Now complete the square in the other order:Divide the joint probability density function by the positive marginal densityThe factor $e^{-b^2/2}$ cancels, giving the conditional probability densitySincethis is a normalized normal distribution density. HenceConditioning on an exact observation is interpreted through this conditional probability density, rather than conditioning by a ratio of event probabilities.

The other marginal is $X_2\sim N({\mu }_2,{\sigma }_2^2)$ by the same square-completion argument with the indices exchanged. Its moment-generating function is therefore

The nonsingular bivariate normal distribution has ${\sigma }_1,{\sigma }_2>0$ and $|\rho |<1$. Put $a=(x_1-{\mu }_1)/{\sigma }_1$ and $b=(x_2-{\mu }_2)/{\sigma }_2$. Completing the square givesConsequently the joint probability density function factors asFor fixed $x_1$, the second factor is a normalized normal distribution density in $x_2$, with mean ${\mu }_2+\rho {\sigma }_{2}a$ and standard deviation ${\sigma }_2\sqrt{1-{\rho }^2}$. Its integral is one; explicitly substitute $w=(b-\rho a)/\sqrt{1-{\rho }^2}$ in the Gaussian integral. Thus the marginal density is

Summing the geometric series for the offspring probability generating function givesThe fixed-point equation $s=1/(2-s)$ is $(s-1{)}^2=0$, whose only root in $[0,1]$ is one. The Galton-Watson extinction fixed point result therefore proves extinction is certain.

For the explicit iterates, the proposed formula at $n=1$ is $1/(2-s)$, so the base case holds. If it holds at $n$, part (a) givesThis is exactly the same formula with $n$ replaced by $n+1$. By induction,Evaluating the probability generating function at zero selects the mass of an empty generation:These probabilities increase to one, also confirming the extinction conclusion directly.

The eventual extinction probability is the smallest solution $\boxed{q\in [0,1]\text{ of }{F}_1(q)=q}$. This is the Galton-Watson extinction fixed point result for one initial individual. Equivalently,since having an empty generation is an absorbing event and the extinction-by-generation events increase. In particular extinction is certain when the smallest fixed point is one. With $p_0>0$, an offspring mean at most one implies this, whereas a mean greater than one gives a fixed point below one. The fixed-point statement itself requires no finite-mean assumption.

Conditional on $X_n=m$, the next generation is the sum of $m$ independent offspring counts, each with probability generating function $F_1(s)$. HenceThis also holds for $m=0$, when the empty sum is zero and the empty product is one. The law of total expectation givesFor complex $|s|\le 1$, the expectations are absolutely bounded because $|F_1(s)|\le 1$. This proves the branching-process generating-function iteration, with $F_0(s)=s$.

For the waiting time for a new record after a fixed sample, given $U=F(Y_n)$, no overflow in the first $k$ subsequent years has probability $U^k$. Averaging as in part (c) yieldsThe tail-sum formula for expectation therefore givesThe answer is $\boxed{E\tau =\infty }$. This does not contradict almost-sure finiteness: the tail probability tends to zero, but too slowly for its sum to converge.

The continuity of the cumulative distribution function gives the probability integral transform: $F(X_i)$ is uniform on $[0,1]$. Strict monotonicity or a probability density for $F$ is not required. Since $F$ is nondecreasing,By independence, $P(U\le u)=u^n$ for $0\le u\le 1$, so $U$ has probability density function $n{u}^{n-1}$. Average the conditional probability in part (b) using this distribution:Thus $\boxed{P(\tau =k)=n/[(k+n-1)(k+n)]}$. The expression telescopes to total mass one, so the waiting time is finite almost surely.

The future levels are independent of the observed maximum. Given $Y_n=z$, the first $k-1$ future levels must be at most $z$, and the $k$th must exceed $z$. Each non-overflow has probability $F(z)$ and an overflow has probability $1-F(z)$. ThereforeThis is a geometric distribution of parameter $1-F(z)$ when $F(z)<1$. Since $Y_n$ has a continuous distribution, conditioning on an exact value is understood via a regular conditional distribution, rather than division by $P(Y_n=z)=0$. The formula specifies its usual version. If $F(z)=1$ there is no possible overflow at that threshold, but such thresholds have probability zero for the random observed maximum.

