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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

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.

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.

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.

For the exponential distribution with rate $\lambda$, the cumulative distribution function is $F(x)=1-e^{-\lambda x}$ for $x\ge 0$, and zero for $x<0$. Its quantile function isFor inverse transform sampling, draw $U$ from the uniform distribution on $(0,1)$ and return $-\log (1-U)/\lambda$. Equivalently return $-\log U/\lambda$, since $1-U$ has the same uniform distribution. Indeed, for $x\ge 0$, $\mathbb{P}(-\log U/\lambda >x)=\mathbb{P}(U<e^{-\lambda x})=e^{-\lambda x}$.

Solving $u=F(x)$ gives $1-(1-p)e^{-\beta x}=p^{1-u}$. Hence the quantile function isFor such $u$, $p<p^{1-u}<1$, making the logarithm's argument lie in $(0,1)$ and the output positive. Its endpoint limits are $F^{-1}(0)=0$ and $F^{-1}(1)=+\infty$. Substitution into (i)'s cumulative distribution function recovers $u$.

Let $r=1-p$. Sinceintegration from zero to $x>0$ gives $F(x)=(C/\beta )[\log (1-r{e}^{-\beta x})-\log p]$. Letting $x\to \infty$ and imposing unit total probability givesThe probability density function is positive for $x>0$ because $0<(1-p)e^{-\beta x}<1$. Consequently the cumulative distribution function is increasing, starts at zero and tends to one. This verifies that the derived normalizing constant defines a probability distribution. The family is the exponential-logarithmic distribution with the complementary mixing parameter $r=1-p$.

The central limit theorem gives $\sqrt{n}({\hat{\theta }}_n^{\mathrm{MC}}-\theta )\Rightarrow N(0,1)$, so, with $\Phi$ the cumulative distribution function of the standard normal distribution,The error has two tails. Using the supplied $2.58$ quantile, whose two-sided coverage is approximately $0.9902$, givesThe supplied $2.33$ quantile would be appropriate for a one-sided $99\%$ calculation; it would give only approximately $98.02\%$ two-sided coverage. Using the exact $0.995$ quantile instead of $2.58$ would give $66,349$ draws; we consistently use the paper's rounded value $N=66,564$ below.

For routine Monte Carlo integration, the central limit theorem calculation is preferable as a practical budget: it uses about one fifteenth as many draws as the Chebyshev's inequality bound. However, its coverage is approximate. Finite variance alone supplies no universal rate of convergence to the normal distribution; an integrand with large skewness or a heavy-tailed distribution can make the approximation poor even at a large sample count. If an unconditional finite-sample $99\%$ guarantee is required under only the stated assumptions, use the Chebyshev's inequality budget.

Let $z={\Phi }^{-1}(1-\alpha /2)$, where $\Phi$ is the standard normal distribution's cumulative distribution function. Statistical consistency and Slutsky's theorem allow replacement of $I_1(\theta )$ by $I_1(\hat{\theta })$:Thus the probability that this standardized estimation error lies between $-z$ and $z$ tends to $1-\alpha$. Rearranging gives the Wald confidence intervalThe coverage probability is approximate, with limiting value $1-\alpha$; it is a repeated-sampling statement about the random confidence interval.

The sample mean $\overline{X}=T/n$ has mean $\theta$ and variance $\sigma (\theta {)}^2=\theta (1-\theta )/n$. The central limit theorem gives a normal approximation for interior $\theta$ when both expected binomial counts are sufficiently large. Under the null hypothesis, a level-$\alpha$ upper-tail cutoff is approximatelyTo obtain the power function, evaluate this same cutoff under the true statistical parameter $\theta$, standardizing with $\sigma (\theta )$:Here $\Phi$ is the standard normal distribution's cumulative distribution function. The formula treats the binomial threshold as continuous; exact integer cutoffs and boundary randomization are finite-sample refinements.

An equal-tailed credible interval allocates equal posterior probability to its two excluded tails:For a continuous strictly increasing posterior cumulative distribution function $F$, the endpoints are $F^{-1}(\alpha /2)$ and $F^{-1}(1-\alpha /2)$.

A highest posterior density interval instead includes the parameter values of greatest posterior density: there is a cutoff $c$ such that the retained region is a superlevel set of $\pi (\theta \mid x)$, of posterior mass $1-\alpha$. Every interior retained value has density at least as large as every excluded value, apart from possible ties. For a unimodal continuous density this is an interval; when both endpoints lie inside the support they have equal posterior density. For a multimodal posterior distribution, a highest-density region may be a union of intervals, rather than one interval. Equal tail probabilities and highest density are different criteria.

Let $F$ be the cumulative distribution function of a gamma distribution with shape $A=a+{\sum }_i{x}_i$ and rate $B=b+n$; explicitly,The posterior distribution is continuous and has positive density on $(0,\infty )$, so its quantiles give the requested equal-tailed credible interval:Equivalently, $2B\theta \mid x\sim {\chi }_{2A}^2$, so the endpoints can be written as the corresponding chi-squared distribution quantiles divided by $2B$. Either form excludes posterior probability $\alpha /2$ from each tail.

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.

