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The marginal probability density function is $f_Y(y)={\int }_{\mathbb{R}}{f}_{X,Y}(x,y)dx$. For $y$ with $f_Y(y)>0$, the conditional density isIf the conditional first absolute moment is finite, the conditional expectation isConditioning here uses a density, since the individual event $Y=y$ normally has probability zero; a conditional law is specified only up to marginal-null values of $y$.

Normalization of the probability density function requiresusing ${\cos }^{2}x=(1+\cos 2x)/2$. Therefore

The independent identically distributed central limit theorem states that for independent identically distributed random variables $U_k$ with finite mean $\mu$ and positive finite variance ${\sigma }^2$,For $0<u<1$, continuity, strict monotonicity and the endpoint limits of the cumulative distribution function give a unique $x=F^{-1}(u)$. Then $P(F(X)\le u)=P(X\le x)=F(x)=u$. Thus $Y=F(X)$ has the uniform distribution on $(0,1)$, by the probability integral transform.

For $W=-\log (1-Y)$, $P(W\le w)=1-e^{-w}$ when $w\ge 0$, and it is zero otherwise. Therefore its probability density function is $e^{-w}{\chi }_{(0,\infty )}(w)$: the exponential distribution of rate one. By integration by parts, $E[W]=1$ and $E[W^2]=2$, so $\mathrm{Var}(W)=1$.

The variables $W_k=-\log (1-F_k(X_k))$ are independent because they are separate measurable functions of independent random variables. The probability integral transform makes all of them identically distributed as $\mathrm{Exp}(1)$, even though the $X_k$ need not be identically distributed. The displayed $Z_n$ is exactly $n^{-1/2}{\sum }_{k=1}^n(W_k-1)$. The central limit theorem therefore gives

By reflection symmetry, $G$ is the midpoint of the cylinder axis. With position $r$ measured from $G$, define $J_{ij}^G=\int (|r{|}^2{\delta }_{ij}-r_i{r}_j)dm$. Symmetry makes the off-diagonal entries zero. Uniform density gives $⟨x^2⟩=⟨y^2⟩=a^2/4$ and $⟨z^2⟩=b^2/3$, by integrating over a disk and the interval $[-b,b]$. HenceFor parallel axes based at $P$, the tensor form of the parallel-axis theorem is $J^P=J^G+M(|d{|}^{2}I-d{d}^T)$, where $d=\overrightarrow{PG}$. This follows by substituting $r+d$ in the defining integral and using $\int rdm=0$. Here $d$ is an axial vector of length $b$, soThe Euler angles $\psi ,\varphi$ are cyclic coordinates. Their conserved conjugate momenta are $p_{\psi }=C(\dot{\psi }+\dot{\varphi }\cos \theta )=Cn$ and $p_{\varphi }=A\dot{\varphi }{\sin }^2\theta +Cn\cos \theta =h$. The Lagrangian has no explicit time dependence, so the conserved energy is kinetic plus gravitational potential, namely the expression given with constant $E$.

Eliminate the angular velocities using $\dot{\varphi }=(h-Cn\cos \theta )/(A{\sin }^2\theta )$ and $\dot{\psi }+\dot{\varphi }\cos \theta =n$. The energy equation becomesThusThe elimination is valid away from the coordinate poles; the multiplied equation extends by continuity to any pole reached by a regular motion.

Assume $a,b>0$, the standard deviations. Substitute the two normal densities into (ii):The last step uses ${\int }_0^{\infty }u{e}^{-A{u}^2/2}du=1/A$. With $c=b/a$ this is $\boxed{f_V(v)=c/[\pi (c^2+v^2)]}$, the centered Cauchy density of scale $c$.

The desired event is $|V|<1$, outside the null event $X=0$. Integrating the density gives

Integrate the joint density in $u$. Symmetry gives $f(-u)g(-uv)=f(u)g(uv)$, so the integrand $|u|f(u)g(uv)$ is even. HenceTonelli's theorem justifies marginalization for these nonnegative densities.

The inverse change of variables is $x=u$, $y=uv$, whose Jacobian determinant is $u$. Since $X,Y$ are independent, the original joint density is $f(x)g(y)$. ThusThe transformation is nonsingular away from $u=0$; this line has probability zero because $X$ has a density, so no extra mass is omitted.

For a rejection function $\phi (x)\in [0,1]$, its power is $\beta (\theta )={\mathbb{E}}_{\theta }\phi (X)$. A test has level at most $\alpha$ if ${\sup }_{\theta \in H_0}\beta (\theta )\le \alpha$; its size is that supremum. In a simple null the significance probability is just the rejection probability under that null.

