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Write the Taylor series of the entire function asThe Cauchy estimate on the circle $|z|=R$ givesIf $k>n/2$, letting $R\to \infty$ gives $c_k=0$. Since $n$ is odd,

For the second question, suppose such an $f$ existed. It never vanishes, so $g=1/f$ is analytic on $\mathbb{C}\setminus \{0\}$ andThe Riemann removable singularity theorem extends $g$ analytically across zero with $g(0)=0$. Applying the result just proved with $n=1$, $a=1$, and $b=0$ makes $g$ a polynomial of degree at most zero. It must then be identically zero, contradicting $g=1/f$ away from zero. Therefore no such function exists.

The softplus functionis convex becausePart (b) therefore shows that every $\beta \mapsto h(a_i^T\beta )$ is convex. The absolute value function is convex, so $\beta \mapsto |{\beta }_j|$ is convex for every coordinate $j$. Finally, part (a) says that a finite sum of convex functions is convex. Thereforeis convex. The second sum is the L1 norm regularizer.

Because $x\mapsto c^{T}x$ is a linear map,Hence composition of a convex function with an affine map is convex, and $g$ is convex.

For $0\le t\le 1$ and $x,y\in {\mathbb{R}}^d$, convexity givesTaking the maximum over $i$ and then bounding each term by the corresponding endpoint maxima yieldsThus the finite pointwise maximum of convex functions is convex. Summing the original inequalities over $i$ proves that ${\sum }_i{f}_i$ is convex as well.

The estimator $U=X_1{X}_2$ is unbiased for $p^2$ because the independent random variables $X_1,X_2$ satisfy $\mathbb{E}[X_1{X}_2]=p^2$. Given $T=t$, all placements of the $t$ successes are equally likely, soThus the Rao-Blackwellized estimator isIt is unbiased by the tower property. Since $n\ge 3$ and $p\in (0,1)$, the event $T=2$ has positive probability, and conditionally on it $X_1{X}_2$ takes both zero and one with positive probability. Hence $\mathbb{E}[\mathrm{Var}(X_1{X}_2\mid T)]>0$, so the new estimator has strictly smaller variance.

The Rao-Blackwell theorem states that if $T$ is sufficient and $U$ is an estimator with finite variance, thenhas the same expectation as $U$ and no larger variance; it preserves unbiasedness. The tower property of conditional expectation givesThe law of total variance givesSufficiency ensures that $U^{\ast }$ is a statistic whose definition does not depend on the unknown parameter. The inequality is strict exactly when the conditional variance is positive with positive probability.

A sufficient statistic $T$ is one for which the conditional distribution of the full sample $(X_1,\dots ,X_n)$ given $T$ does not depend on $p$. TakeFor a binary sample $x$ with ${\sum }_i{x}_i=t$, its likelihood function iswhich depends on the data only through $t$. By the Fisher-Neyman factorization theorem, $T$ is sufficient. Equivalently, conditionally on $T=t$, the sample is uniform over the $\left(\genfrac{}{}{0ex}{}{n}{t}\right)$ binary vectors containing $t$ ones, independently of $p$.

By part (a) and repeated use of the Strong Markov property, the probability of descending from $(2,k)$ to $(2,0)$ isStarting from square 1 with £$m$, heads moves to $(2,m)$, whereas tails lands on square 3, loses £1, and moves to $(2,m-1)$. Therefore the loss probability is

A random time $T$ is a stopping time when the event $\{T\le n\}$ is determined by $X_0,\dots ,X_n$. The Strong Markov property says that, conditionally on $T<\infty$ and $X_T=x$, the process $(X_{T+k}{)}_{k\ge 0}$ is a fresh Markov chain started at $x$, independent of the history before $T$.

Use the state space $\{1,2,4\}\times {\mathbb{Z}}_{\ge 0}$, with $(2,0)$ absorbing. Observe the chain only when it is at square 2. From $(2,k)$, the change $Y$ in wealth by the next return to square 2 has distributionIndeed, heads lands on square 3 and returns to square 2 after losing £1. After tails reaches square 4, the remaining two or three moves give the other cases.

For $r=2/3$,Thus $r^{M_j}$ is a martingale for the embedded wealth random walk $M_j$. Stopping when it first reaches $m-1$ or a large upper level and then letting that level tend to infinity givesThe upper-bound contribution vanishes because $0<r<1$; equivalently, this is the smaller probability solution of the first-step equation.

The normalized density isFor $\theta =1$ against $\theta =0$, the sample likelihood ratio isIt is strictly decreasing in $S$. Under $H_0$, parts (a) and (b) giveIf $q_{\alpha }$ is the lower $\alpha$-quantile of this gamma distribution, the Neyman-Pearson lemma gives the most powerful size-$\alpha$ critical region

For any fixed $\theta >0$,is again strictly decreasing in the same statistic $S$. Thus the same critical region is most powerful against every $\theta >0$ and is consequently a uniformly most powerful test of $H_0:\theta =0$ against $H_1:\theta >0$.

The Neyman-Pearson lemma states that, for testing one simple hypothesis with density $f_0$ against another with density $f_1$, a size-$\alpha$ test that rejects for the largest values of the likelihood ratio $f_1/f_0$ is most powerful among all tests of size at most $\alpha$, with boundary randomization if needed.

For $y\ge 0$ and $X\sim U(0,1)$,The distribution function is zero for $y<0$, so this is exactly the distribution function of $\mathrm{Exp}(1)$. Therefore

The moment-generating function of $W_i\sim \mathrm{Exp}(1)$ isIndependence makes the moment-generating function of the sum equal the product:This is the moment-generating function of the gamma distribution $\Gamma (n,1)$, and moment-generating functions determine distributions in a neighbourhood of zero. Hence

HereFor $0<h\le 1$, the Lipschitz bound on $f$ givesso part (ii) applies with a constant independent of sufficiently small $h$.

The exact solution has the Taylor expansion>Meanwhile,so one numerical step from $y$ isThe local error is therefore $O(h^3)$. Taking $p=2$ in part (ii) proves the required second-order global-error bound. This method is a two-stage Runge-Kutta method.