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The iid central limit theorem states that if $Y_1,Y_2,\dots$ are independent and identically distributed with finite mean $\mu$ and variance $0<{\sigma }^2<\infty$, thenEquivalently, its cumulative distribution functions converge at every real argument to the standard normal cumulative distribution function.

For $w\ge 0$, $\mathbb{P}(W\le w)=\mathbb{P}(R\le \sqrt{w})=1-e^{-w/2}$. Differentiation givesThis is an exponential distribution of rate $1/2$, with mean two and variance four.

The independent variables $W_i$ have mean two and variance four by part (b). Their standardization isThe central limit theorem therefore proves the claimed convergence to the standard normal law, whose continuous cumulative distribution function is $(2\pi {)}^{-1/2}{\int }_{-\infty }^x{e}^{-u^2/2}du$ at every real $x$.

Let ${\overline{X}}_n=n^{-1}{\sum }_i{X}_i$ and write ${\sigma }^2=\mathrm{Var}(X_1)<\infty$. Independence gives $\mathbb{E}{\overline{X}}_n=\mu$ and $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. Apply part (b) to the nonnegative variable $({\overline{X}}_n-\mu {)}^2$, which is the Chebyshev inequality:Thus ${\overline{X}}_n\to \mu$ in probability, proving the finite-variance weak law of large numbers.

The full column rank makes $X^{\mathsf{T}}X$ invertible. The least-squares normal equations from minimizing squared residuals giveThe errors have covariance matrix ${\sigma }^2{I}_n$ by independence and equal variance. Applying the covariance transformation rule to this linear estimator yieldsThe parameter estimator is unbiased because $\mathbb{E}Y=X\theta$.

With independent Gaussian errors, maximizing the likelihood over $\alpha ,\beta$ minimizes the residual sum of squares, so the maximum-likelihood estimators are the same as least squares. The invariance property of maximum likelihood estimation givesThe two estimates have zero covariance by part (ii). ThereforeThis is the uncertainty in the estimated expected response. Estimating an additional noisy observation would add its own error variance, but the quantity requested is the mean response $E_t$.

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.

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

Assume independent survival times and positive rate $\lambda$. With $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$ on the observed nonnegative data. The score and negative expected second derivative are $n/\lambda -S$ and $n/{\lambda }^2$. For $S>0$, which holds almost surely under the model,The usual independent-sample likelihood asymptotics give $\sqrt{n}(\hat{\lambda }-\lambda )\Rightarrow N(0,{\lambda }^2)$. Thus an approximate Wald interval isFor a positive interval with the same first-order justification, apply the delta method to $\log \hat{\lambda }$, obtaining endpoints $\hat{\lambda }\exp (\pm z_{1-\alpha /2}/\sqrt{n})$.

A single individual survives until $T$ with probability $s=e^{-\lambda T}$. Thus $N_T\sim \mathrm{Bin}(n,s)$ and the new likelihood is $\left(\genfrac{}{}{0ex}{}{n}{n_T}\right)e^{-\lambda T{n}_T}(1-e^{-\lambda T}{)}^{n-n_T}$. For $0<n_T<n$, its maximum occurs at $s=n_T/n$, givingIf $n_T=0$, the supremum occurs as $\lambda \to \infty$ and there is no finite maximum. If $n_T=n$, it occurs at the boundary $\lambda =0$ of the extended parameter space, or is only a supremum when the rate is required to be strictly positive.

A binomial experiment has information $n/[s(1-s)]$ for $s$. Since $ds/d\lambda =-Ts$, parameter transformation givesAs quantified by Fisher information in an exponential survival snapshot, the ratio of this information to the complete-data information is $z^2/(e^z-1)$ with $z=\lambda T>0$, strictly less than one by the supplied inequality. Hence every fixed survival snapshot loses information. Under regular large-sample conditions its efficient rate estimator has larger asymptotic variance, and correspondingly wider uncertainty intervals, than the estimator based on all individual survival times. This comparison concerns information and asymptotic uncertainty, not a claim that every realized interval in one experiment is wider.

The variables $(m-1)S_X^2/{\sigma }^2$ and $(n-1)S_Y^2/{\sigma }^2$ are independent chi-squared variables with $m-1$ and $n-1$ degrees of freedom. Independence follows because the two whole samples are independent. Their sum is chi-squared with the sum of the degrees of freedom, soWithin each normal sample the sample mean is independent of its sample variance, and independence of the samples makes the pair of means jointly independent of the pair of variances. Hence this pooled chi-squared variable is independent of the standardized mean difference, which is the independence required for the Student pivot.

Independence makes a difference of the two normal sample means normal, with means subtracting and variances adding. Therefore

For a sample size $m\ge 2$ define $\overline{X}=m^{-1}{\sum }_i{X}_i$ and $S_X^2=(m-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The normal-sample theorem givesIt also states that the mean and sample variance are independent. Combining (i) and (ii), and the independence across samples, the correctly centred pivot isThe source reverses the mean difference inside the subtraction. Its literal statistic is generally noncentral when ${\mu }_X\ne {\mu }_Y$, rather than having a central Student distribution. Under the null hypothesis both differences are zero, so this misprint does not change the requested null test.