For two simple hypotheses with joint sample densities $f_0,f_1$, the Neyman–Pearson lemma states that a test rejecting where $f_1>k{f}_0$, accepting where $f_1<k{f}_0$, and randomizing if necessary on equality to have size $\alpha$, is most powerful among level-$\alpha$ tests. Let ${\phi }_{\ast }$ be that test and $\phi$ any competitor. Pointwise,Integration givessince $k\ge 0$, the first test has size $\alpha$, and the other has size at most $\alpha$. This proves the claimed power bound, allowing null-density-zero regions as well. Equality implies agreement away from the likelihood-ratio boundary, up to null sets, when the inequalities are strict.

Independently draw $Y\sim q$ and $U\sim \mathrm{Uniform}(0,1)$. Accept $Y$ if $U\le p(Y)/(Mq(Y))$; otherwise discard it and repeat with fresh independent draws. Return the first accepted value. The bound makes the acceptance probability at each proposal at most one. As $\int p=1$, the total acceptance probability is $1/M>0$, so the procedure terminates almost surely. Values outside the proposal support must have target density zero for the stated envelope assumption to make sense.

Generate independent standard normal variables $Z_1,Z_2$ and put $\boxed{Y=Z_1/Z_2}$. The event $Z_2=0$ has probability zero. The ratio calculation in M4 with equal standard deviations gives density $1/[\pi (1+y^2)]$, exactly the required $t$ distribution with one degree of freedom. The equivalent form $Z_1/|Z_2|$ has the same law by symmetry.

This relates an thermodynamic equilibrium variance to the local stiffness of an appropriate thermodynamic free energy. At fixed bath temperature, suppose the effective thermodynamic free energy $A(x)$ includes the constraints and the measure for a chosen coordinate. Its thermodynamic equilibrium probability density function is proportional to $e^{-A(x)/(k_{B}T)}$. Near a stable minimum $x_0$, writeThe resulting normal distribution gives the Gaussian fluctuation approximationThis is exact for a quadratic thermodynamic free energy with the stated flat coordinate measure; in general it is a small-fluctuation result. A smaller curvature means a softer restoring response and larger fluctuations. Near vanishing curvature, higher-order terms become important, so the quadratic approximation cannot alone predict an actual infinite variance.

For example optical trap Langevin calibration uses the Brownian displacement of a trapped bead: $A=\kappa (x-x_0{)}^2/2$ gives $\kappa =k_{B}T/⟨\Delta x^2⟩$. A camera or position detector records the fluctuations. Finite exposure averages rapid motion and can bias the measured variance; the measurement protocol matters as well as the thermodynamic equilibrium identity.

The correlation function of a point process is even and tends to zero as $|\lambda |\to \infty$. Its derivative factors asConsequently its local maxima occur at $0,\pm \sqrt{3}$, and its local minima at $\pm 1$. The heights are

For an interval $I$, ${\int }_I{R}_3^{(1)}(\lambda )d\lambda$ is the expected value of the number of eigenvalues in $I$. Thus equally short intervals have the highest expected counts nearwith the central peak highest. Each exchangeably labelled eigenvalue has the same marginal probability density function $R_3^{(1)}/3$. The three peaks should not be interpreted as three separate marginal densities for ordered eigenvalues.

In this case the most likely joint configuration also lies at these three positions. On the ordered chamber, the logarithm of the joint probability density function, up to a constant, isIts stationary equations are ${\lambda }_i=2{\sum }_{j\ne i}({\lambda }_i-{\lambda }_j{)}^{-1}$, satisfied by $(-\sqrt{3},0,\sqrt{3})$. For every nonzero real vector $v$, the second directional derivative isThus $L$ is strictly concave on this convex chamber, proving that the stationary configuration is its unique global maximum. Permuting the labels gives the other maxima on ${\mathbb{R}}^3$.

The difference of the two eigenvalues is $2i\sin U$, hence $\Delta ={\sin }^{2}U$. Its expectation isFor $0\le x\le 1$, put $\alpha =\arcsin \sqrt{x}\in [0,\pi /2]$. The set of angles satisfying $|\sin u|\le \sqrt{x}$ consists of $[0,\alpha ]$, $[\pi -\alpha ,\pi +\alpha ]$ and $[2\pi -\alpha ,2\pi ]$, of total length $4\alpha$. Thus the full cumulative distribution function isDifferentiating on $(0,1)$ gives the probability density functionThis is the arcsine distribution, equivalently the beta distribution with both parameters $1/2$. It has no atoms at either endpoint.

Interpret the source's missing variable in the limit arrow as $t\to \infty$. We prove convergence in distribution by showing that the angle of the Brownian motion becomes uniform on the circle group.

