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Use a force-tilted activation barrier for a generic one-coordinate escape model. A tensile force tilts a free-energy landscape to $G_F(x)=G_0(x)-Fx$. If $x_b$ is the initial minimum and $x_s$ the saddle, define $\Delta x=x_s-x_b>0$ and the zero-force barrier $\Delta G_0^{‡}=G_0(x_s)-G_0(x_b)$. The relevant activation barrier is measured relative to the initial minimum, not an arbitrary absolute transition-state energy. To first order in force,The attempt prefactor $\nu$ is assumed force independent, and the saddle displacement is approximated as constant. This single-barrier approximation need not survive large landscape deformations or a change of escape pathway. TODO: user-deferred biomolecular implementation; the protein-specific shear-force arrangement and attachment chemistry are not solved.

For a generic force-ramp measurement, a calibrated atomic force microscope cantilever pulls an attachment while a stage changes separation. cantilever deflection $d$ gives force $k_{c}d$, and the cantilever compliance correction subtracts $d$ from imposed separation to give tether extension, with further corrections for other apparatus compliance. A prescribed stage speed does not automatically produce constant force loading rate when tether stiffness varies; the assumed $F(t)=ft$ must be measured or controlled. The original schematic below illustrates a generic nonbiological bond/tether geometry. Protein-specific construction is deferred.

The no-rebinding approximation is a physical escape assumption: after the link breaks, separation removes the partners from the capture region before appreciable reassociation. It requires the integrated return/association probability during that separation to be small. If the separated state stays nearby or the force is held without separation, setting $k_{+}=0$ can fail. Under the stipulated approximation, with the link initially intact,This survival under a linear force ramp differs from the probability density for rupture. It starts at one, decreases to zero and shifts to larger force as the loading rate increases. For small $F/F_0$, $p\simeq e^{-k_{0}F/f}$; a slow ramp allows much escape before substantial force is reached.

The median rupture force under a linear ramp satisfies $p=1/2$, givingAt high loading rate, the slope of $F_{1/2}$ against $\ln f$ tends to $F_0$; it is not exactly $F_0$ for a slow ramp. Fit the full expression across loading rates to recover $F_0$ and $k_0$. The rupture-force probability density is $\rho (F)=k(F)p(F)/f$, whose most probable rupture force under a linear ramp is $F_{\ast }=F_0\ln [f/(k_0{F}_0)]$ if $f>k_0{F}_0$, and zero otherwise. median and mode are different observables and must not be interchanged when fitting.

At known temperature, $F_0$ yields the activation distance $\Delta x=k_{B}T/F_0$; $1/k_0$ gives the zero-force mean lifetime in this exponential model. Absolute barrier height additionally requires the attempt prefactor, and neither equilibrium binding free energy nor an association rate follows from dissociation-only data. A loading-rate-dependent rupture force is a kinetic readout, not an intrinsic equilibrium strength. TODO: user-deferred biological conclusions. The protein-specific stability interpretation and cell-adhesion strategy are left unsolved.

First establish the limiting law needed by the delta method. For fixed $t$, put $u_n=1/2+t/\sqrt{n}$ and let $N_n$ count how many of the $U_i$ are at most $u_n$. Then $N_n$ has a binomial distribution with expectation $n/2+t\sqrt{n}$ and variance $n/4-t^2$. The event $M_U\le u_n$ is exactly $N_n\ge (n+1)/2$. Its standardized threshold tends to $-2t$, so the central limit theorem for these binomial counts givesHere $\Phi$ is the cumulative distribution function of the standard normal distribution. Equivalently,Use the usual local regularity for the delta method: $f$ is continuous at the population median $\theta$ and $f(\theta )>0$. The inverse function theorem, applied at $F(\theta )=1/2$, givesThereforeThe asymptotic normality gives the requested leading variance approximationHere the variance parameter for the rescaled limiting law is $1/(4f(\theta {)}^2)$, while the source states the corresponding $1/n$ approximation for the unscaled estimator. Strict increase of $F$ alone does not force $f(\theta )>0$; that local condition is needed for this form of the asymptotic distribution of a sample median.

For the normal distribution, the density at its centre is $f(\theta )=1/(\sigma \sqrt{2\pi })$. The sample mean has exact variance ${\sigma }^2/n$, whereas part (b) gives the sample median the leading variance $\pi {\sigma }^2/(2n)$. Thus the requested asymptotic relative efficiency isAt a fixed confidence level, a confidence interval based on the sample mean has a leading half-width smaller by the factor $\sqrt{2/\pi }$. With known $\sigma$, $\overline{X}\pm z_{1-\alpha /2}\sigma /\sqrt{n}$ also has exact normal coverage. Prefer the sample mean under the normal model.

