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Expand around the expectation $\mu$:Thus variance as the minimum mean squared error of a constant givesThe finite second moment also guarantees the finite first absolute moment needed below.

For a probability density function $f$, splitting at $a$ givesLet $F$ be the cumulative distribution function. Since a density gives no mass at a singleton, differentiating under this integral yields $H^{\prime }(a)=F(a)-(1-F(a))=2F(a)-1$. This derivative is nondecreasing, so $H$ is convex andThis is the median minimizes expected absolute loss property. The minimizer need not be unique: a gap in the density can give an interval of medians, all with the same loss.

The bounded classical solution is the heat-kernel convolutionThe two-dimensional heat kernel has integral one. Boundedness of $g$ permits differentiation under the integral for $t>0$, proving $u_t=\Delta u$ and $|u|\le \Vert g{\Vert }_{\infty }$. Its approximate identity property gives $u(x,t)\to g(x)$, locally uniformly since $g$ is continuous. This bounded solution satisfies the required Gaussian-growth bound.

For heat-equation uniqueness under Gaussian growth, let $w$ be the difference of two solutions, so $w(x,0)=0$ and $|w|\le M_0{e}^{a|x{|}^2}$ on a finite time interval. Choose $b>a$ and a slab $0\le t\le \tau <1/(4b)$. The positive comparison solutionsatisfies $({\Phi }_b{)}_t=\Delta {\Phi }_b$. For any $\epsilon >0$, its faster spatial growth makes $|w|\le \epsilon {\Phi }_b$ on a sufficiently large lateral cylinder boundary. At the initial boundary, $w=0$. The heat equation maximum principle applied to $\pm w-\epsilon {\Phi }_b$ proves the same inequality inside. First allow the cylinder radius to increase, then let $\epsilon \downarrow 0$. Repeating these slabs proves uniqueness on every finite interval with the stated uniform growth bound. No spatial integrability of $g$ is required.

At time $T$, this is Gaussian filtering with covariance $2TI$, hence $\sigma =\sqrt{2T}$ per coordinate. With the angular-frequency Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, the Gaussian filtering Fourier multiplier isFor merely bounded $g$, this identity is understood through tempered distributions. The multiplier leaves the zero frequency unchanged and attenuates large frequencies exponentially: it suppresses rapid noise fluctuations but blurs image edges. In cycles-per-length frequency $k$, the multiplier is $e^{-4{\pi }^{2}T|k{|}^2}$.

For the printed Perona-Malik equation, retain the supplied conductance $c(s)=s{e}^{-s^2/(2{\lambda }^2)}$, which includes an extra factor $s$. In one dimension write $p=u_x$ and $F(p)=c(|p|)p=p|p|e^{-p^2/(2{\lambda }^2)}$. Then $u_t=F^{\prime }(u_x)u_{xx}$, withThe forward-backward threshold for gradient-weighted exponential diffusion is thereforeAt $|u_x|=0$ and $\sqrt{2}\lambda$ the coefficient vanishes, giving degenerate diffusion. Increasing $\lambda$ increases the forward-diffusion range. Small nonzero slopes smooth, while large slopes formally sharpen; a negative coefficient produces short-wave growth and ill-posedness, so this is a dynamics explanation, not a general existence theorem for arbitrary data. Replacing the printed conductance by $e^{-s^2/(2{\lambda }^2)}$ would give threshold $\lambda$, but that is a different equation.

The four-neighbour mean expansion follows by Taylor expansion: opposing first and third derivatives cancel, soIf $u(x_0)={\mathrm{mean}}_h(u)(x_0)$ for every sufficiently small $h$, division by $h^2$ and passage to the limit give $\Delta u(x_0)=0$. This is a pointwise conclusion; it does not assert harmonicity in a whole neighbourhood.

For the area median over the disk, the disk-median curvature expansion, at a point where $|\nabla u|\ne 0$, isConsequently the analogous median fixed-point condition yieldsThis is the vanishing of the level-line curvature at $x_0$: to second order the level line is straight there. It need not be a straight segment. For example $u(x,y)=y+x^3$ has zero disk median at the origin for every radius, by odd symmetry, but its zero level line is the cubic $y=-x^3$. Vanishing curvature on an entire connected regular level arc would force that arc to be straight. The factor $1/6$ is for the disk's uniform area measure; a circle-boundary median has a different scale factor, while producing the same zero-curvature equation.

