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Let $\lambda >0$ be nonintegral. This time $f(t)=O((t-1{)}^{\lambda })$ is integrable at $t=1$, and the endpoint factor behaves as $(t-1{)}^{\lambda +1}\to 0$. Suitable contours are thereforeandwhere the second path makes a fixed detour above or below the branch point $0$ and the branch of $t^{-\lambda -1}(t-1{)}^{\lambda }$ is continued along it. The exponential again controls the endpoint at $-\infty$ for $\mathrm{Re}z>0$.

As in part (i), the finite ${\gamma }_1$ integral is analytic near zero and hence is a constant multiple of the unique power-series solution $y_1$.

After division by $z$, the coefficient functions areBoth $zP(z)=1-z$ and $z^{2}Q(z)=\lambda z$ are analytic at zero, so the regular singular point criterion for a second-order equation shows that $z=0$ is regular singular. Substitution of $y=z^r$ into the leading terms gives the indicial equationThe exponent is repeated. Directly substituting $y={\sum }_{n\ge 0}{a}_n{z}^n$ givesThus $a_0$ determines the entire power series, so there is only one such solution $y_1$ up to scale. The logarithmic solution from a repeated Frobenius exponent predictsso its leading nonanalytic term is proportional to $\log z$.

For the contour ansatz, differentiation under the integral and one integration by parts giveThe integral vanishes whenwhose solution is the Laguerre contour-integral amplitudeThe contour and branches must make $f$ single-valued along the traversed path and must kill the endpoint term

Now suppose $\lambda <0$ is nonintegral. Near $t=0$, $f(t)=O(t^{-\lambda -1})$ is integrable and the endpoint factor is $O(t^{-\lambda })\to 0$. We may therefore choosea finite loop based at the branch point $0$ and avoiding $1$ except by encirclement. A second choice isalong one bank of the negative real axis, with a consistent branch. At $-\infty$, $e^{zt}$ kills the algebraic endpoint factor because $\mathrm{Re}z>0$.

The ${\gamma }_1$ integral is analytic in $z$ because its contour is finite, and it is a nonzero solution. By uniqueness of the analytic local solution, it is a constant multiple of $y_1$. This is the Finite Laguerre contour solution construction.

Here $p=4$ and $n-p=96$. The supplied quantile givesSincethe coefficient estimate after deleting this observation lies outside the $90\%$ confidence ellipsoid centered at the full-data estimate. It also greatly exceeds the common rough reference value $D_i=1$. The observation is therefore highly influential and merits investigation:

Let ${\hat{\beta }}_{(i)}$ be the least-squares estimate after deleting observation $i$, and put $\hat{Y}=X\hat{\beta }$ and ${\hat{Y}}_{(i)}=X{\hat{\beta }}_{(i)}$. Cook's distance isThus $D_i$ is the squared displacement caused by deleting observation $i$, measured in the same $X^{T}X$ metric and scale as the confidence ellipsoid. In particular, ${\hat{\beta }}_{(i)}$ lies outside the $(1-\alpha )$ ellipsoid centered at $\hat{\beta }$ exactly when

LetThe pivotal quadratic form has an $F_{p,n-p}$ distribution, so the $(1-\alpha )$ normal linear-model confidence ellipsoid isHere $F_{p,n-p}(1-\alpha )$ is the $(1-\alpha )$ quantile of the indicated $F$ distribution.

Because $I-H$ is an orthogonal projection of rank $n-p$,It follows that Mallows Cp satisfieswhere the last equality is part (i). Thus $C_p$ is an unbiased estimator of the independent-copy prediction error, as summarized by the unbiased prediction-error identity for ordinary least squares.

Writewhere $\epsilon$ and ${\epsilon }^{\ast }$ are independent, centered, and have covariance matrix ${\sigma }^2{I}_n$. ThenThe deterministic term is orthogonal in expectation to the two centered random terms, and the random terms are independent. Since the hat matrix is a symmetric idempotent projection of rank $p$,while $\mathbb{E}\Vert {\epsilon }^{\ast }{\Vert }^2=n{\sigma }^2$. ThereforeIn this bias-variance decomposition for linear prediction, $\Vert (I-H)\mu {\Vert }^2$ is squared model bias, $p{\sigma }^2$ is variance from fitting $p$ coefficients, and $n{\sigma }^2$ is irreducible noise in the future response. Enlarging the model tends to reduce the first term while increasing the fitted-model variance, which is the bias-variance tradeoff.

Membership is immediate from the supplied characterization of ${\Pi }_2$. Define the partial computable functionThenso $\mathbf{Tot}\in {\Pi }_2$.

Now let $P\in {\Pi }_2$. There is a partial computable $f$ such thatFor each $w$, define a unary partial functionOperationally, its program simulates $f(w,x)$ and returns $x$ if that computation halts. By the S-m-n theorem, a total computable map $h$ produces an index $h(w)$ for $g_w$. ThereforeThus every Pi-2 set many-one reduces to $\mathbf{Tot}$. Combined with membership, this proves that the totality problem is ${\Pi }_2$-complete.

First, $\mathbf{K}\in {\Sigma }_1$: the partial procedure on input $w$ simulates $f_{w,1}(w)$ and halts exactly when that computation halts.