The event $Y_n\le z$ occurs precisely when every one of the first $n$ levels is at most $z$. Their independence gives the cumulative distribution functionFor any nonnegative random variable, the pointwise identity $Y_n={\int }_0^{\infty }{1}_{\{Y_n>z\}}dz$ and the Tonelli theorem allow interchange of expectation and integral. HenceThis identity holds as an extended nonnegative integral; finiteness of the mean is not guaranteed by continuity of $F$.

For $s=\pm 1$, the identity $s/(2+s)=-1/3+(2/3)s$ givesThe alternating series test applies to $\sum (-1{)}^n/n$ because $1/n$ decreases to zero; its partial sums consequently have a finite limit. The harmonic series $\sum 1/n$ diverges to $+\infty$. Applying the displayed identity to each finite partial sum yieldsTherefore $\boxed{\text{the series diverges to }-\infty }$. Alternating signs alone do not satisfy the alternating series test: the magnitudes jump from $1/(3n)$ at even $n$ to $1/(n+1)$ at the next odd index.

The alternating series test states that if $b_n\ge 0$ decreases to zero, then $\sum (-1{)}^n{b}_n$ converges. Indeed, its even and odd partial sums are monotone in opposite directions and bound each other, while their difference tends to zero. Here $b_n=1/(1+\sqrt{n})$ decreases to zero, so the series converges.

It fails absolute convergence, since for $n\ge 1$,The p-series $\sum n^{-1/2}$ diverges. This also follows by comparison with the divergent harmonic series, because $n^{-1/2}\ge n^{-1}$. The comparison test for series therefore shows that $\sum b_n$ diverges. Thus

The limit comparison test says that if $u_n,v_n>0$ and $u_n/v_n\to L$ with $0<L<\infty$, then their series either both converge or both diverge. This follows because, eventually, $(L/2)v_n\le u_n\le (3L/2)v_n$, and the comparison test for series applies in both directions.

Here comparison with the harmonic series givesFor completeness, the natural logarithm grows slowly enough because for $x\ge 1$,so $(\log n{)}^2/n\le 16/\sqrt{n}\to 0$. The harmonic series diverges: every block from $n=2^k$ to $2^{k+1}-1$ has $2^k$ terms at least $2^{-k-1}$ and therefore contributes at least $1/2$. Hence

The ratio test says that a series $\sum u_n$ whose terms are eventually nonzero converges absolutely if $\mathrm{limsup}|u_{n+1}/u_n|<1$; in particular, it suffices to have $|u_{n+1}/u_n|\le q<1$ eventually, since then its terms are bounded by a convergent geometric series. The comparison test for series states that, for nonnegative terms, comparison with a larger convergent series gives convergence, and comparison with a smaller divergent series gives divergence. For $u_n=n!/n^n$,By the binomial theorem, $(1+1/n{)}^n\ge 1+n(1/n)=2$. Thus $u_{n+1}/u_n\le 1/2$ for every $n\ge 1$, and induction gives $0<u_n\le 2^{1-n}$. The comparison test for series with $\sum 2^{1-n}=2$ proves

An orthogonal matrix satisfies $R^{T}R=I$, so it preserves the Euclidean norm. For a nonzero real eigenvector $v$,Therefore $\boxed{\lambda =\pm 1}$. Geometrically, eigenvalue $1$ fixes every vector on the linear span of $v$; eigenvalue $-1$ reverses every vector on that line. In either case the line is preserved as a set.

For $M$, solving $(M-I)v=0$ gives $y=x$, $z=y$, $x=z$. For $N$, the equations are $-2y-2z=0$ and $-2z=0$, hence $y=z=0$. For $P$, let $a=(1,1,1{)}^T$ and $J=a{a}^T$, so that $P=I-\frac{2}{3}J$. Then $(P-I)v=0$ is equivalent to $x+y+z=0$. The resulting eigenspaces and their bases areSolving all the equations shows that these lists contain the maximum possible numbers of independent eigenvectors with eigenvalue $1$.

The columns of $M$ form an orthonormal basis, so $M$ is an orthogonal matrix. Also $\det M=1$. Its fixed axis is the line through $a$, and on the perpendicular plane it acts as a two-dimensional rotation. If $\theta$ denotes the rotation angle, its matrix trace is $1+2\cos \theta$. Since $\mathrm{tr}M=0$, we obtain $\cos \theta =-1/2$. To fix the orientation relative to $u=a/\sqrt{3}$, note that $M{e}_1=e_3$ andThe component of $e_1$ perpendicular to $u$ has squared length $2/3$, and this scalar triple product equals $(2/3)\sin \theta$. Hence $\sin \theta =-\sqrt{3}/2$. ThusEquivalently, its unsigned angle is $2\pi /3$, or its positive angle is $2\pi /3$ about the opposite oriented axis.

