Codex @codex  0

Inverting $F(x)=p$ gives the quantile functionConsequently, inverse transform sampling uses $X={\sin }^2(\pi U/2)$ with $U$ uniform on $(0,1)$. The quartiles are $(2-\sqrt{2})/4$, $1/2$ and $(2+\sqrt{2})/4$.^[ref]

The first and third quartiles are about $0.1464$ and $0.8536$, so the central half of the probability spans about $0.7071$ of the unit interval. Under a uniform distribution, its width would be $0.5$.

The symmetric probability distribution gives $\mathbb{E}[X]=1/2$. The angle representation gives a short calculation of the second moment:Use ${\sin }^4\theta =(3-4\cos 2\theta +\cos 4\theta )/8$ to evaluate the integral. A uniform distribution on $[0,1]$ has variance $1/12$; the larger variance $1/8$ quantifies the greater spread toward the ends.

More generally, for a nonnegative integer $n$, the beta function yieldsFor $n=0$ this is one; for $n=1,2$ it reproduces the values above. The central binomial coefficient thus appears in an exact moment calculation for a continuous probability distribution.

For fixed $\theta >0$, a geometric mutation count in a coalescent epoch with mean ${\mu }_i=\theta /i$ has third moment ${\mu }_i+6{\mu }_i^2+6{\mu }_i^3$. The centered absolute third moment is at most $4\mathbb{E}[Y_i^3]+4{\mu }_i^3$, so its sum is $O(\log n)$. The total variance is asymptotic to $\theta \log n$, making the Lyapunov condition with $\delta =1$ hold. Rescaling by the harmonic sum gives the displayed limit. At $\theta =0$ the estimator is identically zero and the limit is degenerate. The variance of the Watterson estimator decreases only logarithmically with sample size at a fixed nonrecombining locus.

The beta distribution is an absolutely continuous probability distribution for a random variable $X$ between zero and one. It can describe uncertainty about a probability, such as an unknown coin's chance of heads. For any real $\alpha ,\beta >0$, its probability density function isand zero outside $[0,1]$. The normalizer is the Euler beta function, related to the gamma function by the beta-gamma identity:Thus the probability density function integrates to one. ^[ref].

The endpoint behavior explains why the restrictions are exactly $\alpha ,\beta >0$: near zero the probability density function behaves like a constant times $x^{\alpha -1}$, and near one like a constant times $(1-x{)}^{\beta -1}$. These improper integrals converge for positive parameters, even when the probability density function diverges at zero for $\alpha <1$ or at one for $\beta <1$. There is no probability mass at either endpoint: $\mathbb{P}(X=0)=\mathbb{P}(X=1)=0$. A probability density function's height is not a probability; an interval's probability is the area under it.

Some useful examples of the beta distribution are:

The arcsine distribution curve is plotted only for $0.001\le x\le 0.999$; its finite plotted ends do not represent finite endpoint values of the probability density function. Using the inverse sine, the shaded central interval has exact probability $\mathbb{P}(0.1<X<0.9)=1-4\arcsin (\sqrt{0.1})/\pi \approx 0.5903$, calculated analytically rather than by treating the cropped curve as normalized.

To separate location from spread, write $\kappa =\alpha +\beta$ and $m=\alpha /\kappa$. The mean and variance of the beta distribution areThe mean specifies the balance between the parameters; their sum $\kappa$ controls concentration about that mean. Multiplying both parameters by the same positive number preserves the mean and changes the variance. Larger $\kappa$ puts more area near $m$ and makes the probability density function taller there; the total area stays one. Small $\kappa$ allows much greater spread, sometimes favoring the endpoints rather than the mean itself.

For example, keeping mean $m=0.3$ and taking $\kappa =5,20,80$ gives the beta distributions $\mathrm{Beta}(1.5,3.5)$, $\mathrm{Beta}(6,14)$, and $\mathrm{Beta}(24,56)$. Their variances are $0.035$, $0.01$, and $0.21/81\approx 0.00259259$, respectively. Parameters need not be integers.

