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For a fixed true parameter $0<p<1$, the usual asymptotic normality of maximum likelihood estimators givesEquivalently, for a large sample, $\hat{p}$ is approximately normally distributed with mean $p$ and variance $p^2(1-p)/(nr)$. Replacing the unknown variance by its consistent plug-in estimate gives the Wald confidence intervalHere $z_{1-\alpha /2}$ is a standard-normal quantile. Intersect with the physical parameter range $(0,1)$ if needed. This is an interior, large-sample approximation; the degenerate estimate $\hat{p}=1$ from $S=0$ does not justify a zero-width finite-sample interval.

Full column rank gives the least-squares estimatorWrite $A=(X^{T}X{)}^{-1}{X}^T$. Since $AX=I$ and $\mathrm{Cov}(ϵ)={\sigma }^{2}I$, we have $\hat{\theta }=\theta +Aϵ$ andThis calculation needs the stated uncorrelated equal-variance errors, but not normality.

The expected activity is $E_t=\alpha +\beta (e^{-t}-c)$. Under normal errors, least squares is maximum likelihood, and invariance of maximum-likelihood estimation givesThe covariance term vanishes by part (ii), soThis is uncertainty in the estimated mean activity; the variance of a future individual observation would include its additional measurement-error variance.

By the addition rule for independent Poisson random variables, $X_n$ has the same distribution as a sum of $n$ independent variables $Z_j\sim \mathrm{Poisson}(\lambda )$. Each has expectation and variance $\lambda$. The central limit theorem therefore givesFor $\lambda =0$, every variable is zero, and the conclusion is the degenerate normal law with zero variance.

Let ${\ell }_n(\theta )={\sum }_{i=1}^n\log f(X_i;\theta )$ be the log-likelihood. Under the usual regularity conditions, the expected Fisher information in the whole sample isThe equality of the two expressions uses mean-zero score and permitted differentiation of the normalization identity. Independent observations make the information additive. The maximum-likelihood estimator has asymptotic normality:The factor $n$ is included in $I_n$; it must not be inserted a second time into the variance.

For the same sample size $n$, the geometric distribution provides more Fisher information:The corresponding asymptotic variances of the maximum-likelihood estimators areHence choose the geometric sample for more precise inference at equal numbers of observations. Its asymptotic variance is $\theta$ times as large, and its standard error is $\sqrt{\theta }$ times as large. This illustrates Fisher information per observation and per trial. It is a per-observation comparison: one geometric observation records a waiting time containing, on average, $1/\theta$ Bernoulli trials. A constraint on the total number of underlying trials would be a different sampling-budget comparison.

Independence and common error variance give $\mathrm{Var}(ϵ)={\sigma }^2{I}_n$. For the fixed matrix $A=(X^{T}X{)}^{-1}{X}^T$, the covariance transformation rule givesThus the covariance is determined by the error variance and the inverse Gram matrix of the design columns.

The future mean response is $E_{\ast }=\alpha +\beta x_{\ast }+\gamma z_{\ast }$. By invariance of maximum likelihood estimation, its estimator is the fitted meanIts sampling variance is the quadratic form of the coefficient covariance matrix with $c=(1,x_{\ast },z_{\ast }{)}^T$:This estimator is unbiased because each fitted coefficient is unbiased. The future error ${ϵ}_{\ast }$ does not enter the estimator of the mean, so its independent variance is not part of this expression.

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.

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 $S={\sum }_{j=1}^n|X_j|$. 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 $\mathrm{Var}(S)=n{\theta }^2$.

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 $s=1+it$. 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.

Linearity of expectation and the vanishing of cross covariances for independent variables giveThe expectation formula does not require independence; independence is used for the variance sum.

The moment-generating function of a random variable $X$ is $M_X(t)=\mathbb{E}[e^{tX}]$ wherever this expectation is finite. Existence on an open interval containing zero is the condition needed for the usual uniqueness and convergence theorems. For the exponential distribution with rate $\lambda >0$,For $t\ge \lambda$ the integral diverges. Differentiating at zero gives mean $1/\lambda$ and variance $1/{\lambda }^2$.

The independent identically distributed central limit theorem states that for variables of common finite mean $\mu$ and finite positive variance ${\sigma }^2$,For the exponential distribution, $\mu =1/\lambda$ and $\sigma =1/\lambda$. Thus the event in question is exactlyThe limiting normal distribution has a continuous distribution function at every real $x$, so convergence in distribution givesHere $\Phi (x)=(2\pi {)}^{-1/2}{\int }_{-\infty }^x{e}^{-u^2/2}du$, including the strict-inequality convention because the limiting distribution is continuous.