For the Laplace distribution, symmetry puts both the mean and median at $\theta$. Its density there is $1/(2\sigma )$, andConsequently the sample mean has variance $2{\sigma }^2/n$, while the sample median has leading variance ${\sigma }^2/n$. HenceA large-sample confidence interval based on the sample median has leading half-width smaller by $1/\sqrt{2}$; with known scale it is $M_X\pm z_{1-\alpha /2}\sigma /\sqrt{n}$. Prefer the sample median under the Laplace model. Consistent estimation of an unknown scale preserves these leading comparisons. The reason for the reversal is quantitative: the median's precision depends on the central density $f(\theta )$, while the sample mean's precision depends on the whole population variance.

The cumulative distribution function is continuous and strictly increasing for $x>0$, so the median is the unique positive $c$ satisfying $F_X(c)=1/2$. Consequently $e^{-c^2/2}=1/2$, givingThere is no atom at $c$, so $\mathbb{P}(X>c)=1-F_X(c)=1/2$ as required.

The arithmetic mean is $\bar{x}=n^{-1}{\sum }_{i=1}^n{x}_i$, the total measured concentration divided by the number of observations. The median is the central value of the sorted observations, or an interpolation between the two central values under the chosen even-sample convention. It is the 50th percentile.

The table's “75th quartile” means the 75th percentile, or upper quartile $Q_3$, with approximately three quarters of observations at or below it. Its “25th quartile” means the 25th percentile, or lower quartile $Q_1$. There are four quarters of a distribution, rather than 75 different quartiles. Quantile interpolation rules can differ for a finite sample; their intended population interpretation is unchanged.

Let $T_n=\inf \{t\ge 0:N_t\ge n\}$ be the $n$th arrival time, and set $T_0=0$. In a Homogeneous Poisson process of rate $\lambda >0$, the gaps $W_n=T_n-T_{n-1}$ are independent and identically distributed random variables with an exponential distribution:Here $\mathrm{Exp}(\lambda )$ uses the rate convention. Under the scale convention $\theta =1/\lambda$, the same probability density function is ${\theta }^{-1}{e}^{-w/\theta }$; a simulation function accepting a scale must receive $1/\lambda$. Conversely, summing independent rate-$\lambda$ exponential random variables to form $T_n$ and counting those sums constructs a Homogeneous Poisson process.^[ref]

The first-gap formula follows immediately from $\{W_1>w\}=\{N_w=0\}$ and the Poisson distribution. Independence of all successive gaps is the additional interval-theorem assertion; deterministic-time independent increments alone should not simply be substituted at random arrival times without justification. The memorylessness of the exponential distribution givesAt the toy rate of two per minute, the mean gap is half a minute, the median is $(\log 2)/2\approx 0.346574$ minutes, and the probability of waiting longer than half a minute is $e^{-1}\approx 0.367879$. Here $\log$ is the natural logarithm. Already having waited a minute without an arrival leaves this same conditional probability of waiting another half minute. The exponential distribution describes a model with no aging of the remaining wait; a scheduled arrival with a fixed gap has no such property.

The quantile function converts a cumulative probability level into a value. For a cumulative distribution function $F$ and $0<p<1$, use the generalized inverseIt selects the first threshold reaching the requested level. The infimum handles jumps and flat portions, so this construction does not require an ordinary inverse function. Other endpoint conventions are possible at $p=0$ or $p=1$; the open probability interval avoids those issues. NIST introduces the inverse-direction question as the percent point function. ^[ref].

For the finite example under probability distribution,Thus $Q_X(0.1)=0$, $Q_X(0.5)=1$, $Q_X(0.9)=3$. The value at $p=0.2$ is still zero, because $F_X(0)=0.2$ already reaches that level. For $F_Y(y)=y^2$ on $[0,1]$, the quantile function is $Q_Y(p)=\sqrt{p}$, giving $Q_Y(0.5)=1/\sqrt{2}\approx 0.70710678$.

A median is a central threshold; $Q(1/2)$ supplies one, though a median need not be unique. The same quantile function becomes a sampling algorithm through inverse transform sampling. Jumps in $F$ become flat bands in $Q$, precisely the mechanism needed to assign positive probability to one outcome.