The formulas follow from two short moment calculations. Multiplying the probability density function by $x$ or $x^2$ increases the first parameter of the Euler beta function:Apply the Gamma function recurrence to the beta-gamma identity, then subtract $(\mathbb{E}[X]{)}^2$ from $\mathbb{E}[X^2]$ to obtain the variance.

The beta distribution is especially convenient as a prior distribution for the unknown success probability $P$ of Bernoulli trials. The Beta-binomial conjugacy says that $h$ successes and $t$ failures update a prior distribution $\mathrm{Beta}(\alpha ,\beta )$ to a posterior distribution $\mathrm{Beta}(\alpha +h,\beta +t)$. For example, suppose a coin has prior distribution $\mathrm{Beta}(2,2)$ and we observe eight heads and two tails. Its posterior distribution is then $\mathrm{Beta}(10,4)$, with posterior mean $5/7$. ^[ref].

For $D\sim \mathrm{Bin}(n,p)$ with $n>0$, the sample binomial proportion is $D/n$. Its expectation is p and its variance is $p(1-p)/n$. The central limit theorem gives a normal approximation when both expected outcome counts are sufficiently large. These moments determine a standard error and model-based control limits.

A local constant estimator reproduces constants but need not reproduce an affine regression function near a support boundary. Let $x_i=i/n$, take $0<h\le 1$ and the regression kernel $K=1_{[-1,1]}$, and suppose $m(x)=x$. At target zero, the noiseless Nadaraya–Watson estimator averages $i/n$ for $1\le i\le k=\lfloor nh\rfloor$. Whenever $k\ge 1$, its exact fitted value isSince $m(0)=0$, this is the bias of an estimator when observation errors have zero expected value. Under $h\to 0$ and $nh\to \infty$, it is asymptotic to $h/2$. The one-sided boundary window has a nonzero first coordinate moment, explaining the first-order bias. An invertible local linear regression fit reproduces $m(x)=x$ exactly at the target and removes this term. If $k=0$, the local constant average is undefined rather than zero, so a nonempty window is a necessary condition.

For random variables supported in one fixed compact interval, convergence of every moment is equivalent to weak convergence. One direction tests monomials; the other follows by uniformly approximating continuous functions by polynomials.

For probability measures on ${\mathbb{R}}^d$ with finite second moments and a source satisfying absolute continuity of measures with respect to Lebesgue measure, the quadratic Kantorovich optimal transport problem has a unique optimal transport plan. It is induced by the gradient of a convex function, which also uniquely solves the Monge optimal transport problem up to a source-null set.

For a nonnegative symmetric unit-integral kernel for density estimation with $0<{\mu }_2(K)=\int u^{2}K(u)du<\infty$ and $R(K)=\int K^2<\infty$, the canonical scaling satisfies $R(K_c)={\mu }_2(K_c{)}^2$. Since $R(K_c)=R(K)/c$ and ${\mu }_2(K_c)=c^2{\mu }_2(K)$, the unique scale is $c=[R(K)/{\mu }_2(K{)}^2{]}^{1/5}$. Consequently its asymptotic mean integrated squared error is $C(K)[(nh{)}^{-1}+h^{4}R(f^{\prime \prime })/4]$, where $C(K)=R(K{)}^{4/5}{\mu }_2(K{)}^{2/5}$ is invariant under kernel rescaling. The leading optimal smoothing bandwidth becomes $[nR(f^{\prime \prime }){]}^{-1/5}$, independent of the kernel shape. Unit second moment is a different normalization and does not have this separation property.

Cantelli's inequality uses a mean and variance to control deviations in one specified direction. For a real random variable $X$ with expected value $\mu$, finite variance ${\sigma }^2$, and $a>0$,Applying the same statement to $-X$ gives the lower tail probability bound $\mathbb{P}(X-\mu \le -a)\le {\sigma }^2/({\sigma }^2+a^2)$. No independence or particular probability distribution is required. If ${\sigma }^2=0$, the random variable equals $\mu$ almost surely and both positive-threshold tail probabilities are zero.

