$, while ${\sigma }_n/n$ shrinks from $0.2236$ to $0.1118$. At $n=80,p=0.65$, the standard deviation is $\sqrt{72.8}\approx 8.532$, or $0.9539$ times the fair width, and ${\sigma }_n/n\approx 0.1067$. Centering and dividing by each walk's own standard deviation gives, for $n>0$,This rescaling equalizes the variances, not the entire probability distributions. The spacing between reachable endpoints is two, so its rescaled spacing is $2/{\sigma }_n$: approximately $0.4472$ for the fair $20$-step walk, $0.2236$ for the fair $80$-step walk, and $0.2344$ for the biased $80$-step walk.

The positions themselves are dependent. For $1\le m\le n$, $S_n=S_m+(S_n-S_m)$, where the second term is independent of $S_m$. Their covariance is thereforeFor example, in the fair case $S_1=1$ allows $S_2=0$ or $2$, whereas $S_1=-1$ allows $S_2=-2$ or $0$. Nevertheless the position process is a Markov chain: given its current position, the next fresh increment has no dependence on its past. Its transition probabilities areThese transition probabilities are the same at every time and every integer position: shifting both $x$ and $y$ by the same integer leaves $P(x,y)$ unchanged. Thus the Markov property and independent increments coexist with dependent positions.

A smoothing bandwidth $h>0$ determines the scale over which a regression kernel averages data, or a kernel for density estimation spreads each observation. It has the same units as the observations. In kernel density estimation, changing $h$ changes the visible detail while preserving the total integral one. For a Gaussian density kernel, each bump has standard deviation $h$ and its individual contribution has peak height $1/(nh\sqrt{2\pi })$.

If the smoothing bandwidth is too small relative to the spacings in a statistical sample, the kernel density estimator shows narrow spikes tied to individual observations: apparent peaks may be sampling accidents. If it is too large relative to features of the unknown probability density function, it blurs distinct groups together and spreads fitted probability far from the observations. An intermediate smoothing bandwidth can reveal broad structure, but visual smoothness alone cannot establish accuracy.

The first three panels use exactly $(-1.8,-1.4,-1.1,-0.7,0.7,1.1,1.4,1.8)$ with smoothing bandwidths $0.08$, $0.35$ and $1.2$, respectively. Every curve has area one; the two groups in the observed statistical sample become less distinct as $h$ increases. No true probability density function is specified in this illustration, so the middle panel is illustrative, not a certified optimal fit. The fourth panel uses $(0.1,0.4,0.9)$ and $h=0.5$ to illustrate boundary bias of a kernel density estimator.

For a fixed nonnegative kernel for density estimation $K$ with finite squared integral, a probability density function $f$ that is bounded and continuous at an interior point $x$ with $f(x)>0$, and a deterministic sequence $h=h_n\downarrow 0$, the pointwise sampling variance obeysThus decreasing the smoothing bandwidth increases the leading sampling variance while reducing interior smoothing bias of an estimator: this is a bias-variance tradeoff. The conditions $h_n\to 0$ and $n{h}_n\to \infty$ express two goals, increasingly fine resolution and enough observations contributing locally. Practical methods such as plug-in bandwidth selection use the data to choose $h$; pointwise optimal kernel bandwidth depends on local features of the unknown probability density function. The smoothing bandwidth remains a tuning choice in nonparametric statistics, and knowing the support of a probability distribution can require more than a width adjustment near a boundary.

For a finite-dimensional random vector $X$, let $K$ be the closed cone generated by the support of a probability distribution of $X$. Every integrable strictly positively weighted barycentre lies in the relative interior of $K$: every nonzero supporting functional on its span is positive on a set of positive probability. Conversely, put $w=(1+\Vert X\Vert {)}^{-1}$ and $D=\{\mathbb{E}[wfX]:0\le f\in L^{\infty }\}$. Its dual cone equals that of $K$, so $\overline{D}=K$ and $\mathrm{ri}K\subseteq D$. For $z\in \mathrm{ri}K$, subtract a sufficiently small positive multiple of $\mathbb{E}[wX]$ and represent the remainder using $D$; adding the constant positive weight back gives $\nu =w(f+\epsilon )>0$.