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PutThe call payoff is $(S_T-K{)}^{+}$, and it is positive whenWriting $\varphi$ for the standard normal density, the given risk-neutral pricing formula becomesCompleting the square givesso the two tail integrals are $\Phi (d_{+})$ and $\Phi (d_{-})$, where $\Phi$ is the standard normal distribution function. Therefore the Black-Scholes formula is

The payoff identityreplicates the call-minus-put position by one stock and borrowing the present value of $K$. Hence put-call parity givesand therefore

The standard normal density isIt is nonnegative, and the Gaussian integral gives ${\int }_{-\infty }^{\infty }\varphi (z)dz=1$, so it is a probability density function. Completing the square gives the moment-generating function of a standard normal variableThus $\mathbb{E}Z=M_Z^{\prime }(0)=0$ and

Write $F=(1-ϵ)\Phi +ϵH$ and assume $0<ϵ<1/2$. If $m$ is its median, thenMoreover its density satisfies $f(m)\ge (1-ϵ)\varphi (m)$. Since the standard normal density decreases with $|m|$, the smallest possible density at the median occurs at either endpoint. PutThe two endpoints are $\pm q_{ϵ}$ by normal symmetry. The bound is attained by choosing a contaminating density supported strictly to the right of $q_{ϵ}$, or symmetrically to the left of $-q_{ϵ}$, with zero density at the selected median. Therefore

Complete the square in the integral against the standard normal density:The remaining integral is one because its integrand is the probability density function of the normal distribution $N(t,1)$. Hence the moment-generating function of a standard normal variable is finite on the whole real line.

Integrate against the standard normal density by completing the square:Thus the moment-generating function of a standard normal variable is

For $\lambda >1$, use $x/\lambda \ge 1$ on the tail:This proves the requested Gaussian tail bound.

The subsequent printed maximum inequality is false as stated. For $n=1$ its integrand never reaches one, and its right-hand side isBut for two independent absolute standard normal variables, using their common distribution function $F(x)=2\Phi (x)-1$ and the standard normal density $\phi$,The middle step is integration by parts with ${\phi }^{\prime }=-x\phi$. Thus no proof of the displayed source inequality can hold for all its stated indices.

The correct tail integral formula for expectation isA valid union bound givesIn particular, for $x>1$ this is at most $\min (1,C_n{e}^{-x^2/2})$, where $C_n=2^{n+1}/\sqrt{2\pi }$. This corrected estimate proves the intended summability in part (c), without silently replacing the source formula.

Differentiate the cumulative distribution function for $y>0$ using the chain rule and the standard normal density:Thus a probability density function isThe value assigned at zero is immaterial. The probability density function is unbounded near zero, but $F_Y(0)=0$ and continuity of the cumulative distribution function there show that there is no point mass. This is a gamma distribution with shape $1/2$ and rate $1$.

A pulse-contrast denominator acceptance rule decides which shots may be normalized by their measured central signal. For background-corrected side signal $S$ and central signal $C$, considerThe rule prevents normalization by an arbitrarily small accepted denominator. It also changes the estimand: averaging accepted ratios estimates $\mathbb{E}[R\mid A]$, a conditional expectation. An unconditional expected value is a different target and need not exist. The pulse-contrast ratio denominator noise explains why even the unfiltered ratio mean can differ from the ratio of mean signals.

The rule provides a precise stability guarantee when its moment assumptions hold. Interpret the indicator-weighted integrand below as zero outside $A$. If $p_A=\mathbb{P}(A)>0$ and $\mathbb{E}{S}^2<\infty$, thenThis proves finite ratio moments under a positive denominator cutoff. The accepted ratio has finite variance and finite first absolute moment, the latter by the Cauchy-Schwarz inequality. No independence of $S$ and $C$ was used. The bound concerns mathematical stability, not whether accepted shots represent the full pulse population.

For an original brightness-selection example, suppose half the shots have $(C,S)=(2,0.02)$ and half have $(10,0.02)$. The two ratios are $0.01$ and $0.002$, so their expected value over all shots is $0.006$. A cutoff $c_{\min }=5$ retains only the bright group, giving accepted conditional expectation $0.002$. Meanwhile the ratio of overall mean signals is $0.02/6\approx 0.003333$. These three quantities answer different questions. Physical dependence between brightness and temporal shape can make the choice consequential even without readout noise.

Selection bias from a noisy denominator threshold arises even at fixed true brightness. Consider another original model with constant $S=s>0$ and measured $C\sim N(\mu ,{\sigma }^2)$, where $\mu >0$. Write $z=(c_{\min }-\mu )/\sigma$, with $\varphi$ the standard normal density and $\Phi$ the standard normal distribution function. The accepted central measurements have a truncated normal distribution, withTo derive the second identity, standardize $C$ and use ${\varphi }^{\prime }(x)=-x\varphi (x)$, so ${\int }_z^{\infty }x\varphi (x)dx=\varphi (z)$. The density-to-tail quotient is an inverse Mills ratio.

