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For standard one-dimensional Brownian motion and a fixed $h>0$, the Brownian reflection principle gives the survival probability $2\Phi (h/\sqrt{t})-1$. The standard normal distribution function has derivative $1/\sqrt{2\pi }$ at zero, giving the displayed asymptotic. Thus the persistence exponent is one half, with leading constant $h\sqrt{2/\pi }$.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

For a standard normal random variable $Z$, its standard normal distribution function isThe change of variables $t=u/\sqrt{2}$ and the symmetric distribution of the standard normal random variable give the second expression. For $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization of a normal random variable then givesIn particular, the intervals within one, two, and three standard deviations have calculated probabilities about $0.682689$, $0.954500$, and $0.997300$. These values depend on a normal distribution; they are not universal coverage guarantees for an arbitrary random variable. This is the direct connection between a Gaussian function area and the familiar central intervals.

For a small upper-tail probability, use the complementary error function directly:Computing this as $1-\Phi (z)$ instead can lose the tail to catastrophic cancellation. The Gaussian tail bound gives another useful estimate without requiring an exact error function value.

In the dividend-free Black-Scholes model, let $\tau =T-t>0$, $K>0$, $S>0$, $\sigma >0$ andThen the displayed finite sum prices the European option with power put option payoff. Here $\Phi$ is the standard normal distribution function. Indeed, $\log S_T$ conditionally has a normal distribution with mean $\log S+(r-{\sigma }^2/2)\tau$ and variance ${\sigma }^2\tau$. Expand $(K-S_T{)}^n$ by the binomial theorem on $S_T<K$ and apply the lower truncated lognormal moment to each term. Equivalently,This expectation representation proves monotonicity and convexity in $S$. It also gives $P_n(0,t)=K^n{e}^{-r\tau }$ and $P_n(S,t)\to 0$ as $S\to \infty$ by the dominated convergence theorem. For positive $S$ the lognormal distribution smoothing makes the price smooth and strictly positive.

For $X\sim N(m,v)$ with $v>0$ and $0<\epsilon <1$, $P(X\ge C)\le \epsilon$ is equivalent to the displayed chance constraint, because the standard normal distribution function is continuous and strictly increasing. The coefficient is a standard normal quantile, which can be negative. If $v=0$, equality is an atom and the exact condition becomes $C>m$ instead.

For a normal distribution with mean $m$, positive variance $v$, and $\gamma >0$, the target $\mathbb{P}(X\ge C)\le e^{-\gamma }$ is equivalent to the displayed constraint, where $\Phi$ is the standard normal distribution function. This follows from $\mathbb{P}(X\ge C)=1-\Phi ((C-m)/\sqrt{v})$ and strict monotonicity of $\Phi$. Unlike the sufficient Chernoff bound margin $\sqrt{2\gamma v}$, this constraint is necessary as well as sufficient. The exact quantile can be negative when $e^{-\gamma }>1/2$.

For $Z$ with a standard normal distribution and $a\ge 0$, split the Gaussian integral at zero and complete the square to obtain $\mathbb{E}{e}^{a|Z|}=2e^{a^2/2}\Phi (a)\le 2e^{a^2/2}$, where $\Phi$ is the standard normal distribution function. This converts a bound on an absolute Gaussian random variable into a two-sided Gaussian tail bound using Markov inequality.

The Gaussian error linear unit activation is $\mathrm{GELU}(x)=x\Phi (x)$, where $\Phi$ is the standard normal distribution function.

Take the original synthetic model $T\sim N(0,0.6)$ and ${\epsilon }_1,{\epsilon }_2\sim N(0,0.4)$, mutually independent random variables, with the second argument denoting variance. The pair $(X,Y)$ has a multivariate normal distribution with mean zero, unit marginal variances, and Pearson correlation coefficient $0.6$. Its conditional multivariate normal distribution isFor a first reading $x=2$, the second reading has conditional expectation $1.2$ and conditional variance $0.64$, hence standard deviation $0.8$. There is still conditional probability $1-\Phi (1)\approx 0.1587$ of $Y>2$, a movement farther from zero, where $\Phi$ is the standard normal distribution function. The line $0.6x$ describes a conditional mean, not an individual trajectory.

A short derivation isolates $R=Y-0.6X$. Its covariance with $X$ is zero and its variance is $0.64$. Because $(X,R)$ has a multivariate normal distribution, zero covariance implies independence: $R\sim N(0,0.64)$ independently of $X$. This also explains why reversing the two measurements gives $\mathbb{E}[X\mid Y=y]=0.6y$. The statistical association does not identify a direction of causal effect.