Continuity and strict increase make the quantile function $F^{-1}$ well-defined on $(0,1)$. The endpoints of the uniform distribution occur with probability zero and do not affect the construction. For every real $x$,Thus each $X_i$ has cumulative distribution function $F$. Applying a fixed measurable function separately to independent random variables preserves their independence; explicitly, joint lower-tail probabilities factor asConsequentlyThis is inverse transform sampling. Since $F^{-1}$ preserves order, it also sends the central order statistic to the central order statistic, justifying $M_X=F^{-1}(M_U)$.

Under the null hypothesis, the full model has two free positive statistical parameters and the constrained model has one. The Wilks theorem therefore givesas both sample sizes tend to infinity, with $n/(n+m)$ tending to a number strictly between zero and one. For the observed $t$, calculateThe large-sample p-value is the upper-tail probabilitywhere $\Phi$ is the cumulative distribution function of the standard normal distribution. The last equality uses the fact that the square of a standard normal random variable has a chi-squared distribution with one statistical degree of freedom. Values of $T$ on either side of $n/(n+m)$ are already included by the transformation to $W$.

Put $A=a+s>1$ and $B=b+n-s>1$ as in part (a), and let $F_{A,B}$ be the cumulative distribution function of the beta distribution $\mathrm{Beta}(A,B)$. The two endpoint equations areThe first equation fixes the posterior probability, and the second makes the two endpoint posterior densities equal. Their admissible ordering isTo justify sufficiency and uniqueness, the derivative of the logarithm of the posterior density iswhich is positive below $m=(A-1)/(A+B-2)$ and negative above it. Thus the posterior density increases strictly to its unique mode and decreases strictly thereafter, vanishing at both ends. Each density threshold $k$ between zero and the maximum determines exactly one endpoint on each side of $m$. The probability between these endpoints decreases continuously and strictly from one to zero as $k$ increases, so exactly one threshold gives 0.95. This is the beta highest posterior density endpoint characterization.

A likelihood-ratio test rejects for small $\lambda$. The derivative of the natural logarithm of its variable factor isThus $t^n(1-t{)}^m$ increases up to $n/(n+m)$ and decreases thereafter, tending to zero at both endpoints. Every nontrivial lower level set of $\lambda$ therefore consists of two tails. For a prescribed test size $0<\alpha <1$, the exact critical region isWriting $F_{n,m}$ for the cumulative distribution function of the beta distribution with parameters $(n,m)$, the required two equations areThe first makes both cutoffs have the same likelihood ratio; the second calibrates the null hypothesis rejection probability. Equivalently, the second left-hand side isAs the common likelihood ratio threshold increases from zero to one, this probability increases continuously and strictly from zero to one; hence the equations determine unique cutoffs. The two tail probabilities need not each equal $\alpha /2$ when $n\ne m$. Exact calibration holds for every common rate because the given null distribution of $T$ is independent of that rate. Boundary equalities have zero probability.

Condition on the total of 400 complaints. Assume each complaint independently has the same probability $p$ of being received on Monday; in particular, repeated complaints from one incident must not create material dependence. The Monday count $S$ then has a binomial distribution with parameters $(400,p)$. Test $H_0:p=0.2$ against $H_1:p>0.2$, rejecting for large $S$ as suggested by the monotone likelihood ratio.

Under the null hypothesis, ${\mathbb{E}}_{0}S=80$ and ${\mathrm{Var}}_0(S)=400(0.2)(0.8)=64$. Both expected successes and expected failures are large, so the central limit theorem gives a suitable normal approximation:Using the paper's supplied rounded table, the upper-tail p-value is approximatelyA more accurate evaluation of the standard normal distribution cumulative distribution function gives about 0.0062. For example, choose significance level 0.05 before examining the data; the supplied upper 5% standard normal quantile is approximately 1.6, and $2.5>1.6$. Reject $H_0$ at 5%, and also at 1%, using this approximation. The observed binomial proportion is $100/400=0.25$, providing evidence that the Monday probability exceeds 0.2 under the sampling assumptions.

If applying a continuity correction, approximate ${\mathbb{P}}_0(S\ge 100)$ by $1-\Phi ((99.5-80)/8)=1-\Phi (2.4375)\approx 0.0074$; this gives the same 5% and 1% conclusions. The uncorrected value 0.005 is approximate, so a decision exactly at the 0.5% boundary is not warranted by the supplied rounded table.

Here the gamma distribution uses a rate statistical parameter $\theta$, with known shape $\lambda$. Write $S={\sum }_i{X}_i$ and $a=n\lambda$. For each fixed ${\theta }_1>{\theta }_0$, the alternative-to-null likelihood ratio isstrictly decreasing in $S$. The Neyman-Pearson lemma therefore gives the same lower-tail critical region $S<c_{\alpha }$ for every larger rate. Since $S$ has a continuous gamma distribution under the null hypothesis, there is a unique positive $c_{\alpha }$ satisfyingThus reject when $S<c_{\alpha }$ is a uniformly most powerful test of exact test size $\alpha$. If $G_a$ denotes the unit-rate gamma distribution cumulative distribution function with shape $a$, thenThis is the uniformly most powerful test for a gamma rate.

When $\lambda =1/n$, the sum has shape one and hence an exponential distribution with rate $\theta$. ThereforeIn particular $\pi ({\theta }_0)=\alpha$, and the power function is strictly increasing, exceeding $\alpha$ for $\theta >{\theta }_0$ and tending to one as $\theta \to \infty$.