In a regular one-parameter model with joint density $f_{\theta }(x)$, write the score function as $U_{\theta }={\partial }_{\theta }\log f_{\theta }(X)$ and the Fisher information as $I_n(\theta )={\mathbb{E}}_{\theta }[U_{\theta }^2]$. Assume a parameter-independent support, differentiability and sufficient domination to interchange parameter differentiation and integration, and $0<I_n(\theta )<\infty$. If $T$ has finite variance and is unbiased for a differentiable target $g(\theta )$, the Cramér-Rao lower bound isIndeed, differentiating total mass gives ${\mathbb{E}}_{\theta }{U}_{\theta }=0$, and differentiating ${\mathbb{E}}_{\theta }T=g(\theta )$ gives ${\mathrm{Cov}}_{\theta }(T,U_{\theta })=g^{\prime }(\theta )$. Cauchy-Schwarz inequality proves the bound. For $n$ independent identically distributed observations, information adds: $I_n=n{I}_1$. In particular for the parameter itself, $g(\theta )=\theta$, the bound is $1/I_n$. The assumptions are material: the same formula cannot simply be imposed on models with parameter-dependent support.

Put $S={\sum }_i{X}_i$. Independence and the exponential sampling model give the gamma distribution with densityFor $n>k$, substitution $u=\lambda s$ yieldsAt $k=1$, $n\ge 2$, this gives $\mathbb{E}({\bar{X}}^{-1})=n\lambda /(n-1)$. Thereforeis an unbiased estimator of the rate. For $n>2$, the same formula with $k=2$ givesFor the source-allowed value $n=2$, the first moment is still finite, but the second moment contains an integral proportional to ${\int }_0^1{s}^{-1}{e}^{-\lambda s}ds$ and diverges. Hence the estimator has infinite variance when $n=2$; the finite formula applies only to $n>2$.

The sample score is $n/\lambda -S$, and its Fisher information is $I_n(\lambda )=n/{\lambda }^2$. The Cramér-Rao lower bound for unbiased rate estimation is thus ${\lambda }^2/n$. For every $n>2$, the actual variance is strictly larger by the factor $n/(n-2)$. The ratio approaches one as $n$ grows, but equality is not attained at finite $n$. At $n=2$ the finite-variance form of the bound is inapplicable to this estimator; its infinite variance provides no efficiency equality.

Minimize the residual sum of squares $\Vert Y-X\theta {\Vert }^2$. Differentiating gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ positive definite and invertible, so the unique minimizer isSince $\mathbb{E}Y=X\theta$, it is unbiased. The independent common-variance errors give $\mathrm{Cov}(Y)={\sigma }^{2}I$, henceIndependent Gaussian errors with common variance are a sufficient condition for this estimator also to be the maximum likelihood estimator: their log-likelihood differs from $-\Vert Y-X\theta {\Vert }^2/(2{\sigma }^2)$ by terms independent of $\theta$. Profiling an unknown common variance gives the same minimizer. Mean and variance assumptions alone do not specify a likelihood.

For the seven design points, $\bar{x}=0$ and $\overline{x^2}=4$. In increasing time order the design matrix isIts columns are mutually orthogonal: the odd sums vanish and $\sum (x_i^2-4)=0$. Their squared lengths are $7,28,84$, respectively. ThereforeIn particular $\hat{\gamma }={\sum }_i(x_i^2-4)Y_i/84$. Testing for a quadratic contribution means testing $H_0:\gamma =0$ against $\gamma \ne 0$.

The cooling measurements are stated to have zero-mean errors and known common variance, but normality is not explicitly assumed for them. Under the independent-error linear regression model as stated, Chebyshev's inequality gives a valid conservative level-$\delta$ test: reject whenIts null rejection probability is at most $\delta$ because the estimator has mean zero and the calculated variance. If Gaussian errors are additionally assumed, the sharper exact test uses $Z=\sqrt{84}\hat{\gamma }/\sigma \sim N(0,1)$ under the null and rejects for $|Z|>z_{1-\delta /2}$. With only seven observations, exact normal calibration cannot be deduced from the printed moment assumptions alone.

Minimize $\Vert Y-X\theta {\Vert }^2$. Differentiation gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ positive definite, soThe expectation uses the zero error mean; the covariance uses $\mathrm{Cov}(ϵ)={\sigma }^{2}I$. With independent Gaussian errors of common variance, maximizing likelihood minimizes exactly the same residual sum of squares, so this is also the maximum likelihood estimator, whether the common variance is known or estimated.

For the quadratic design, decompose each group's residual sum of squares into its within-group constant plus the group size times the squared residual of its sample mean. The three parameter combinations $\alpha$, $\alpha +\beta +\gamma$ and $\alpha -\beta +\gamma$ can fit all three group means exactly. ThereforeThe group means are independent normal variables, with variances ${\sigma }^2/m,{\sigma }^2/n,{\sigma }^2/n$. Hence their two stated linear combinations are jointly normal, andJoint normality therefore proves the two estimators are independent. Their variances are ${\sigma }^2/(2n)$ and ${\sigma }^2/(2n)+{\sigma }^2/m$, respectively. Equal replication at the two off-center temperatures is what cancels the covariance.