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For any $h$,If the normal equation $A^{T}A{c}^{\ast }=A^{T}y$ holds, the last two terms reduce to $\Vert Ah{\Vert }^2\ge 0$, so $c^{\ast }$ minimizes. Conversely, at a minimizer the directional derivative in every $h$ vanishes, forcing $A^T(A{c}^{\ast }-y)=0$.

On $|z|\ge \rho$, put $w=1/z$, so $|w|\le 1/\rho$. The Taylor polynomialsconverge uniformly there to $e^w$. MoreoverFor all sufficiently large $n$, the uniform error is below half this lower bound, so $|f_n(z)|>0$ whenever $|z|\ge \rho$. Every zero therefore lies in $|z|<\rho$.

The Laurent series isThus zero is an essential singularity.

In the positive-recurrent case, let $q=p_0+p_1<p$. Detailed balance for the length chain requiresso, with $r=q/p<1$,Normalization gives ${\pi }_0=1-r$. The self-loop at zero makes the length chain aperiodic, so the convergence theorem for irreducible positive-recurrent aperiodic Markov chains gives

PutThe length process $L_n=\ell (X_n)$ is a nearest-neighbour chain on the nonnegative integers: away from zero it moves up with probability $q$ and down with probability $p$, while at zero it moves up with probability $q$ and stays put with probability $p$.

Returns of the original chain to the root are exactly returns of $L_n$ to zero, so their recurrence classifications agree. The reflected biased random walk on the nonnegative integers is transient when $q>p$, null recurrent when $q=p$, and positive recurrent when $q<p$. Since $p=1-q$, the conditions are respectivelyIrreducibility transfers the classification from the root to every state.

From any vertex, following the parent edge repeatedly reaches the root with positive probability $p^k$. From the root, any prescribed binary string $(b_1,\dots ,b_k)$ can be reached by following its bits, with positive probability $p_{b_1}\cdots p_{b_k}$. Thus every state communicates with every other state and the chain is irreducible.

At the root there is a self-loop of probability $p>0$, so the root has period one. All states of an irreducible chain have the same period; hence the chain is aperiodic.

The Rao-Blackwell theorem suggests conditioning $S$ on $T$. Part (d) givesIt remains unbiased. Since $T\sim N(2\mu ,2)$ and a normal variable with mean $m$ and variance $v$ satisfies $\mathrm{var}(T^2)=2v^2+4m^{2}v$,This is strictly below $2+4{\mu }^2$ for every $\mu$. The calculation is recorded as the Rao-Blackwell estimator of a squared normal mean.

The pair $(X_1,T)$ is jointly normal, withThe conditional distribution of a bivariate normal variable is thereforeEquivalently, its conditional density is

Writing $X_1=\mu +Z$ with $Z\sim N(0,1)$ givesso $S$ is unbiased. Its mean square error is therefore its variance. Sincewe obtain

Expanding the exponent givesThe Fisher-Neyman factorization theorem therefore shows that $T=X_1+X_2$ is sufficient.

It is also minimal sufficient. For two sample points $x,y\in {\mathbb{R}}^2$,where $C$ is independent of $\mu$. This ratio is independent of $\mu$ exactly when $T(x)=T(y)$. The likelihood-ratio criterion for minimal sufficiency applies, proving the claim and the general normal sample sum with known variance result in this case.

Independence gives

Yes. The Euler recurrence isWrite $y_0=mh+r$, where $m=\lfloor y_0/h\rfloor$ and $0\le r<h$. For $n\le m$, the numerical values agree with the exact linear descent. If $r=0$, the method reaches zero and stays there. If $0<r<h$, thenand the recurrence thereafter alternates between $r$ and $r-h$. The exact solution is then zero, while both numerical values have magnitude at most $h$. Thus, uniformly for $0\le n\le N$,This is the explicit Euler method for the sign-decay equation.

The functionis continuous, satisfies the equation away from $\tilde{t}=y_0$, and has unequal one-sided derivatives there.

It is unique in the stated class. While a solution is positive its derivative is $-1$, so continuity forces it to follow $y_0-t$ until it first reaches zero at $t=y_0$. It cannot subsequently make a positive excursion: on any connected interval $(a,b)$ where $y>0$ and $y(a)=0$, integration of $y^{\prime }=-1$ gives $y(t)=-(t-a)<0$. A negative excursion similarly contradicts $y^{\prime }=1$. Hence the solution remains zero. This is the sign-decay differential equation.

The function is not Lipschitz on $\mathbb{R}$. For $\epsilon >0$,which is unbounded as $\epsilon \downarrow 0$.