Quartiles are the quantiles at probability levels $1/4,1/2,3/4$; $Q_2$ is a median. For a population we can use its quantile function, $Q_r=Q(r/4)$ for $r=1,2,3$. Finite-sample conventions differ. Our box plot uses linear interpolation between order statistics: for $n\ge 2$ and $0<p<1$, setThen $Q_r=\tilde{Q}(r/4)$. This is R's type-7 convention, described in the documentation authored by Ivan Frohne and Rob J Hyndman. It differs from the generalized-inverse sample quantile defined elsewhere in this wiki: for the twelve toy values, that rule selects ranks $3,6,9$ and gives $2,4,6$, rather than $2.75,4.5,6.25$. These are different stated conventions, not arithmetic disagreements. ^[ref]

Using $u=r^2/(2{\sigma }^2)$ in the defining expected value integral givesThe last integral is the gamma function. Near zero, the integrand before substitution behaves as $r^{q+1}$, explaining the threshold $q>-2$; at infinity the Gaussian tail controls every power. In particular,^[ref]

For the chosen $\sigma =2\mathrm{cm}$, the mean radius is about $2.5066\mathrm{cm}$, its median is $2\sqrt{2\log 2}\simeq 2.3548\mathrm{cm}$, and its mode is $2\mathrm{cm}$. Its root mean square is $2\sqrt{2}\simeq 2.8284\mathrm{cm}$. Different summaries answer different questions; the component scale $\sigma$ is neither the radial mean nor the radial standard deviation.

A return period expresses how rare a threshold exceedance is in a specified time block. Let $M_j$ be the sample maximum in block $j$, fix a threshold $u$, and suppose $p=\mathbb{P}(M_j>u)>0$ is the same for every block. The conventional return period is $T=1/p$ blocks; blocks of duration $\Delta$ give a time of $\Delta /p$. Thus an annual exceedance probability $p=0.01$ means a 100-year return period. This names a frequency scale, not a schedule or a guarantee of one exceedance in each century. The annual convention and the misleading “once every 100 years” interpretation are discussed in USGS: the 100-year flood.

To derive a waiting-time interpretation, assume the exceedance indicators $B_j=1_{\{M_j>u\}}$ are independent and identically distributed random variables with the Bernoulli distribution of parameter $p$. Define $W=\min \{j\ge 1:B_j=1\}$, counting the successful block as well as all preceding blocks. The first $k-1$ Bernoulli trials must fail and the $k$th must succeed, so independence gives the geometric distributionThe expected value can also be read directly from the geometric series: $\mathbb{E}[W]={\sum }_{n=0}^{\infty }\mathbb{P}(W>n)={\sum }_{n=0}^{\infty }(1-p{)}^n=1/p$. This counts blocks from the starting boundary, not the exact time of an exceedance within its block. Counting only failed blocks instead would give expected value $T-1$. The waiting law and its memorylessness of the geometric distribution are developed in Kyle Siegrist, Random Services: the geometric distribution.

For a horizon of $n$ complete years, at least one exceedance is the complement of no exceedances. Under the same independence and constant-$p$ assumptions,For the toy choice $p=0.01$, this gives $9.6\%$ in 10 years, $26.0\%$ in 30 years, $39.5\%$ in 50 years, and $63.4\%$ in 100 years. Even after 100 years there is a $36.6\%$ chance of no exceedance. These numbers are exact model calculations rounded to one decimal place, not measured frequencies. The median waiting count is 69 years, since $1-0.99^{68}<1/2\le 1-0.99^{69}$; the expected value is 100 years. A mean and a median answer different questions.

The exceedance count $N_n={\sum }_{j=1}^n{B}_j$ has the binomial distribution under independence, with expected value $np$. Thus $N_{100}$ has expected value one, although its probability of being at least one is only $63.4\%$: some realizations contribute several exceedances. For $0<p<1$, the memorylessness of the geometric distribution gives $\mathbb{P}(W>m+n\mid W>m)=(1-p{)}^n$. A long quiet spell does not make the next independent year's exceedance overdue, and an exceedance this year does not force a quiet interval. Two specified consecutive years both exceed with probability $p^2=0.0001$ in this toy model.