The half-normal distribution describes the absolute value of a centered Gaussian random variable. Its unit-scale version is $H=|Z|$ for a standard normal random variable $Z$. It folds the negative half of the standard normal distribution onto the positive half, retaining the size of a deviation while discarding its sign. If a centered Gaussian random variable models measurement error, its magnitude therefore has a half-normal distribution.Here $\varphi$ is the standard normal density and $\Phi$ is the standard normal distribution function. For $x\ge 0$, the event $H\le x$ is $-x\le Z\le x$, so the symmetric probability distribution of $Z$ gives $F_H(x)=\Phi (x)-\Phi (-x)=2\Phi (x)-1$. The upper-tail probability is therefore $\mathbb{P}(H>x)=2(1-\Phi (x))$. For example, $\mathbb{P}(H\le 1)\approx 0.6826894921$. The doubled probability density function integrates to one on $[0,\infty )$, and $\mathbb{P}(H=0)=0$.

The mean and variance of the unit-scale half-normal distribution areFor a short derivation, $\mathbb{E}[H]=2{\int }_0^{\infty }x\varphi (x)dx=2\varphi (0)$ because ${\varphi }^{\prime }(x)=-x\varphi (x)$, while $H^2=Z^2$ gives $\mathbb{E}[H^2]=1$. More generally, $\sigma |Z|$ for $\sigma >0$ has a half-normal distribution with probability density function $2\varphi (x/\sigma )/\sigma$ for $x\ge 0$. Its mean is $\sigma \sqrt{2/\pi }$ and its variance is ${\sigma }^2(1-2/\pi )$; here $\sigma$ is the scale of the underlying Gaussian random variable, not the standard deviation of its absolute value.

Fix a scale $s>0$ and cutoff $k>0$. Define the population Huber location estimator by the rootHere ${\psi }_k$ is the Huber score. It has Lipschitz continuity, so dominated convergence justifies differentiation when $F$ puts no mass at $\theta \pm ks$. Then $A={\mathbb{P}}_F(|X-\theta |<ks)$, and the nonzero derivative identifies the local population root. The implicit function theorem applies away from the contaminating score's kinks; its Lipschitz continuity also gives the first-order expansion at those kinks. Differentiating its contaminated equation yieldsThe normalization matters: the Huber score alone is not the influence function. The gross-error sensitivity is $sk/A$, while the rejection point of an influence function is infinite because the clipped tails stay nonzero. The original clipped estimating criterion and its derivative normalization also appear in the population analysis of M-estimators.^[ref]

Scale sets what counts as a large residual. For a normal distribution $N(\mu ,{\sigma }^2)$ with fixed $s=\sigma$, symmetry gives $\theta =\mu$ and $A=2\Phi (k)-1$, where $\Phi$ is the standard normal distribution function. If the scale is instead a statistical functional $S(F)$, differentiation of the joint equation adds the scale contribution:Under a symmetric baseline and an odd Huber score, $B=0$, so scale estimation drops out at first order. This cancellation needs regularity of $S$ and does not protect finite contamination changes: an inflated standard deviation enlarges the physical cutoff $ks$. A robust choice, such as a suitable Scale M-estimator, needs its own influence function and breakdown point checked. Fixing a numeric cutoff in raw units also sacrifices scale equivariance when units change.^[ref]

If a cumulative distribution function is constant between consecutive lattice points spaced by $h$ near zero, then its Kolmogorov distance from the standard normal distribution function is at least half the increase of that function across one such interval. This gives a lower bound of order $h$.

The Lilliefors test calibrates the Kolmogorov-Smirnov statistic against a normal distribution whose mean and variance are estimated from the tested observations. In the original procedure, $s^2$ is the sample variance with denominator $n-1$, and $\Phi$ is the standard normal distribution function. Its reference law includes this fitting step. The original paper's opening procedure and Table 1 explain the changed calibration.^[ref]

For $a>0$, real $b$ and standard Brownian motion, the first-passage time to a linear boundary hasHere $\Phi$ is the standard normal distribution function. Differentiation gives $a\exp (-(a+bt{)}^2/(2t))/\sqrt{2\pi t^3}$, the drifted Brownian first-passage density. Its Laplace transform is $\exp [-a(b+\sqrt{b^2+2\theta })]$ for $\theta >0$. There is mass $1-e^{-2ab}$ at infinity when $b>0$; for $b\le 0$ the hitting time is finite almost surely.

For $v,m\ge 0$, define $F(v,m)=\mathbb{E}[(e^{-v/2+\sqrt{v}Y}-m{)}^{+}]$, where $Y$ has the standard normal distribution. For $v,m>0$, completing the square in the standard normal density giveswhere $\Phi$ is the standard normal distribution function. The boundary values are $F(0,m)=(1-m{)}^{+}$ and $F(v,0)=1$. Multiplying by the current stock price produces a zero-interest Black-Scholes formula with total variance $v$ and relative strike $m$.