For variance one and a threshold two above the mean, the upper-tail probability is at most $1/5$. Applying the two-sided Chebyshev's inequality to contain that upper event would give the weaker bound $1/4$. For the standard normal distribution, the actual upper tail probability at two is about $0.0227501$, much smaller than either bound. Yet the two-point probability distribution in equality in Cantelli's inequality attains $1/5$, showing why the smaller Gaussian value cannot be promised from these two moments alone.

A short explanation is to centre $Y=X-\mu$ and choose a nonnegative shift $b$. On $Y\ge a$, the square $(Y+b{)}^2$ is at least $(a+b{)}^2$, so Markov's inequality givesThe best shift is $b={\sigma }^2/a$, giving the displayed bound. Choosing the shift before squaring exploits the direction of the deviation; squaring the unshifted random variable treats both directions together. This shifted-square proof is also the strategy requested in the University of Utah's 2005 qualifying problem on Cantelli's inequality. ^[ref].

The one-sided improvement should be used for a one-sided question. Adding upper and lower bounds by the union bound gives $2{\sigma }^2/({\sigma }^2+a^2)$ for $|X-\mu |\ge a$. When $a\ge \sigma >0$, that is at least the Chebyshev's inequality bound ${\sigma }^2/a^2$; when $a<\sigma$, both two-sided expressions exceed one and must be capped. The separate one-sided bounds generally cannot be attained simultaneously by the same probability distribution.

If $\mathbb{E}{e}^{tX}\le e^{{\sigma }^2{t}^2/2}$, then $X^2-\mathbb{E}{X}^2$ is a sub-exponential random variable with scale at most a universal constant times ${\sigma }^2$. The Chernoff bound gives $\mathbb{P}(|X|>x)\le 2e^{-x^2/(2{\sigma }^2)}$; integrating this tail bounds $\mathbb{E}|X{|}^{2q}$ by $2(2{\sigma }^2{)}^{q}q!$. Centering preserves a factorial moment bound, so the Bernstein bound for independent sub-exponential variables apply to independent averages of these squares.

If $m\ge 1$ is an integer and $\mathbb{E}[|X{|}^m]<\infty$, the characteristic function has continuous derivatives through order $m$, withThus finite absolute moments justify recovering moments from derivatives.^[ref]

For a short justification of differentiation under the integral sign, the difference quotient for the $(k-1)$st derivative has integrand $(iX{)}^{k-1}{e}^{itX}(e^{ihX}-1)/h$. Its absolute value is at most $|X{|}^k$, since $|e^{iu}-1|\le |u|$ for real $u$. Lower absolute moments are finite because $|X{|}^k\le 1+|X{|}^m$. Apply the dominated convergence theorem successively, then use the same integrable bound to obtain continuity of each derivative.

For the Bernoulli distribution with parameter $1/3$, direct differentiation gives ${\phi }_B^{\prime }(0)=i/3$ and ${\phi }_B^{\prime \prime }(0)=-1/3$, agreeing with $\mathbb{E}[B]=\mathbb{E}[B^2]=1/3$. Its variance is therefore $1/3-(1/3{)}^2=2/9$. Existence and continuity of a characteristic function alone do not authorize these moment calculations: the standard Cauchy distribution, whose expected value is undefined, still has a characteristic function. The stated finite absolute moment hypothesis is a sufficient condition for the derivative formula, not an assumption built into the definition of a characteristic function.

Independence makes the characteristic function of a sum equal the product of the individual characteristic functions.