Put $Q(u)=\inf \{x\in \mathbb{R}:F(x)\ge u\}$ for $0<u<1$. This is the quantile function, rather than an ordinary inverse requiring strict increase. The limits of the distribution function at infinity make $Q(u)$ finite, and the fact that $F$ is a right-continuous function ensures $F(Q(u))\ge u$. ConsequentlyIncreasing $u$ shrinks the set in the infimum, so the quantile function is a monotone function. To prove the left continuity of the quantile function, fix $u$ and let $L={\sup }_{v<u}Q(v)\le Q(u)$. If $L<Q(u)$, choose $L<y<Q(u)$. Then $F(y)<u$, so some $v\in (F(y),u)$ satisfies $Q(v)>y$, a contradiction. Thus $Q(v)\uparrow Q(u)$ as $v\uparrow u$. At a flat stretch of the distribution function, the quantile function may jump immediately to the right, consistent with this left-continuous convention.

For the uniform order statistic, count the observations at most $u$. That count has the binomial distribution with parameters $n,u$, givingDifferentiating makes adjacent terms telescope: use $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ and $(n-r)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$. The probability density function is thereforeIn particular, $U_{(j)}$ has the Beta distribution with parameters $j,n-j+1$.

Write $m=Q(1/2)$ for the specified median. Continuity of the distribution function gives $F(m)=1/2$ and no mass at $m$, even if the distribution function has a flat stretch there. Thus $B={\sum }_i{1}_{\{X_i<m\}}$ has the binomial distribution with parameters $n,1/2$. Except on a null event, the order-statistic confidence interval for a median covers $m$ exactly when $j\le B\le n-j$. Symmetry of the binomial distribution makes its two failure probabilities equal. Alternatively, the probability integral transform givesHence the coverage is exactly $1-\alpha$, withThe asymmetric open/closed endpoint convention does not change the coverage because the continuous probability distribution assigns no mass to the median.

Write $f(x)=\mathrm{TS}\{x_1,\dots ,x_n\}$. Besides Talagrand's convex distance inequality, we use the precise squared edge bound for tours in the unit square stated above and the triangle inequality, which permits shortcutting a closed walk without increasing its length. We first prove the geometric comparison that connects these results.

For each fixed $x$, choose a cyclic tour $C$ with edge lengths ${\ell }_i$ satisfying ${\sum }_i{\ell }_i^2\le 4$. At vertex $x_i$, let $a_i(x)$ be the sum of the lengths of its two incident edges in $C$. The Cauchy-Schwarz inequality givesWe claim the weighted mismatch bound for Euclidean tours:To prove it, keep the vertices whose coordinates agree. In the cyclic ordering $C$, the deleted vertices split into consecutive runs between kept vertices. For one run, let $a,b$ denote its two boundary edge lengths and $S$ the sum of its internal edge lengths. Starting at the left kept endpoint, visit the run and return along the same path, at cost $2(a+S)$; starting at the right endpoint costs $2(b+S)$. Choose the cheaper detour. Its cost is at mostwhich is exactly the sum of $a_i(x)$ over that run. Attach these detours to a shortest tour through $y$, which already visits every kept point. The resulting closed walk visits all points of $x$; shortcut it using the triangle inequality. Summing the detour costs proves the claimed weighted mismatch bound for Euclidean tours. If there is just one kept vertex, the same argument uses that vertex as both endpoints of the unique run. If there are no kept vertices, $f(x)\le {\sum }_i{\ell }_i\le {\sum }_i{a}_i(x)$ proves the comparison directly. Repeated point locations cause no difficulty, since their connecting edges may have length zero.

Let $M$ be any median of $f(X)$ and set $A=\{y:f(y)\le M\}$. For $f(x)\ge M+t$, the weighted mismatch bound for Euclidean tours impliesThe weights have squared sum at most one, so $d_T(x,A)\ge t/4$. Since $\mathbb{P}(A)\ge 1/2$, Talagrand's convex distance inequality yieldsFor the lower tail, put $B=\{x:f(x)\le M-t\}$. If $\mathbb{P}(B)=0$ there is nothing to prove. For every $y$ with $f(y)\ge M$, apply the same weighted mismatch bound for Euclidean tours with its weights $a_i(y)$ and all $x\in B$. It gives $d_T(y,B)\ge t/4$. The set of such $y$ has probability at least $1/2$, so Talagrand's convex distance inequality instead givesCombining both tails,The bound has a constant fluctuation scale, independent of $n$. Only independence of the point coordinates and their support in the square are needed here; their uniform distribution was needed for the earlier lower bound on the expected value.

