The further observation is independent of the earlier observations conditional on $\theta$. Integrate its conditional success probability against the posterior:The given factorial formula for the beta function gives this ratio as $A/(A+B)$. Thus the posterior predictive probability isIt is the posterior mean of the Bernoulli success parameter. Independence here is conditional: sharing the uncertain parameter generally makes past and future observations dependent before conditioning on it.

Let $F_{A,B}$ denote the distribution function of the posterior beta distribution, with $A=a+s$ and $B=b+n-s$. This distribution is continuous with positive density on $(0,1)$. Its equal-tailed credible interval isCompute the two quantiles by solving ${\int }_0^{{\theta }_L}\pi (\theta \mid x)d\theta =\alpha /2$ and ${\int }_{{\theta }_U}^1\pi (\theta \mid x)d\theta =\alpha /2$. Each excluded tail has mass $\alpha /2$, leaving posterior mass $1-\alpha$ between the endpoints.

A $100(1-\alpha )\%$ credible set is a measurable subset $C(x)\subseteq (0,1)$ with posterior probability $1-\alpha$:In ordinary language, given the observed data and the chosen prior model, the probability that the parameter lies in this set is $1-\alpha$. Such a set need not be an interval or be unique; its defining requirement is posterior mass.

Let $s={\sum }_{j=1}^n{x}_j$. The Bernoulli distribution likelihood is proportional to ${\theta }^s(1-\theta {)}^{n-s}$. Multiplication by the beta distribution prior givesNormalizing with the beta function therefore givesThe parameters remain positive. This is Beta-binomial conjugacy: successes add to the first shape parameter and failures to the second.

Let $z=z_{1-\alpha /2}={\Phi }^{-1}(1-\alpha /2)$, where $0<\alpha <1$. By the preceding asymptotic normality and Slutsky's theorem, $\sqrt{n}(\hat{\theta }-\theta )/\hat{\theta }$ is approximately standard normal. Inverting $|\sqrt{n}(\hat{\theta }-\theta )/\hat{\theta }|\le z$ gives the Wald confidence intervalIts coverage approaches $1-\alpha$ as $n\to \infty$. The interpretation of the confidence interval is repeated-sample coverage of the fixed parameter; it does not assign a posterior probability to $\theta$.

The absolute observations are independent exponential random variables with mean $\theta$ and variance ${\theta }^2$. Applying the central limit theorem directly to their average provesThus the large-sample approximation is $\hat{\theta }\approx N(\theta ,{\theta }^2/n)$, consistent with inverse Fisher information. Also $\hat{\theta }\to \theta$ in probability by the weak law of large numbers, so $\theta$ in the standard error can be replaced by $\hat{\theta }$ through Slutsky's theorem.

Put $S={\sum }_{j=1}^n|X_j|$. The log-likelihood for the Laplace distribution isFor $S>0$ the derivative is positive for $\theta <S/n$ and negative for $\theta >S/n$, so the unique maximum-likelihood estimator isThe exceptional data set $S=0$ has probability zero; there the likelihood has its supremum at the excluded boundary $\theta =0$. For the Fisher information, $|X_j|$ has exponential density ${\theta }^{-1}{e}^{-y/\theta }$ for $y>0$, hence mean $\theta$. Since ${\ell }^{\prime \prime }(\theta )=n/{\theta }^2-2S/{\theta }^3$,This also follows from the score variance, because $\mathrm{Var}(S)=n{\theta }^2$.

For the uniform distribution on $[0,1]$, the mean is $1/2$ and the variance is $1/12$. The central limit theorem for independent identically distributed variables with finite nonzero variance givesChoose $\boxed{\alpha =1/2,\beta =1/2}$. ThenThe standard normal distribution function is continuous and strictly increasing, so its value at the finite positive number $\sqrt{12}$ is strictly between $1/2$ and one, as required.

Put $Y_n=n(1-M_n)\ge 0$. For fixed $x\ge 0$ and sufficiently large $n>x$,For $x<0$ its distribution function is zero. The limiting distribution function is therefore $F(x)=0$ for $x<0$, $F(x)=1-e^{-x}$ for $x\ge 0$, the exponential distribution of rate one. Since this distribution function is continuous everywhere,

For $0\le z\le 1$, independence and the uniform distribution giveSince $M_n\le 1$, for $0<\epsilon <1$,For $\epsilon \ge 1$ the probability is zero. This is exactly convergence in probability of $M_n$ to one.

