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A standardized mean difference divides a group mean difference by a pooled standard deviation.

This gives a dimensionless effect size, so mean differences can be compared across measurement scales.

When $Y_p=0$, $\mathcal{F}=X_{H^{+}}$, and

The flux of the local Couette-Poiseuille profile is

For $U_0=0$, $U_1=U$, and constant $G=p_x$, zero shear at $y=0$ requiresThe velocity becomes $u=U{y}^2/h^2$, the flux per unit width is $Uh/3$, and the fluid exerts shear $-2\mu U/h$ on the moving upper wall.

A beta plane retains the leading meridional variation $f=f_0+\beta y$ of the Coriolis parameter.

An f-plane approximates the Coriolis parameter by a constant $f_0$ over a limited horizontal region.

The dual sequence consists of the coordinate functionals $(e_n^{\ast })$ in $X^{\ast }$. It is always a basic sequence, though it need not span the entire dual space.

For ${\omega }_m=e^{-2\pi i/m}$ and length-$m/2$ transforms $E_k,O_k$ of the even and odd samples,

For $F(x)=e^{-x}{1}_{x\ge 0}$, induction in the convolution integral givesIts Fourier transform in the angular-frequency convention is $(1+ik{)}^{-n}$.

For nonzero terms, ifthen $\sum a_n$ converges absolutely. If the ratio has a limit greater than one, the terms do not tend to zero and the series diverges.

For positive terms, if the ratio of two sequences tends to a finite positive number, their series either both converge or both diverge.

The map $(x_n)\mapsto (\lim x_n,x_1-\lim x_n,x_2-\lim x_n,\dots )$ proves $c\cong c_0$, equivalently $c_0\oplus \mathbb{F}\cong c_0$.

An increasing real sequence converges exactly when it is bounded above. For a bounded sequence, its limit is the supremum of its set of terms.

For a positive sequence converging to $L\ge 0$, its cumulative geometric means converge to the same limit.

The Cesàro mean of the first $n\ge 1$ terms of a sequence is their arithmetic mean. It is a cumulative average: increasing $n$ includes the whole initial segment, rather than a fixed-length sliding window. For $a_k=(-1{)}^k$, the even-length averages are zero and the odd-length averages are $-1/n$, so they tend to zero even though the sequence does not converge.

For a periodic sequence with period $p\ge 1$ and one-cycle arithmetic mean $M$, split an initial segment into full cycles and a remainder of fewer than $p$ terms. The full cycles contribute exactly $M$ per term; the remainder contributes a fixed bounded error divided by $n$. Thus the Cesàro means tend to $M$. The cycle $0,2,1$ in the bounded partial sums need not be Cesàro summable illustration has arithmetic mean one. Treating it as a sequence of partial sums gives a series with first-order Cesàro sum one; averaging a sequence and summing its terms remain different operations.

Independent indicators with success probabilities $1/n$ converge to zero in probability, but successes occur infinitely often almost surely by the second Borel--Cantelli lemma.

From $X_n\to X$ in probability, choose $n_k$ with
$\mathbb{P}(|X_{n_k}-X|>2^{-k})<2^{-k}$. The first Borel--Cantelli lemma then gives $X_{n_k}\to X$ almost surely.

Convergence in probability is equivalent toThe forward implication splits at a fixed error threshold; the reverse implication is Markov's inequality.

If $X_n$ converges in probability to both $X$ and $Y$, then $X=Y$ almost surely. Indeed, the triangle inequality and a union bound show that $\mathbb{P}(|X-Y|>\epsilon )=0$ for every $\epsilon >0$.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$ and expected value $\mu =\mathbb{E}[X_1]$. The weak law of large numbers states that their sample mean has convergence in probability to $\mu$: for every fixed $\epsilon >0$,Thus a large sample is unlikely to give an average with a specified fixed-size error. The strong law of large numbers makes the stronger statement of almost sure convergence along an entire sequence. Under the displayed assumptions both laws apply; their names distinguish the conclusions, not two competing predictions.

When the variance ${\sigma }^2$ is finite, variance additivity for independent random variables gives $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. The Chebyshev inequality immediately supplies a useful quantitative form and a brief proof:For independent trials with a Bernoulli distribution of parameter $p=0.2$, the expected value is $0.2$ and the variance is $0.16$. At $n=1000$, the chance of an error of at least $0.1$ is therefore at most $0.016$. The exact binomial distribution gives a much smaller probability, as the figure under the strong law of large numbers shows. The bound is conservative, but it is valid for every sample size under these assumptions and does not require a normal approximation.

The weak law of large numbers addresses a fixed tolerance $\epsilon$. The central limit theorem instead resolves errors on the shrinking scale $\sigma /\sqrt{n}$ and gives their approximate normal distribution. In particular, the weak law of large numbers alone supplies neither a bell-shaped error law nor a universal rate of approach. For example, a Pareto distribution with minimum one and shape $3/2$ has expected value three but infinite variance: both the weak law of large numbers and strong law of large numbers apply, while the finite-variance statement of the central limit theorem does not.