The useful form is Talagrand's convex distance inequality. Let $({\Omega }_i,{\mu }_i)$ be standard probability spaces, let $\Omega ={\prod }_{i=1}^n{\Omega }_i$ carry the product measure $\mu ={⨂}_i{\mu }_i$, and let $A\subset \Omega$ be a measurable event with $\mu (A)>0$. For $x\in \Omega$, form the mismatch vectorsThe Talagrand convex distance is the distance from zero to their closed convex hull:Equivalently, by the Hahn-Banach separation theorem for two convex sets,Nonnegative weights suffice because every mismatch vector has nonnegative coordinates. Talagrand's convex distance inequality statesIn particular, Markov's inequality gives the product-space concentration inequalityThe same conclusion is interpreted with outer probabilities if a distance set is not measurable; the continuous geometric function used below has measurable level sets. A median $M$ of a real random variable $Z$ means that $\mathbb{P}(Z\le M)\ge 1/2$ and $\mathbb{P}(Z\ge M)\ge 1/2$. Applying the displayed concentration inequality to suitable level sets will control deviations around a median.

For a nonnegative continuous event time, the survivor function is $F(t)=\mathbb{P}(T>t)$; here the printed $F$ denotes survival, not the usual cumulative distribution function. A lower median isFor a continuous strictly decreasing survivor function this is the unique solution of $F(m)=1/2$. If the curve has a flat stretch at $1/2$, the distribution has a nonunique median and the infimum specifies the lower endpoint.

The tail integral formula for moments follows from $T={\int }_0^{\infty }1\{T>t\}dt$ and Tonelli theorem. Therefore, when survival vanishes beyond $t^{\ast }$,If $F(t)>0$ at every finite time, the mean is finite precisely when ${\int }_0^{\infty }F(t)dt<\infty$. For a concrete sufficient condition, require $F(t)\le C{t}^{-1-\epsilon }$ for all $t\ge M>0$, with $C<\infty$ and $\epsilon >0$. The integral over $[0,M]$ is at most $M$, and the tail integral is at most $C{M}^{-\epsilon }/\epsilon$. An exponential tail bound also suffices. Merely having $F(t)\to 0$, or even $tF(t)\to 0$, does not suffice: a properly completed decreasing tail proportional to $1/(t\log t)$ for $t\ge e$ has infinite mean.

Let $a_1<a_2<\cdots$ be the distinct observed event times, $r_j$ the risk set size immediately before $a_j$, and $d_j$ the events there. Under independent censoring, the Kaplan–Meier estimator isIt starts at one, jumps down at events, and is unchanged by a pure censoring time; right censoring removes people from subsequent risk sets. If events and censorings coincide, the convention here processes events before censoring, so those censored at the recorded time are included just before the event. The estimated median is the first time the curve reaches or falls below $1/2$. If it never does during observation, the median is not estimable from the observed curve without tail assumptions.

With one remaining individual censored at $t_1$, no event factor is introduced. HenceThis is the terminal observed step value. With an empty risk set after $t_1$, continued horizontal plotting is a convention and supplies no information about the true later survivor function. The tail rests on one person and is imprecise; this is terminal censoring and survival-mean identifiability. If ${\varphi }_0>0$, the unrestricted mean cannot be obtained nonparametrically from this censored tail: extending the last positive step to infinity would give an infinite area, which does not establish an infinite population mean. A restricted mean survival time up to a suitably supported finite cutoff remains estimable.

Normalization of the probability density function givessoThe expected value isThe second moment isTherefore the variance is

For the median $m$,so

Expanding around the expected value $\mu$,because $\mathbb{E}[X-\mu ]=0$. Thereforewith equality exactly when $a=\mu$.

For the absolute loss,Leibniz differentiation gives, with $F(a)={\int }_{-\infty }^{a}f(x)dx$,Thus $H$ decreases while $F(a)<1/2$ and increases while $F(a)>1/2$. It is minimized at any median, characterized in the continuous case by

Let $m=F^{-1}(1/2)$ be the unique median. The influence function of the sample median isaway from $m$. Its second moment, equivalently the asymptotic distribution of a sample median, givesNo symmetry assumption is used: the density is evaluated at the actual median of $F$.

Choose a fixed $\alpha <1/2$. On the stated high-probability event, strictly more than half of the block means satisfy, simultaneously for every $\Vert v{\Vert }_2\le 1$,The median of a collection with a strict majority in an interval lies in that interval. Hence the same bound holds for $|{\mathrm{MOM}}_k(Xv)-{\mu }_0^{T}v|$ uniformly in $v$. Part b then giveswith probability at least $1-e^{-k/c_{\alpha }}$. Renaming the constant proves the claim.

Complementing every friendship sends every degree $d$ to $n-1-d$, and hence sends $S=M+L$ to $2(n-1)-S$. Since the random graph and its complement have the same law, the distribution of $S$ is symmetric about $n-1$, giving $\mathbb{E}S=n-1$.

The same symmetry gives equal probabilities strictly below and above $n-1$. There is positive probability at $n-1$: for odd $n$ use a regular graph of degree $(n-1)/2$; for even $n=2m$, use $K_{m,m}$ with a perfect matching removed and then restore one matching edge. Thus the lower median as defined is also $n-1$.