For a two-sided level-$\alpha$ test reject when $|T|>t_{m+n-2,1-\alpha /2}$. In the numerical example the pooled variance is $0.038$, the standard error is $\sqrt{0.038(2/5)}\approx 0.12329$, and $T\approx -2.190$. There are eight degrees of freedom and the supplied critical value is $2.31$. Thus do not reject equal mean cell volume at the 5% level. This does not establish equality of the means; it states that the observed difference does not cross this test's two-sided threshold.

Age matching introduces a shared block factor: the two observations in a pair can be associated, and their age-dependent means also undermine the simple unblocked independent-samples model. A paired t-test on the within-pair drug-minus-placebo differences accounts for that design, assuming independent pairs and the corresponding normal-difference model. Shared age variation cancels in those differences; positive within-pair association reduces their variance and can increase power, although the reduction in degrees of freedom means improvement is not automatic. The paired test cannot be computed from the two marginal means and variances alone: the within-pair differences or covariance are needed.

The total is the independent random sum $Z_t={\sum }_{j=1}^{N_t}{Y}_j$, with the empty sum zero. Conditional on $N_t=n$, its moment generating function is $M_Y(\theta {)}^n$. The Poisson probability generating function therefore givesAssuming a moment generating function in a neighborhood of zero, write $\mu =\mathbb{E}[Y]$ and $s^2=\mathrm{Var}(Y)$. Conditional expectation and variance yieldThe second contribution to the variance is due to randomness in the number of arriving coaches, not just randomness in the occupancy of each coach.

For fixed $\theta$, Taylor expansion at zero gives $e^h-1-h=h^2/2+O(h^3)$. With $h=\theta /\sqrt{\lambda t}$, the logarithm of the moment generating function tends to ${\theta }^2/2$. Thus $M_{U_t}(\theta )\to e^{{\theta }^2/2}$ on a neighborhood of zero. The continuity theorem for moment generating functions identifies the limit asThis is the centered Poisson normal limit with its exact variance scaling.

Since a Poisson count has mean and variance $\lambda t$, substitute the centering and scaling into its moment generating function:This is finite for every real $\theta$ when $t>0$.

Put $D^{\ast }=D-A$. By (i), $D^{\ast }X=0$. Thus $D^{\ast }{A}^T=D^{\ast }X(X^{T}X{)}^{-1}=0$, and its transpose vanishes too. Expanding $D{D}^T=(D^{\ast }+A)(D^{\ast }+A{)}^T$ yieldsThe covariance difference is positive semidefinite. In particular its $i$th diagonal entry is ${\sigma }^2{\sum }_j(D_{ij}^{\ast }{)}^2\ge 0$, proving the requested componentwise variance inequalities. Prefer $\boxed{{\hat{\theta }}_1}$ within the class of linear unbiased estimators, since it has the least variance and the same expectation. A tie is possible if the first row of $D^{\ast }$ is zero. This is the Gauss-Markov theorem conclusion; no normal-distribution assumption on the errors is required.

Under the null put $p=q=r$. The likelihood becomes the same two binomial constants times $r^{m+n}(1-r{)}^{2M+2N-m-n}$. Maximization gives the pooled estimateUnder an interior null, (ii) has zero mean and variance $r(1-r)(1/(2M)+1/(2N))$. Statistical consistency of $\hat{r}$ and Slutsky's theorem permit replacement of $r$ by $\hat{r}$ in the denominator, giving the stated $Z\Rightarrow N(0,1)$. For a two-sided level-$\alpha$ test, reject when $\boxed{|Z|>{\Phi }^{-1}(1-\alpha /2)}$. The large-sample approximation requires both expected counts of each type to grow; if all observed alleles are identical the stated denominator degenerates and this approximation is not applicable.

A linear combination of independent normal random variables is normal, with variances adding when the coefficients are $1,-1$. Applying the joint central limit approximation from (i) givesEquivalently the difference minus $p-q$, divided by the displayed standard deviation, converges in distribution to a standard normal under the same regular large-sample regime.

The normal-error log-likelihood, for fixed variance, is a constant minus $\Vert Y-X\theta {\Vert }^2/(2{\sigma }^2)$. Differentiating the squared residual gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Since $X$ has full column rank, $X^{T}X$ is positive definite and the squared residual is strictly convex in $\theta$. Hence the unique minimizer, and thus the maximum-likelihood estimator, isProfiling over the unknown positive variance does not change the residual-minimizing value of $\theta$.

Let $\mathrm{RSS}=\Vert Y-X\hat{\theta }{\Vert }^2$. The usual unbiased error-variance estimator isIndeed the residual is the orthogonal projection of $ϵ$ onto the $(n-p)$-dimensional complement of the design column space, giving $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$ and expected RSS equal to $(n-p){\sigma }^2$. The maximum-likelihood estimator instead divides RSS by $n$; the unbiased divisor above is the one used for the exact t-interval in part (d).