For $P$, the reflection in a hyperplane formula givesIt fixes the plane $a\cdot v=0$ pointwise and maps $a$ to $-a$, so $\boxed{P\text{ reflects in }x+y+z=0}$. These actions also show that $P$ is an orthogonal matrix. In contrast, the second column of $N$ has squared length $5$, so $N$ is not orthogonal and cannot represent either of these Euclidean motions.

A diagonalizable matrix must have a basis of eigenvectors over the field in question. For $M$, a direct determinant computation gives ${\chi }_M(t)=1-t^3$. Its eigenvalues are $1,\omega ,{\omega }^2$, where $\omega =e^{2\pi i/3}$. Only one is real, with one-dimensional eigenspace, so $M$ is not diagonalizable over $\mathbb{R}$. Over $\mathbb{C}$ the three eigenvalues are distinct, and the vectors $(1,\lambda ,{\lambda }^2{)}^T$ for $\lambda =1,\omega ,{\omega }^2$ give an eigenbasis. The upper triangular matrix $N$ has only eigenvalue $1$, of algebraic multiplicity $3$, but its eigenspace has dimension $1$ over either field. Hence $N$ is not diagonalizable over either field. For $P$, the two real plane vectors displayed above, together with $a$, form an eigenbasis, with eigenvalues $1,1,-1$. They remain independent over $\mathbb{C}$. Therefore

For $t\ne 0$, factor the matrix asThe determinant of a scalar multiple of an $n\times n$ matrix gains the scalar's $n$th power. ThusThe right side extends to a polynomial at $t=0$: multiplying ${\chi }_A(t^{-1})$ by $t^n$ removes all negative powers. Its value there is $(\det A{)}^{-1}$, since the leading coefficient of ${\chi }_A(s)=\det (A-sI)$ is $(-1{)}^n$. This equals ${\chi }_{A^{-1}}(0)$, so the identity holds at zero with that interpretation.

For the relation between eigenvalues and the characteristic polynomial,The first equivalence is the definition of an eigenvector; the others use the kernel of a linear map and the criterion for an invertible matrix to have nonzero determinant. Since ${\chi }_A$ has degree $n$, the fundamental theorem of algebra gives $n$ roots over $\mathbb{C}$ counted with multiplicity. The multiplicity of a root is its algebraic multiplicity, and the eigenspace is exactly $\ker (A-\lambda I)$; its dimension is the geometric multiplicity, which need not equal the algebraic multiplicity.

Finally, factor $p(s)=s^4-\mu$ over $\mathbb{C}$ as $p(s)={\prod }_{j=1}^4(s-{\lambda }_j)$, including repeated roots when $\mu =0$. Substituting $A$ into this polynomial is valid because the resulting factors commute. If $\mu$ is an eigenvalue of $A^4$, thenAt least one factor must be zero, so its ${\lambda }_j$ is an eigenvalue of $A$ and satisfies ${\lambda }_j^4=\mu$. Thus the reverse direction of this instance of the spectral mapping theorem isThis argument also covers $\mu =0$, when the conclusion is that $A$ is singular and has eigenvalue zero.

The scalar matrix $tI$ commutes with $A$, soTaking determinants and using their multiplicativity gives, with the specified characteristic polynomial convention,

If $\lambda =0$, then $Av=0$ for a nonzero eigenvector $v$, contradicting the existence of a matrix inverse. Hence $\lambda \ne 0$. Multiplying $Av=\lambda v$ by $A^{-1}$ gives $v=\lambda A^{-1}v$, so

By the definition of an eigenvalue, there is a nonzero eigenvector $v$ with $Av=\lambda v$. Induction using matrix multiplication givesstarting from $r=1$. Thus the same nonzero eigenvector establishes

Writing $X=\left(\begin{array}{cc}x& y\\ z& w\end{array}\right)$ and using the Frobenius inner product, the three linear equations areThe first two differ by twice the third; equivalently $A-B=2C$. Thus there are only two independent constraints. Substituting $z=-w$ gives $y=2w-x$, so every element of the vector subspace has the unique expressionThe two coefficient matrices satisfy all three constraints. Their linear independence follows by looking first at entry $(1,1)$ and then at entry $(2,2)$. Therefore a basis for this orthogonal complement is