Let $Z_t=e^{iB(t)}$. For every integer $m$, its angular Fourier coefficients satisfyThese are precisely the Fourier coefficients of the uniform distribution on the unit circle. Thus expectations of every trigonometric polynomial converge to its uniform angular average. By the Stone-Weierstrass theorem, trigonometric polynomials approximate continuous functions on the unit circle uniformly. Given an approximation error $\epsilon$, the discrepancy of expectations is at most the polynomial discrepancy plus $2\epsilon$. Taking $t\to \infty$ and then $\epsilon \to 0$ proves weak convergence of probability measures to uniform angular measure.

The function $z\mapsto (\mathrm{Im}z{)}^2$ is continuous on the unit circle, so the continuous mapping theorem and part (a)(ii) implyEquivalently, the limiting arcsine distribution has cumulative distribution function $2\arcsin \sqrt{x}/\pi$ for $0\le x\le 1$, and probability density function $1/(\pi \sqrt{x(1-x)})$ for $0<x<1$. This proves the limiting law rather than inferring it merely from the limiting mean.

With $H>0$, setThen the first integral gives $1/\sqrt{1+y^{\prime 2}}=s$. On the descending and ascending halves of the physical track choose respectively $0<\varphi <\pi$ and $\pi <\varphi <2\pi$, so that $y^{\prime }=\cot (\varphi /2)$. Solving for $y$ givesThe unsquared equation requires $1-bs>0$. At $s=1$ this says $b<1$; it is not permissible to retain other values merely because the displayed formula for $y$ has a squared denominator.

For completeness, the Euler-Lagrange equation also gives $y^{\prime \prime }=(A^{\prime }(y)/A(y))(1+y^{\prime 2})<0$. Thus $y^{\prime }$ decreases through zero exactly once, so the chosen parameter runs through the whole single arch without an extra branch. Differentiating the formula for $y$ and dividing by $y^{\prime }=\cot (\varphi /2)$ gives, continuously also at $\varphi =\pi$,Using $x(0)=0$ yieldsSince $\sqrt{1+y^{\prime 2}}=1/\sin (\varphi /2)$, the arc length is obtained by cancelling one factor of $\sin (\varphi /2)$ in $dx/d\varphi$. The conditions $x(2\pi )=1$ and $S=L$ becomeThese determine a unique parameter pair for each $L>1$. To see uniqueness, regard $L$ as the expected value of $1/s$ under a probability density function proportional to $s^2/(1-bs{)}^3$ on $0<u<2\pi$. Differentiation under the integral sign gives the covariance identityThe first argument decreases with $s$ and the second increases. The strict sign follows by applying $(f(s)-f(t))(g(s)-g(t))<0$ to two independent unequal values and averaging; the integrable endpoint singularity of $1/s$ causes no problem. As $b\uparrow 1$, the weight concentrates at $s=1$, so $L\downarrow 1$. As $b=-k\to -\infty$, endpoint estimates give numerator $\sim 2/k^2$ and denominator $\sim 4\log k/k^3$, so $L\sim k/(2\log k)\to \infty$. Continuity then covers every $L>1$. At $b=0$ one recovers the cycloid and $L=4/\pi$.

Smoothness qualification. Near $\varphi =0$,and the other endpoint has $y^{\prime }\to -\infty$. These are smooth graphs on $0<x<1$, not $C^1$ graphs on the closed interval. Indeed a finite endpoint slope would make the conserved $H$ diverge as $y\to 0$. The travel time remains finite becauseThus the printed phrase “smooth solution” must be understood on the interior, with improper endpoint limits. The variational calculation supplies necessary stationary arches; it does not by itself prove global minimality in every larger class of tracks.

Let $S_1(t)$ be the probability that an ATP-waiting state has not yet bound a substrate molecule. A constant bimolecular binding coefficient $k_1$ at fixed ATP concentration $c$ gives a pseudo-first-order hazard $\lambda =k_{1}c$. Thus ${\dot{S}}_1=-\lambda S_1$, $S_1(0)=1$, and the waiting-time distribution isThe probability density function starts at $k_{1}c$ and decreases exponentially. Increasing $c$ shortens the waiting time in inverse proportion. This supports one rate-limiting ATP-binding event before the associated motion, rather than a concentration-independent chemical delay.

For the second delay, consider sequential independent first-order transitions $A{⟶}^{k_a}B{⟶}^{k_b}C$. Their dwell times have exponential distributions. The total waiting time is their sum, so convolution givesIt is normalized, symmetric under exchange of $k_a,k_b$, and hasUnlike one exponential delay, it vanishes at $t=0$, rises initially as $k_a{k}_{b}t$, reaches a maximum at $t_{\mathrm{mode}}=\log (k_a/k_b)/(k_a-k_b)$ and eventually decays at the slower rate. For equal rates the continuous limit is $P_2(t)=k^{2}t{e}^{-kt}$, an Erlang distribution with mode $1/k$. If one rate is much faster than the other, the observed long-time behavior approaches one exponential. The sketch in the preceding section displays these shapes using arbitrary specified rates.