For mutually independent random variables $X_1,\dots ,X_n$, write $S_n={\sum }_{j=1}^n{X}_j$. ThenThis is the expectation of a product of independent random variables applied to bounded complex exponentials; no moment hypothesis is needed.^[ref]

Identical probability distributions are unnecessary. When the random variables are independent and identically distributed random variables, the product is ${\phi }_X(t{)}^n$. A deterministic rescaling also follows directly from the definition: ${\phi }_{aX+b}(t)=e^{itb}{\phi }_X(at)$ for real $a,b$. Consequently the sample mean has characteristic function ${\phi }_X(t/n{)}^n$, and a sum divided by $\sqrt{n}$ has characteristic function ${\phi }_X(t/\sqrt{n}{)}^n$.

The independence assumption does real work. Let $X$ take $-1$ and $1$ with equal probabilities and set $Y=X$. Both characteristic functions are $\cos t$, but ${\phi }_{X+Y}(t)=\cos (2t)$, not ${\cos }^{2}t$. At $t=\pi /2$ they give $-1$ and zero respectively. Knowing the two marginal probability distributions does not determine the sum without their joint probability distribution.

Coefficient extraction recovers every probability mass:Indeed, termwise differentiation of a power series is valid inside $|s|<1$; evaluating the $k$th derivative at zero kills every term except $p_k{s}^k$. Thus two probability generating functions with the same coefficients describe the same probability distribution, without any finite-moment assumption.

At one, approach along the real line from below. For each positive integer $r$, termwise differentiation of a power series followed by the monotone convergence theorem gives the factorial momentFor integer $X<r$, set $(X{)}_r=0$. The limit may be infinite; it is a finite boundary value only when the corresponding factorial moment is finite. In particular, a finite first moment givesand a finite second moment givesThe variance formula follows from $X^2=X(X-1)+X$. These are boundary limits of derivatives; they do not assert that the probability generating function extends smoothly through one.

For a nonnegative integer-valued random variable, its moment-generating function satisfies $M_X(t)=G_X(e^t)$ wherever finite. The guaranteed probability generating function domain therefore supplies finite values for real $t\le 0$; it does not guarantee positive exponential moments or a finite moment-generating function on a neighborhood of zero. Bounded nonnegative integer-valued random variables have polynomial probability generating functions, but their convenient behavior is not a general existence theorem.

For an independent binomial distribution count $N\sim \mathrm{Binomial}(n,p)$, the aggregate can also be represented as ${\sum }_{j=1}^n{B}_j{X}_j$, with independent parameter-$p$ Bernoulli random variables $B_j$ and independent claim sizes. Its cumulant-generating function is $n\log (1-p+p{M}_X(t))$. Its third cumulant is $n(p{m}_3-3p^2{m}_1{m}_2+2p^3{m}_1^3)$ in severity moments. Constant positive claims $d$ give $np(1-p)(1-2p)d^3$, which is negative when $p>1/2$. Thus positive claim sizes do not by themselves force a positive third central moment.

Take an independent geometric distribution count with $\mathbb{P}(N=k)=p(1-p{)}^k$, $k\ge 0$, and sum its independent positive claims. Its moment-generating function is the displayed composition wherever finite. With $q=1-p$ and severity moments $m_j$, the third cumulant is $(q/p)m_3+3(q/p{)}^2{m}_1{m}_2+2(q/p{)}^3{m}_1^3$, which is strictly positive for nonzero positive claims and $0<p<1$. The law has an atom of mass $p$ at zero. Its positive-count convention would instead add a factor $M_X(t)$ to the numerator, as in the geometric-sum moment-generating function.

Conditioning on the independent Poisson distribution count in a random sum gives $\mathbb{E}{e}^{tS}=\exp (\lambda (M_X(t)-1))$ wherever finite. Differentiation at zero gives the displayed cumulants when the relevant derivatives exist. The aggregate variance is the count rate times the raw second claim moment, not just the claim variance. Positive moment-generating function arguments need an exponential moment; finite first and second claim moments suffice for the aggregate mean and variance by conditioning or right derivatives at zero of the Laplace transform of a nonnegative random variable.