Choose $\mu =5$, $\sigma =1$, $c_{\min }=6$ and $s=0.05$ in arbitrary signal units. The true-signal ratio is $s/\mu =0.01$. Only $15.87\%$ of measured shots pass, and the conditional expectation of their measured denominator is $6.5251$. Since $c\mapsto 1/c$ is a convex function for $c>0$, Jensen's inequality and the cutoff implyThe accepted ratios are necessarily biased downward relative to this model's true-signal ratio. The cutoff selects upward central-readout errors; increasing the accepted-shot count does not remove that conditioning.

In the historical single-shot correlator, nonlinear mixing encoded delays spatially and an imaging lens recorded the third-harmonic signal on a filtered CCD detector.^[ref] Selecting a trace by its central pixels conditions subsequent normalization on the same recorded data. Preserve the threshold, background treatment, number attempted and number accepted alongside the shot-resolved pulse-contrast estimator. For independent shots sharing the same acceptance probability, the accepted count has a binomial distribution, carrying information that disappears if only normalized accepted traces are retained.

Select units with first reading $X>1$, keep that membership fixed, and measure the same units again. The first reading now has a truncated normal distribution. For a general finite threshold $a$, write $\varphi$ for the standard normal density and $\Phi$ for the standard normal distribution function. Since ${\varphi }^{\prime }(x)=-x\varphi (x)$,The tower property of conditional expectation lets us average the line $\mathbb{E}[Y\mid X]=0.6X$ over the selected group.^[ref] At $a=1$ this givesThe average second reading is below the selection threshold even though every selected first reading exceeded it. This is a group expectation; any finite generated group's sample mean fluctuates around it. Selecting again on $Y>1$ would define a different group and answer a different question.

The latent variable decomposition makes the apparent drop especially clear. Gaussian conditioning gives $\mathbb{E}[T\mid X=x]=0.6x$ and $\mathbb{E}[{\epsilon }_1\mid X=x]=0.4x$. ThusThe first measurement error has zero unconditional mean, but positive selected mean. The second measurement error stays centered because it is independent of selection on $X$. The persistent signal $T$ never changed.

If $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, then $Z=(X-\mu )/\sigma$ has the standard normal distribution. The change-of-variables formula for a probability density gives $f_Z(z)=\sigma f_X(\mu +\sigma z)=\varphi (z)$, the standard normal density. Thus standardization of a normal random variable removes both parameters without changing the probabilities of corresponding events.

The standard normal density is

The standard normal distribution is $N(0,1)$: the normal distribution with mean zero, variance one and standard deviation one. Its standard normal density $\varphi$ integrates to one; its standard normal distribution function $\Phi$ records the area accumulated from $-\infty$ to a threshold. The distinction matters: $\varphi (0)=1/\sqrt{2\pi }\approx 0.398942$, $\Phi (0)=1/2$, and $\mathbb{P}(Z=0)=0$ for a standard normal random variable $Z$.

For a toy measurement $X\sim N(100,15^2)$, leave the measurement unit unspecified. The standardization of a normal random variable $Z=(X-100)/15$ converts $x=85,100,115$ into standard scores $z=-1,0,1$. Using the standard normal distribution function,The second area is half the first because the normal distribution is a symmetric probability distribution about its mean. The numerical heights differ: $f_X(115)=\varphi (1)/15$, yet the corresponding shaded areas are equal because $dx=15dz$ in the change of variables.

The standard normal distribution function is the cumulative distribution function of a standard normal random variable $Z$:Here $\varphi$ is the standard normal density and $\mathrm{erf}$ is the error function. The symmetric probability distribution gives $\Phi (-x)=1-\Phi (x)$ and $\Phi (0)=1/2$, while differentiation gives ${\Phi }^{\prime }(x)=\varphi (x)$. For $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization of a normal random variable gives $\mathbb{P}(X\le x)=\Phi ((x-\mu )/\sigma )$; subtracting two values calculates an interval probability.

Stein's lemma relates Gaussian multiplication to differentiation. Let $Z\sim N(0,1)$. One sufficient set of conditions is that $g$ is locally absolutely continuous, both $\mathbb{E}|Zg(Z)|$ and $\mathbb{E}|g^{\prime }(Z)|$ are finite, and $g(x)\varphi (x)\to 0$ as $x\to \pm \infty$, where $\varphi$ is the standard normal density. ThenSince ${\varphi }^{\prime }(x)=-x\varphi (x)$, integration by parts on a finite interval moves the derivative from $\varphi$ to $g$; the boundary condition and integrability justify the infinite-interval limit.