A reproducible tabletop experiment makes the assumptions tangible. Put 100 identical slips numbered 1 to 100 in a bag; after thoroughly mixing, draw one and record whether it is 100. Replace the slip and mix again before every draw. Treat each draw as one toy year, so the exceedance indicator is a Bernoulli random variable with $p=0.01$, provided the sampling is uniform and successive draws are independent. Repeat complete 30-draw runs and record the fraction containing at least one 100; separately repeat runs stopped at the first 100 and record their waiting counts. The first fraction targets $1-0.99^{30}$; the waiting-count histogram targets $0.99^{k-1}(0.01)$. Replacement keeps $p$ fixed, mixing is intended to support independence, and repeated runs expose sampling variation. This is a proposed demonstration of the toy model; no experimental results are asserted here.

Stationarity supports a time-invariant annual probability distribution, but does not imply independence. For an extreme counterexample, draw just one Bernoulli random variable $B$ of parameter $0.01$ and set $B_j=B$ for every year. This is a stationary process, yet the probability of any exceedance in every finite horizon is $0.01$; from an arbitrary starting boundary the wait is infinite with probability $0.99$. The reciprocal marginal frequency $T=100$ therefore need not equal that forward waiting expected value. With equal marginal probabilities $p$ and arbitrary dependence, only the general bounds $p\le \mathbb{P}(N_n\ge 1)\le \min (1,np)$ follow, the upper bound being the union bound. Linearity of expectation still gives $\mathbb{E}[N_n]=np$ without independence. Mere zero correlation does not restore the product formula; even pairwise independence does not imply joint independence. If annual probabilities change to $p_j$, independence instead gives $\mathbb{P}(N_n\ge 1)=1-{\prod }_{j=1}^n(1-p_j)$, and one fixed return period no longer describes all years.

The sample median is a central value of an ordered statistical sample. The usual numerical convention selects the middle order statistic for odd sample size and the arithmetic mean of the two central order statistics for even sample size:NIST uses this convention. ^[ref]. For $(0.7,0.1,0.9,0.4,0.2)$, the sample median is $0.4$; for $(0.1,0.2,0.4,0.7)$, it is $(0.2+0.4)/2=0.3$, which need not be an observed value.

More generally, every value between the two central order statistics in an even statistical sample is a median of its empirical distribution function. The generalized-inverse sample quantile at $p=1/2$ selects the lower central order statistic in that case, so it can differ from the midpoint convention. Ties are allowed in all these definitions.

The sample median measures the center using sorted positions. Changing the largest observation from $0.9$ to $9$ in the five-value example leaves the sample median at $0.4$, while the sample mean rises from $0.46$ to $2.08$. Its sampling distribution still depends on the population probability distribution: the mean $1/2$ found for five observations from a uniform distribution is a special calculation, and a sample median's expected value need not equal the population median in general.

For a concrete counterexample, three independent and identically distributed random variables with rate-one exponential distribution have sample median $X_{(2)}$ with expected value $1/3+1/2=5/6$, by independent exponential order-statistic spacings. The population median instead solves $1-e^{-m}=1/2$, giving $m=\log 2\approx 0.693147$. Thus even the middle order statistic can have mean different from the population median.

For a finite list of real numbers $x_i$ and nonnegative weights of positive total $W$, the weighted mean uniquely minimizes the weighted sum of squared deviationsExpanding around ${\bar{x}}_w$ and using ${\sum }_i{w}_i(x_i-{\bar{x}}_w)=0$ givesEvery displacement from the weighted mean increases the objective. This is the simplest one-parameter weighted least squares problem: represent all entries by one common value.

For the weighted values $90$ and $60$ with weights $2$ and $8$,Thus $S(66)=1440$, while $S(75)=2250$. The minimum is defined by this squared-error criterion; a different loss can select a different summary, such as a median under unweighted absolute deviations.

A median of a weighted probability distribution. With positive weights summing to one, at most half the weight lies strictly below a weighted median and at most half lies strictly above it. The lower weighted median chooses the first observed value whose cumulative weight reaches one half; the convention matters when the cumulative mass is exactly one half.

For independent paired observations, form differences $D_i=U_i-V_i$. Assume the differences are identically distributed, continuous and symmetric about a location $\delta$. Under the zero-location null hypothesis, the signs are independent fair signs conditional on their absolute values. Rank the absolute differences and add ranks with positive signs. A negative-location alternative rejects for small $W_{+}$. Its exact conditional distribution is that of a sum of an independently selected subset of ranks, each of the $2^n$ subsets equally likely. Thus its mean is $n(n+1)/4$ and its variance is $n(n+1)(2n+1)/24$. Symmetry is essential: a zero median alone does not justify this signed-rank null law. Zero differences are omitted in the usual version, and tied absolute differences receive average ranks.