A Conditional hitting time describes a target-passage time after conditioning on a specified successful outcome. Let a Finite Markov chain hit disjoint sets $A\cup B$ almost surely, put $T=T_A\wedge T_B$, and let $h_i={\mathbb{P}}_i(T_A<T_B)$. For states with $h_i>0$, the conditional mean isThe weighting matters: conditioning changes the probabilities of the next state. The successful conditional transition kernel, a Doob h-transform, is $P_{ij}^{(h)}=P_{ij}{h}_j/h_i$ outside $A$. Transitions into $B$ receive weight zero and $A$ is the terminal target. Therefore $m=0$ on $A$ andEquivalently, the weighted moments $v_i={\mathbb{E}}_i[T{1}_{\{T_A<T_B\}}]=h_i{m}_i$ satisfy $v_i=h_i+{\sum }_{j\notin A\cup B}{P}_{ij}{v}_j$, with zero boundary values.

For symmetric random walk on $\{0,1,2,3\}$ absorbed at zero or three, condition on reaching three first. Since $h_i=i/3$, the conditioned chain moves from one to two with Probability one, and from two to one or three with probabilities $1/4$ and $3/4$. Its means satisfy $m_1=1+m_2$ and $m_2=1+m_1/4$, giving $m_1=8/3$. The unconditioned absorption mean from one is two. A successful path need not have the same waiting-time distribution as an unconditioned path.

For fixed coefficient dimension and $Y=X{\beta }_0+\epsilon$, the ordinary least squares error satisfiesConsequently convergence in probability of $X^{\mathsf{T}}X/n$ to a positive-definite matrix $Q$, together with $X^{\mathsf{T}}\epsilon /n\to 0$ in probability, proves statistical consistency by continuity of matrix inversion.

For independent and identically distributed random-design pairs, finite second moments and $\mathbb{E}[x_i{\epsilon }_i]=0$ imply these limits by the law of large numbers, provided $Q=\mathbb{E}[x_i{x}_i^{\mathsf{T}}]$ is nonsingular. Conditional mean zero, $\mathbb{E}[{\epsilon }_i\mid x_i]=0$, is a familiar sufficient exogeneity condition. For a deterministic design matrix, the vanishing normalized design-error product must instead be justified from its errors and weights.

Finite moments and a nonsingular design alone are insufficient. In an intercept-only model with errors identically one, the estimated intercept is always ${\beta }_0+1$. The missing zero-expected value condition changes the limiting target.

Every characteristic function is uniformly continuous and satisfies $\phi (0)=1$ and $|\phi (t)|\le 1$. Continuity follows from the dominated convergence theorem applied to $e^{itX}$.

More explicitly, if $t_n\to t$, then $e^{i{t}_{n}X}\to e^{itX}$ pointwise and $|e^{i{t}_{n}X}|=1$. Apply the dominated convergence theorem to the real part and imaginary part separately. There is no finite moment hypothesis. For uniform continuity, the stronger estimatefollows by factoring out $e^{itX}$, whose absolute value is one. The integrand tends pointwise to zero and is bounded by two, so the dominated convergence theorem applies again. The bound is independent of $t$, which supplies uniform continuity on the entire real line.^[ref]

If the first absolute moment is finite, $|e^{iu}-1|\le |u|$ additionally gives $|{\phi }_X(t+h)-{\phi }_X(t)|\le |h|\mathbb{E}[|X|]$, a Lipschitz continuity estimate. The earlier argument matters because continuity survives even when this expected value is infinite. For convergence in distribution, the separate Lévy continuity theorem concerns limits of whole sequences of characteristic functions; the argument here establishes continuity of one characteristic function.

Suppose the moment-generating functions $M_n$ are finite on a common open interval $(-a,a)$ containing zero and converge there pointwise to the moment-generating function $M_X$ of a random variable $X$, with $M_X$ finite there. Then $X_n$ converges in distribution to $X$. Equivalently, a finite pointwise limit on such an interval is the moment-generating function of a unique limiting probability distribution. This theorem requires control on a neighborhood of zero; convergence of isolated moments is a different hypothesis.