For the same normal distribution model, a two-sided test compares the null hypothesis $H_0:\mu ={\mu }_0$ with the alternative hypothesis $H_1:\mu \ne {\mu }_0$. Under the null hypothesis, the standardized test statistic has standard normal distribution. Write $\Phi$ for the standard normal distribution function. With $\bar{x}=1.2$, $\sigma =2$, $n=16$ and ${\mu }_0=0$, the observed test statistic is $z=1.2/0.5=2.4$. Its p-value isThus the p-value is a tail probability computed under the null hypothesis, including departures in both directions. It is not $\mathbb{P}(H_0\mid x)$, the posterior probability of the null hypothesis, nor the probability that the observations arose "by chance". The normal distribution test statistic and its calibration appear in NIST's teaching reference for testing a process mean; the tail interpretation is stated in Penn State's hypothesis-testing notes.

Choose the significance level $\alpha =0.05$ before observing the sample. The rejection region is $|Z|>z_{0.975}$, whose probability under $H_0$ is exactly 0.05. Here $2.4>1.9599639845\dots$, so reject $H_0$. This agrees with zero lying outside $[0.22,2.18]$. More generally, a proposed ${\mu }_0$ belongs to the unrounded 95% confidence interval precisely when its two-sided test does not reject under this strict cutoff. The equality boundary has probability zero in this continuous statistical model.

The significance level and statistical power describe different repeated-experiment probabilities. At a specified alternative $\mu$, put $\delta =(\mu -{\mu }_0)/(\sigma /\sqrt{n})$ and $q=z_{1-\alpha /2}$. ThenThe first is the Type I error probability; the second is the Type II error probability at that alternative. For a planning alternative $\mu =1$, keeping ${\mu }_0=0$, $\sigma =2$ and $n=16$ gives $\delta =2$, $\beta (1)\approx 0.483995$ and statistical power $0.516005$. At $n=64$, $\delta =4$ and statistical power rises to $0.979327$. These are model calculations for specified alternatives, not posterior probabilities inferred from a nonsignificant result. They explain why failure to reject cannot establish equality and why choosing sample size before an experiment is useful.

Historically, Ronald Fisher's 1925 book discusses the normal cutoff $1.96$ and tail probability 0.05 as a convenient convention on page 47, rather than a mathematically compulsory threshold: ^[ref].

Under the risk-neutral measure, set $a=\sigma \sqrt{\tau }$ and $m=\log (S/K)+(r-{\sigma }^2/2)\tau$. Then $\log (S_T/K)=m+aZ$ with $Z$ a standard normal random variable, and $d_{-}=m/a$. The log-call option is the discounted Gaussian positive-part expectation:since ${\varphi }^{\prime }(z)=-z\varphi (z)$. ThereforeHere $\varphi$ and $\Phi =N$ are the standard normal density and standard normal distribution function. At expiry the continuous terminal limit is $(\log (S/K){)}^{+}$.

Under the risk-neutral measure, write $Z\sim N(0,1)$, $a=\sigma \sqrt{\tau }$, and $m=\log (S/K)+(r-{\sigma }^2/2)\tau$. Then $\log (S_T/K)=m+aZ$ and $d_{-}=m/a$. By risk-neutral pricing, this log-call option has value $e^{-r\tau }\mathbb{E}[(m+aZ{)}^{+}]$.

Let $\varphi$ and $\Phi$ denote the standard normal density and standard normal distribution function. The Gaussian positive-part expectation follows by integration:because ${\varphi }^{\prime }(z)=-z\varphi (z)$. ThereforeFor $\tau =0$ take the continuous terminal limit $(\log (S/K){)}^{+}$.

Source: original released examination PDF, bundle pages 66–69; paper pages 1–4. The bundle filename says 2020, but the printed paper is dated 2 June 2021.

Solutions to all three questions. Write $\tau =T-t$, $\Phi$ for the standard normal distribution function, and $\varphi$ for the standard normal density. Formulas with $\tau$ in a denominator are for $t<T$; terminal values are specified separately. Barrier formulas refer to a contract whose barrier has not previously been hit, unless stated otherwise.

Write $\tau =T-t>0$ and $g_n(s)=(K-s{)}_{+}^n$. The European power put formula isHere $Y>0$ has a lognormal distribution. Since $g_n$ is decreasing and is a convex function, its discounted expected value is also decreasing and convex as a function of $S$. For $S>0$ it is smooth and strictly positive: the normal distribution assigns positive probability to $SY<K$ whatever the current positive spot price. Its endpoints areThe dominated convergence theorem justifies the second limit, because $0\le (K-SY{)}_{+}^n\le K^n$. For a quantitative sketch, the binomial theorem and truncated lognormal moment givewhere $\Phi$ is the standard normal distribution function. This finite sum is useful for drawing the curve; its positive expectation form makes its qualitative shape clearer.

The solid curves use $K=1$, $r=0.05$, $\sigma =0.30$ and $\tau =1$; the dashed curves show their terminal financial payoffs for comparison. Each price remains positive at $S=K$ and tends smoothly towards zero beyond it. At small $S$, discounting gives a price below the undiscounted financial payoff; European exercise is permitted only at expiry.