Independent marking with probability $5/6$ is Poisson thinning and gives a Poisson process of chip-buying arrivals, of rate $12(5/6)=10$ per hour. Its first arrival time therefore satisfiesFor example, its survival probability is $\mathbb{E}[(1/6{)}^{N(t)}]=\exp (12t(1/6-1))=e^{-10t}$, which also derives the thinning conclusion directly.

Let $I_2=T_2-T_1$ and $I_3=T_3-T_2$ be the interarrival times adjacent to the second arrival. They are independent exponential random variables of rate twelve per hour. The second customer meets the first if $I_2<1/12$ hour, and meets a later customer if $I_3<1/12$ hour. If the third arrives too late, every subsequent customer does also. Thus the only way to meet nobody is for both gaps to be at least five minutes:Hence $\boxed{\mathbb{P}(\text{meeting})=1-e^{-2}}$. Exact equality of arrival and departure times has probability zero. The calculation includes both customers already present and customers arriving during the visit; it is a two-sided interarrival test for overlap.

The first arrival time $T_1$, measured in hours after opening, has exponential distribution of rate twelve. ThereforeEquivalently, there must be no arrivals in the first quarter hour and at least one in the next; independent increments give $e^{-3}(1-e^{-3})$.

A Poisson process has a Poisson distribution for the count in an interval of length $t$, with mean equal to rate times length. Fifteen minutes is $1/4$ hour, so

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus isIt changes width and height with time, so this is not a stationary state. Its integral is the constant $\sqrt{\pi \hslash /m}$; multiplying the wavefunction by $(m/(\pi \hslash ){)}^{1/4}$ normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance $\hslash (1+t^2)/(2m)$. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

The Time-dependent Schrödinger equation isA stationary state of energy $E$ has a wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, whereIts time dependence is only a global phase, and its probability density is independent of time.

The polar Euler-Lagrange equation isFor a constant angle with $0<{\theta }_0<\pi$, conservation of $h$ makes $\dot{\varphi }$ constant, say $\Omega$. The remaining equation isIf $g>0$, every nonvertical constant-angle motion is a conical pendulum motion:The rod sweeps out a cone about the downward vertical, in either azimuthal direction. Equivalently, $h^2=(3g/(2a)){\sin }^4{\theta }_0/\cos {\theta }_0$. The exactly downward and exactly upward vertical rods are also static solutions; azimuth has no physical meaning at either pole. No static tilted rod is possible under nonzero gravity.

If $g=0$, the condition is $\sin {\theta }_0\cos {\theta }_0{\Omega }^2=0$. Thus a rod may remain at rest in any orientation, or rotate uniformly in the horizontal plane with ${\theta }_0=\pi /2$ and arbitrary constant $\Omega$. The vertical cases are included among static orientations; changing only their azimuth does not move the rod.

Away from the coordinate poles, use $\dot{\varphi }=h/{\sin }^2\theta$ in the conserved energy:Rearranging gives the first-order equationWhere the right side is positive, either sign of its square root is possible, according to the direction of motion. Its zeros are turning points. For $h\ne 0$, the centrifugal term excludes the poles; if $h=0$, the polar-coordinate formula is interpreted through the original rod equations.

For an autonomous Lagrangian, the Euler-Lagrange equations implyIndeed, differentiating the expression cancels the acceleration terms, leaving ${\sum }_q\dot{q}(d{L}_{\dot{q}}/dt-L_q)-L_t$, which vanishes. Here $T$ is quadratic in the velocities, so $\sum \dot{q}{L}_{\dot{q}}=2T$. Thus the conserved energy isThe azimuth $\varphi$ is a cyclic coordinate. Its conjugate momentum is constant:Therefore $\boxed{{\sin }^2\theta \dot{\varphi }=h}$ is constant as well. Physically, $p_{\varphi }$ is the vertical component of angular momentum.

The centre of mass is at height $-a\cos \theta /2$. With gravitational potential energy zero at the pivot level,Hence the Lagrangian isAdding a constant to the potential would change the numerical energy reference but would not change the motion.