Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / Section I

Substitution of $y_n={\lambda }^n$ into the linear recurrence relation gives ${\lambda }^{n}p(\lambda )=0$. Since $\lambda \ne 0$, division by ${\lambda }^n$ proves the sequence is a solution exactly when $p(\lambda )=0$.

For a nonzero triple root of a polynomial $\mu$, the repeated characteristic root of a linear recurrence rule gives the three independent solutions ${\mu }^n,n{\mu }^n,n^2{\mu }^n$. One way to see the polynomial factors is to write $y_n={\mu }^n{z}_n$: the recurrence becomes the third forward difference operator applied to $z_n$ equal to zero, so $z_n$ is a quadratic polynomial. ThusThese three constants give arbitrary initial values $y_0,y_1,y_2$, hence the full third-order solution. If $\mu =0$, the recurrence instead reduces to $y_{n+3}=0$: the first three values are arbitrary and every subsequent value is zero. The nonzero-root formula must not be used to discard those initial values.

Eliminate the free constant by comparing consecutive values:Therefore an appropriate inhomogeneous linear recurrence isIts homogeneous part has solution $a{2}^n$, and $-n$ is a particular solution. A first-order linear recurrence relation has one arbitrary initial value, so this gives exactly the desired general solution.

An equilibrium of an autonomous differential equation is a constant $y_{\ast }$ with $f(y_{\ast })=0$. Put $y=y_{\ast }+\eta$. For differentiable $f$, the linear stability analysis givesThus the linear perturbation is proportional to $e^{f^{\prime }(y_{\ast })x}$. More directly, continuity of $f^{\prime }$ makes $f(y)$ point toward $y_{\ast }$ on both sides when $f^{\prime }(y_{\ast })<0$, and away from it when $f^{\prime }(y_{\ast })>0$. HenceThe sign refers to evolution toward increasing $x$. If $f^{\prime }(y_{\ast })=0$, linear stability analysis is inconclusive: the leading nonzero nonlinear term or the signs of $f$ on either side are needed. For example ${\eta }^{\prime }=-{\eta }^3$ is attracting whereas ${\eta }^{\prime }={\eta }^3$ is repelling, despite the same zero linear coefficient.

Factor the right-hand side as $f(y)=y(y-2)(y+1)$. Its equilibrium points are $-1,0,2$, and $f^{\prime }(y)=3y^2-2y-2$ givesIndeed $f^{\prime }(-1)=3$, $f^{\prime }(0)=-2$ and $f^{\prime }(2)=6$. The phase line has downward motion for $y<-1$, upward motion for $-1<y<0$, downward motion for $0<y<2$, and upward motion for $y>2$. The uniqueness of a smooth autonomous differential equation prevents trajectories from crossing an equilibrium point. Consequently trajectories starting between $-1$ and $2$ approach zero as $x\to \infty$, while those outside that interval move away from the three equilibrium points. The constant trajectories and representative nonconstant trajectories are shown below.

For a nonnegative integer-valued random variable, the probability generating function isinitially for $|s|\le 1$, with $0^0=1$. For the Poisson distribution, insert its probability mass function and sum the exponential series:Thus

The moment-generating function is $M_Y(t)=\mathbb{E}[e^{tY}]$ for real $t$ where this expectation is finite; a finite neighborhood of zero is usually required when speaking of an existing moment-generating function. For a standard normal distribution, completing the square givesThe translated Gaussian density integrates to one, so

Construct independent copies $Y_1,Y_2,\dots$ of $Y$, also independent of $X$, and define the random sum $Z={\sum }_{j=1}^X{Y}_j$, taking the empty sum to be zero. Independence gives, conditional on $X=n$,The law of total expectation gives the transform composition for an independent random sumThe summands are nonnegative, so this equality is valid also as an extended-valued expectation. Where the right-hand side is finite, it is the ordinary moment-generating function of the constructed random sum. The construction does not assert a finite moment-generating function near zero for arbitrary $X,Y$: heavy tails can prevent this. For example a nonnegative $Y$ with no positive exponential moments and $X\equiv 1$ already shows the necessary qualification.

Pairwise independent events satisfy $\mathbb{P}(A_i\cap A_j)=\mathbb{P}(A_i)\mathbb{P}(A_j)$ for every distinct pair. Independent events satisfyfor every nonempty subset $I$ of the indices. For three events this includes the triple intersection, not just the three pairwise intersections.

Write $A_i=B_i\cup C$. Disjointness gives $\mathbb{P}(A_i)=p+q$, while every intersection of two or three distinct $A_i$ is exactly $C$. Thus pairwise independence is equivalent to $q=(p+q{)}^2$. Since $p+q\ge 0$,For $0\le q\le 1/16$, this is an admissible nonnegative $p$: putting $u=\sqrt{q}\in [0,1/4]$ gives $3p+q=3u-2u^2\le 5/8<1$. The four disjoint events can therefore have the stated probabilities, with the remaining probability assigned to their complement. This also covers $q=0$.

Full independence would additionally require $q=(p+q{)}^3$. Together with $q=(p+q{)}^2$, this forces $p+q$ to be zero or one. The first case gives $p=q=0$; the second gives $q=1,p=0$. Neither allows positive $p,q$. There are no such mutually independent events with $p>0$ and $q>0$.