For hardness, let $A\in {\Sigma }_1$. Choose a partial computable function $g$ withFor each fixed $w$, define a unary program $P_w$ as follows: on any input $x$, simulate $g(w)$; if that simulation halts, halt and output $x$, and otherwise run forever. The S-m-n theorem gives a total computable function $h$ that maps $w$ to a code for $P_w$. Its accepted language isIn particular,Thus $h$ is a many-one reduction $A{\le }_m\mathbf{K}$. Since $A$ was arbitrary, $\mathbf{K}$ is ${\Sigma }_1$-hard; together with membership this proves the many-one completeness of the halting problem.

For an effective enumeration $(W_w{)}_{w\in \mathbb{W}}$ of the computably enumerable languages, the diagonal halting problem isEquivalently, if $f_{w,1}$ is the unary partial function coded by $w$, thenThus $\mathbf{K}$ is the diagonal halting set.

A set $A\subseteq \mathbb{W}$ is a Sigma-1 set when it is the domain of a partial computable function: there is a partial computable $f$ such thatEquivalently, $w\in A$ iff $\exists tR(w,t)$ for some computable predicate $R$; one may take $R(w,t)$ to say that the truncated computation function observes a halt within $t$ steps.

A set $A$ is $\mathcal{C}$-complete under many-one reductions whenThe first condition is membership in the class; the second is $\mathcal{C}$-hardness. This is a many-one complete set.

The relation $A{\le }_{m}B$ means that there is a total computable function $g:\mathbb{W}\to \mathbb{W}$ such thatfor every word $w$. Such a $g$ is a many-one reduction.

Every row of the channel matrix is a permutation of $(1-2\alpha ,\alpha ,\alpha )$, and every column sum is one. It is therefore a weakly symmetric channel. The uniform input makes the output uniform, with entropy ${\log }_{2}3$, while the conditional output entropy is the entropy of one row:The weakly symmetric channel capacity formula now gives the ternary symmetric channel capacitybits per channel use, for $0\le \alpha \le 1/2$, with the convention $0{\log }_{2}0=0$.

Let $X\in \{0,1\}$ denote whether $00$ or $11$ was sent, and writewhere $E_1,E_2$ are independent Bernoulli variables of parameter $p$. Because $X$ is uniform, $Y_1$ is uniform, while $H(Y_1\mid X)=h_2(p)$. Thus the mutual information in the first received digit is

By the chain rule for mutual information, the extra information in the second digit is $I(X;Y_2\mid Y_1)$. Conditional on $X$, the second channel error is independent of the first output, soMoreover,Given either value of $Y_1$, the second output therefore differs from it with probability $2p(1-p)$, andConsequently the incremental information in a twofold binary repetition code isbits.

A binary symmetric channel independently flips each input bit with probability $p$, so, with rows indexed by the input and columns by the output, its channel matrix isIf $p>1/2$, complementing every received bit converts the channel into one with crossover probability $1-p<1/2$ without changing its information-carrying ability. The case $p=1/2$ has identical rows and capacity zero, so it suffices to study $p<1/2$.

The Shannon second coding theorem states that for every rate below the channel capacity there are arbitrarily long codes whose decoding error tends to zero, whereas no sequence of codes with rate above capacity can have vanishing error. It also identifiesFor the present channel,The output entropy is at most one bit, with equality for a uniform input. Hence the binary symmetric channel capacity isbits per channel use.

Suppose first that the stated unitary and environment states exist. Preservation of the inner product givesThe environment states are normalized, so the Cauchy-Schwarz inequality gives $|⟨e_0|e_1⟩|\le 1$. Therefore

Conversely, writeand assume $|a|\le |b|$. If $b\ne 0$, put $c=a/b$, so $|c|\le 1$, and chooseThen $⟨e_0|e_1⟩=c$ and the desired output vectors have inner product $bc=a$, exactly matching the input vectors. If $b=0$, the inequality forces $a=0$, and one may take $|e_0⟩=|e_1⟩=|0⟩$.

In either case the input pair and output pair have the same Gram matrix. The isometry taking one pair to the other extends to a unitary matrix, so the required $U$ exists. Hencewhich is the environment-assisted two-state pure-state transformation criterion.

Run the perfect discriminator on the supplied state and record whether the result is $i=0$ or $i=1$. For each $i$, choose a unitary $V_i$ satisfyingwhich is possible by extending $|{\varphi }_i⟩$ to an orthonormal basis. Prepare two fresh systems in $|0⟩|0⟩$ and, controlled by the classical outcome $i$, apply $V_i\otimes V_i$. The output isThe discriminator may destroy its input; the two freshly prepared outputs still implement cloning on this known pair. This is the construction that perfect discrimination implies cloning for a known state family.

Apply the hypothetical quantum cloning device repeatedly to obtain $N$ copies of the unknown state. If $c=|⟨\varphi |\psi ⟩|<1$, the two possibilities arewhose overlap has modulus $c^N$. For equal prior probabilities, the Helstrom-Holevo bound gives optimal success probabilityThus, for any $\epsilon >0$, sufficiently many clones permit success probability above $1-\epsilon$. Orthogonal states already have $c=0$ and can be distinguished perfectly. This is clone-assisted asymptotic state discrimination.

Here distinct states means distinct rays. Vectors differing only by a global phase represent the same quantum state and cannot be distinguished.

Let $u(i)$ be the expected number of visits to state $3$ before absorption, counting the present state when $i=3$. Its reward equations areThe last equation gives $u(3)=2+u(2)$, so the second nontrivial equation gives $u(2)=2+u(1)$. Hence $u(1)=1+u(1)/2$, andThus the expected occupation count is two; together with the previous parts, this is the four-state reflecting absorbing random walk calculation.