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Write $X=\mu +\sigma Z$, where $Z$ has the standard normal distribution. Apply Gaussian integration by parts to $g(z)=U^{\prime }(\mu +\sigma z)$:Since $X-\mu =\sigma Z$,The polynomial growth assumption and decay of the normal distribution density justify the integration by parts.

Condition on $S$ and use the characteristic function of a standard normal distribution:The integrand has modulus one, so Fubini's theorem is immediate. Taking expectation over $S$ yields

The intended monotone rearrangement isHere $G^{-1}$ is the quantile function of $\nu$, agreeing with the ordinary inverse when $G$ is continuous and strictly increasing. With an atomless measure $\mu$, its cumulative distribution function $F$ is continuous, and the probability integral transform makes $F(X)$ uniform on $(0,1)$ for $X\sim \mu$. Thus $G^{-1}(F(X))\sim \nu$. The one-dimensional monotone rearrangement theorem says this transport map minimizes the cost $d(x-y)$ for convex continuous $d$, whenever the cost integrals are well defined. Values at exceptional endpoints may be chosen arbitrarily.

The printed assumptions omit an essential source condition. Invertibility of $G$ alone does not ensure an admissible transport map. For example, $\mu ={\delta }_0$ and $\nu$ a standard normal distribution satisfy the stated condition on $G$, but $T_{\#}{\delta }_0$ is always a Dirac measure. There is no solution to the Monge optimal transport problem in this example. The boxed answer therefore requires the additional assumption that $\mu$ is an atomless measure, or an equivalent condition making the displayed map admissible. For arbitrary sources the always admissible monotone transport plan is $(F^{-1},G^{-1}{)}_{\#}\mathcal{U}(0,1)$, which need not be induced by a map.

Since $Y_1=X_1^2$ and a standard normal variable has second and fourth moments $1$ and $3$,Thus $T_n$ has the chi-squared distribution with $n$ degrees of freedom and

Convergence in distribution $W_n\Rightarrow W$ means that the cumulative distribution functions satisfy $F_{W_n}(x)\to F_W(x)$ at every continuity point of $F_W$. The Lévy continuity theorem states that this is equivalent to pointwise convergence of characteristic functions to the characteristic function of $W$, provided the limiting function is continuous at zero.

For fixed $t$, independence and a Gaussian integral giveUsing $\log (1-z)=-z-z^2/2+O(z^3)$,Therefore ${\phi }_{V_n}(t)\to e^{-t^2/2}$, the characteristic function of the standard normal distribution. Lévy's theorem now givesThis derives the limit directly, without assuming the central limit theorem.

Let $L(t)=f(x,t)$ be the likelihood function. Since the prior distribution is $N(0,I_p)$, the posterior density has the formwhere $Z$ is the normalizing constant.

Choose $0<\delta \le 1/2$ and use the Gaussian autoregressive proposal reversible with respect to a standard normal distributionThuswhose covariance matrix is the required $2\delta I_p$. Reversibility with respect to the standard normal distribution sayssoThe Metropolis–Hastings acceptance probability therefore reduces toThis is the Preconditioned Crank–Nicolson algorithm with proposal scale $\beta =\sqrt{2\delta }$.

Let $m=F^{-1}(1/2)$. Since $F$ has an everywhere positive density, it is continuous and strictly increasing, so $F(m)=1/2$. Membership in the Kolmogorov neighborhood of a distribution givesBy the symmetry of the standard normal distribution,The stated asymptotic-bias formula for the sample median therefore yields

If $X\sim N(0,{\theta }^2)$, then $Y=|X|$ has a scaled half-normal distribution withPut $c={\Phi }^{-1}(3/4)$. The population median is $t_0=\theta c$, and the asymptotic distribution of a sample median gives

Under $H_0:{\theta }^2=1$, positivity of the scale means $\theta =1$. If $z_{1-\alpha }$ is the $(1-\alpha )$-quantile of the standard normal distribution, an asymptotically level-$\alpha$ test rejects forLarger scale makes the population median $\theta c$ larger, so this is the appropriate one-sided rejection region.

Let $z_{1-\gamma /2}$ be the corresponding standard normal distribution quantile. Slutsky theorem permits replacing $\beta$ by the consistent estimator $\hat{\beta }$, giving the confidence intervalAfter standardization, the preceding convergence in distribution and continuity of the normal distribution show that its coverage tends to $1-\gamma$.

The commands choose $n=100$ observations and $p=2$ predictors; generate a $100\times 2$ matrix $X$ of independent standard-normal values; generate independent binary random variables $Y_i$ with success probability $1/2$; and report the observed number of successes, ${\sum }_i{Y}_i=48$. The first glm call then fits the Bernoulli logistic-regression modelThe prediction command returns the fitted response probabilities ${\hat{p}}_i$ and sums them.

The likelihood isAt the maximum-likelihood estimator, the intercept component of the score function isThereforeexactly, by the fitted-mean balance for logistic regression with an intercept.