For $g(z)=z$, the identity gives $\mathbb{E}{Z}^2=1$. For $g(z)=z^3$, it gives $\mathbb{E}{Z}^4=3\mathbb{E}{Z}^2=3$. More generally, for $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization yieldsunder the corresponding conditions. The factor ${\sigma }^2$ is essential.

This integration-by-parts rule converts Gaussian error cross terms into derivatives of an estimator, which is the mechanism of the Stein risk identity for normal mean estimators. Boundary behavior matters for singular estimators: formal differentiation alone does not justify applying the identity across a nonintegrable singularity.

The Student's t-distribution describes a standard normal random variable divided by an independent estimate of its scale. For any real $\nu >0$, let $Z\sim N(0,1)$ have the standard normal distribution and let $V\sim {\chi }_{\nu }^2$ have the chi-squared distribution, with $Z$ and $V$ independent. ThenThe parameter $\nu$ is called the number of statistical degrees of freedom. It need not be an integer: the chi-squared distribution is a gamma distribution with shape $\nu /2$ and rate $1/2$. The independence assumption is essential; specifying only the two marginal distributions does not determine the probability distribution of their ratio.

Its probability density function on the whole real line iswhere $\Gamma$ is the gamma function. Every member is a symmetric distribution about zero. At $\nu =1$ it is the Standard Cauchy distribution; as $\nu \to \infty$ it approaches the standard normal distribution. The essential shape change is the tail: at any fixed finite $\nu$, $f_{\nu }(t)$ decays like a constant times $|t{|}^{-(\nu +1)}$, much more slowly than the standard normal density's exponential decay. Thus it is a heavy-tailed distribution. ^[ref]

The left panel shows the centers; the right panel uses a logarithmic scale to reveal the tails that a linear plot hides. All four curves have location zero and scale one. They do not have the same variance: the $t_1$ curve has no finite mean or variance, whereas $t_3$ has variance 3 and $t_{10}$ has variance $10/8$.

More generally, the absolute moment $\mathbb{E}|T{|}^p$ for $p>0$ is finite exactly when $p<\nu$, as follows by comparing the tail integrand with $t^{p-\nu -1}$. ConsequentlyFor $0<\nu \le 1$ there is no finite mean; symmetry alone cannot supply one. For $1<\nu \le 2$ the mean exists but the variance is infinite. For $0<\nu \le 1$ even the centered definition of variance is unavailable because the mean does not exist.

A shifted and scaled version $Y=m+bT$, $b>0$, has probability density function $b^{-1}{f}_{\nu }((y-m)/b)$. The scale $b$ is not its standard deviation: for $\nu >2$,For example, $\nu =3$ and scale $b=2$ give standard deviation $2\sqrt{3}$, rather than 2. The distinction matters when using a Student's t-distribution to model data with occasional extreme values, as well as when interpreting a Student t confidence interval.

Two successive integrations by parts of the standard normal density tail give positive remainder integrals establishing these strict inequalities for every $x>0$. The lower bound can be negative at small $x$ without invalidating the inequality.

Window-capture sensitivity to mass offset describes how the extracted-ion window capture fraction changes when an ion peak shifts relative to a fixed extracted ion mass window. The shift is in the reported mass coordinate; it can change the selected signal even if the target's total peak area is constant.

For a normalized peak shape $p(m-\delta )$ with offset $\delta$ and window boundaries $a,b$, writeDifferentiation givesThe first-order response is determined by the signal density at the two boundaries. A symmetric peak centered in a symmetric window has equal boundary densities and hence zero first derivative. Zero first derivative does not mean perfect immunity: finite offsets can still reduce capture.

For an original Gaussian function model with standard deviation $\sigma$ and half-width $w$, let $\varphi$ and $\Phi$ be the standard normal density and standard normal cumulative distribution function. ThenFor small offsets the leading loss is consequently quadratic:With $w=2\sigma$, the exact capture is 0.9545 at zero offset and 0.9270 at $\delta =0.5\sigma$. The quadratic approximation gives 0.9275, close for this stipulated displacement. At the boundary $\delta =2\sigma$, capture is about 0.5000 and a small-offset approximation is no longer appropriate.

An experimental check can re-extract the same recorded spectra with deliberately shifted windows while retaining the same time limits and extracted-ion chromatogram baseline. This isolates a software-selection effect. IUPAC's selected-mass profile definition supplies the measurement setting, rather than a claim that all instruments have Gaussian peaks.^[ref] Wider windows may reduce target losses but admit interfering ions; mass resolving power and instrument calibration remain separate instrumental considerations.