The simplest catalytic cycle compatible with the fits therefore contains an ATP-dependent binding delay and at least two resolvable ATP-concentration-independent kinetic events. Hydrolysis, product release and conformational changes are possible microscopic assignments, but the two-rate fit alone does not uniquely identify them or the order of $k_a$ and $k_b$. If each complete $120^{\circ }$ cycle contains these three waits and step durations are negligible, its mean duration is $(k_{1}c{)}^{-1}+k_a^{-1}+k_b^{-1}$ and the mean revolution rate is the inverse of three times this duration. Multiple catalytic sites can cooperate during that cycle.

TODO: the source PDF's stepping plot is absent. Its individual substep angles, numerical timings and the mapping of plotted dwell labels onto the larger or smaller angular movement cannot be read. The surviving text establishes only the $120^{\circ }$ total step and the existence of a larger and a smaller substep. No numerical rate, missing angle or plotted label is inferred from a different experiment. The distribution derivations and the concentration-dependence inference above are independent of that illustration.

Under the simplest interpretation, each spot contains two equally bright, independently bleaching green fluorescent protein emitters. Losing one halves the fluorescence and losing the other removes it. A stable protein dimer carrying one active label per subunit is consistent with this result. The conclusion is two active emitters in the observed spot; it does not by itself rule out unresolved coincident proteins, incomplete fluorescent maturation or blinking. Permanent membrane binding makes loss by dissociation less plausible, but lateral motion out of the observation region would still need experimental control.^[ref]

Let their independent photobleaching times be $T_A,T_B\sim \mathrm{Exp}(k_b)$. Then $t_1=\min (T_A,T_B)$ has rate $2k_b$, while the memoryless remaining lifetime $t_2-t_1$ has rate $k_b$ and mean $1/k_b$, matching the stated single-emitter calibration. The two-emitter photobleaching completion time is the maximum, not one single exponential and not a sum of two equal-rate exponential variables. Its cumulative distribution function isDifferentiating gives its probability density function:with zero density for $t<0$. It integrates to one, and its mean is $1/(2k_b)+1/k_b=3/(2k_b)$. This uses two active fluorophores at the start of observation, constant illumination and independent irreversible bleaching with the calibrated rate.

Write $s={\sigma }^2\ge 1$. The two multivariate normal distribution probability density functions have ratioHere $\Vert x\Vert$ is the Euclidean norm. Consequently the smallest rejection sampling envelope constant, and its acceptance probability, areThe supremum is attained at $x=0$; hence no smaller constant can dominate the ratio. Draw $Y\sim N(0,s{I}_d)$ and an independent $U$ from the uniform distribution on $(0,1)$. Accept $Y$ whenotherwise repeat with fresh independent random variables. One can generate $Y=\sqrt{s}G$ using a random vector $G$ with independent standard normal distribution coordinates.

For any measurable set $B$, the joint acceptance probability is ${\int }_{B}q(y)p(y)/(M_{\ast }q(y))dy=M_{\ast }^{-1}{\int }_{B}p(y)dy$. Dividing by $1/M_{\ast }$ proves that the returned random vector has probability density function $p$. The number of proposals has a geometric distribution on the positive integers, with expected value $M_{\ast }$. When $s=1$, every proposal is accepted.

The integrand is positive, so the minimum-variance importance distribution is its target-weighted probability density function, normalized by $\theta =\sqrt{e}$:By completing the square,Thus $q^{\star }$ is the multivariate normal distribution $N(-e_1,I_d)$, where $e_1=(1,0,\dots ,0)$ is the first standard basis vector. Its importance weight and weighted integrand areA variance cannot be negative, and this proposal makes every summand constant, proving optimality directly. It is admissible under the arbitrary-proposal condition even though its mean vector is not zero. Although optimal proposals usually involve an unknown normalizing integral, here both that integral and the resulting normal distribution are explicit.

Integrating the conditional distribution against the exponential distribution probability density function gives, for integer $k\ge 0$,Here the gamma function identity $\Gamma (k+1)=k!$ evaluates the integral. Henceon $\{0,1,2,\dots \}$. Put $r=1/(1+\lambda )$ and $\rho =1-r$. The geometric series gives $F(k)=1-r^{k+1}$. Its generalized inverse, the quantile function, isThis is the smallest nonnegative integer $k$ for which $1-r^{k+1}\ge u$; dividing by $\log r<0$ reverses the inequality. Draw $U$ from the uniform distribution on $(0,1)$ and use this formula. The commonly used expression $\lfloor \log (1-U)/\log r\rfloor$ has the same probability distribution, but differs from the exact generalized inverse when $U$ is precisely a jump value of the cumulative distribution function. Such values have probability zero for an ideal continuous uniform draw.