The second fit uses probit regression,where $\Phi$ is the standard normal distribution function. Its intercept score isa weighted residual equation. It does not imply ${\sum }_i{\hat{p}}_i=48$, so exact equality is not expected. The output should nevertheless be close to $48$: the data were generated with constant success probability $1/2$ independently of $X$, so the fitted slopes should be small, the linear predictors should cluster near a common intercept, and the score weights should be nearly constant.

Put $U=X/\sigma$. The standardized efficient score function for scale isAn optimal B-robust scale M-estimator solveswhere the optimal bounded score has the clipped-score formThe constants $A,z$ enforce Fisher consistency, the chosen normalization, and the clipping bound $b$ on the influence function.

For the standard normal distribution, $f^{\prime }(u)/f(u)=-u$, hence $\Lambda (u)=u^2-1$. The score therefore simplifies towith $z$ chosen so that $\mathbb{E}\psi (Z)=0$ for $Z\sim N(0,1)$.

Let ${\xi }_{1,n},{\xi }_{2,n}$ be independent random variables with the standard normal distribution. The unconstrained Milstein method step isThe $X$ noise is additive, so its Milstein correction vanishes. The diagonal, independent noise fields commute, so no cross iterated stochastic integral is required.

Implement each reflecting boundary condition for a diffusion by folding the proposed coordinate back into $[-1,1]$. One formula that also handles multiple overshoots isThe complete update isUnder the usual smoothness assumptions, the Milstein discretization has strong order of convergence one and weak order of convergence one. The reflection enforces the boundary pathwise.

The probability density function must integrate to one, soAlso, linearity of expectation and the mean and variance of the standard normal distribution giveUsing the probability density function to compute this expected value givesThese are the values $\Gamma (1/2)$ and $\Gamma (3/2)$ of the gamma function.

Since $Y=Z^2/2$ is nonnegative, its cumulative distribution function is zero for $y<0$. For $y\ge 0$,The last equality uses the symmetry of the standard normal distribution. Thus

Apply the central limit theorem to the independent and identically distributed random variables $X_i^2$, whose mean and variance are $\theta$ and ${\theta }^2$. It gives the approximate standard normal distributionThe supplied standard normal quantile $c=1.96$ leaves probability $0.025$ in each tail. HenceSince $\theta >0$, this event is equivalent toFor $n>1.96^2$, equivalently integer $n\ge 4$, both denominators below are positive. Inverting the inequalities gives the approximate $95\%$ confidence intervalThe approximation is justified for large $n$; positivity at $n\ge 4$ by itself does not guarantee accurate coverage. This construction is a confidence interval from a relative normal pivot: the unknown scale remains inside the pivotal ratio until the inequalities are inverted. Its meaning is that the random interval covers the fixed statistical parameter $\theta$ with probability approximately $0.95$ under repeated sampling.

The sample mean is exactly $N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The given quantile gives probability $0.95$ to the interval $[-1.96,1.96]$. Inverting this pivot yields the exact $95\%$ confidence interval

Let $A=\{|Z|>2\}$ and $B=\{Z>0\}$. The standard normal distribution is symmetric and assigns no probability mass to zero. Thus $\mathbb{P}(B)=1/2$, and its equal two-sided tails giveThereforeso the two events are independent. The permitted approximation gives $\mathbb{P}(A)\approx 0.05$ and $\mathbb{P}(A\cap B)\approx 0.025$, but independence is exact.

An expected value of a signed random variable is finite precisely when its absolute value is integrable. With the standard normal distribution probability density function,For $\alpha <1/2$, this integral equals $1/[\sqrt{2\pi }(1/2-\alpha )]<\infty$. For $\alpha \ge 1/2$, its integrand does not decay sufficiently and the integral diverges. ThereforeThe value is zero by the odd function and the symmetric distribution. For the other $\alpha$, both positive and negative parts have infinite integrals; formal cancellation would give only a Cauchy principal value, not an expected value. This distinction is the integrability threshold for exponential powers of a Gaussian.

The standard normal distribution function is strictly increasing. Thus $p>\alpha$ impliesRearranging givesThis is the confidence-interval/test duality for the matching one-sided procedures. The strict inequalities agree with the printed open interval.

Let $\mu =\mathbb{E}[X_1]$ and ${\sigma }^2=\mathrm{Var}(X_1)\in (0,\infty )$. For independent and identically distributed random variables, the central limit theorem statesThe limit is the standard normal distribution. If the variance is zero, then $X_i=\mu$ almost surely; the centered sums are zero, and division by $\sigma$ is not defined.

By part (c) with $\lambda =1$, let $T_n=E_1+\cdots +E_n\sim \mathrm{Gamma}(n,1)$ for independent and identically distributed exponential random variables. The requested integral is $\mathbb{P}(T_n\le n)$. Each summand has expectation one and variance one, so the central limit theorem givesZero is a continuity point of the standard normal distribution, and consequently