Given $M=m$, there are exactly $m$ independent opportunities to stay awake, each with probability $p$. HenceThe standard binomial distribution formulas are $\mathbb{E}[W\mid M]=pM$ and $\mathrm{var}(W\mid M)=Mp(1-p)$. The Poisson distribution formulas are $\mathbb{E}[M]=\lambda$, $\mathrm{var}(M)=\lambda$, and therefore $\mathbb{E}[M^2]={\lambda }^2+\lambda$. By the law of total expectation,Using the definition of covariance,Thus $W$ and $M$ are not independent, even though the separate awake and asleep counts produced by Poisson thinning are independent.

For each student, the probability of both falling asleep and dreaming about mathematics is $(1-p)q$. These indicators are independent across students, so the same Poisson thinning calculation gives, for the requested count $D$,Explicitly $G_D(s)=\exp (\lambda (1-p)q(s-1))$, the probability generating function of that Poisson distribution.

Let $I_i$ indicate that student $i$ stays awake. These indicators have independent Bernoulli distributions with parameter $p$, and $W={\sum }_{i=1}^M{I}_i$. The Poisson distribution has probability generating functionThe probability generating function of an independent random sum now givesRecognizing the Poisson distribution,This is Poisson thinning.

Using the two masses of the Bernoulli distribution,This probability generating function is a polynomial, so it exists for every finite $s$.

For $0\le s\le 1$, condition on $N=n$. Its independence from the $X_i$ preserves their independent common distribution, so part (ii) givesFor $n=0$, this is $1$, agreeing with $Y=0$. By the law of total expectation,This proves the probability generating function of an independent random sum. For nonnegative $s$ outside $[0,1]$, the same argument is valid whenever the right side is finite, by Tonelli's theorem for nonnegative sums. For signed or complex $s$, a sufficient absolute-convergence condition is $G_N(G_X(|s|))<\infty$; it holds automatically on $|s|\le 1$. Outside this domain, mere evaluation of an analytically extended expression is not a substitute for existence of the probability generating function.

Let $S_n={\sum }_{i=1}^n{X}_i$. By independence, the expectation of a product of independent random variables factors, and the identical distributions giveThis is valid for $|s|\le 1$ and wherever the required absolute expectations exist. It is the probability generating function of an independent sum.

The probability generating function of a nonnegative integer-valued random variable isusing $s^0=1$ also at $s=0$. It is always defined for $0\le s\le 1$, and indeed the power series converges absolutely for all real or complex $s$ with $|s|\le 1$, becauseIt may be defined beyond the unit disk where this power series converges absolutely; such a larger domain depends on the distribution. In particular $G_X(0)=\mathbb{P}(X=0)$ and $G_X(1)=1$.

Put $D=1-(1-p)(1-q)=p+q-pq$. By the law of total probability,For $0\le t<1$, termwise differentiation of the geometric series, valid inside its radius of convergence, givesTaking $t=(1-p)(1-q)$ therefore yieldsSince $D-p=(1-p)q$, this is the zero-based geometric distribution with parameterThis illustrates Bernoulli thinning of a geometric count. Here $0<p_{\ast }<1$, as $0<p,q<1$.

For the reverse conditional distribution, Bayes' theorem gives, for $n\ge k$,ThusFor $n<k$ the conditional probability is zero. The conditional distribution of $X-k$ is the failure-count form of a negative binomial distribution, with $k+1$ successes and success parameter $r$: the geometric-binomial posterior is shifted negative binomial.

To obtain the $(k+1)$st head on toss $n+1$, the first $n$ tosses must contain exactly $k$ heads, and the last toss must be a head. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ choices of the first $k$ head positions. By independence, each such arrangement followed by a head has probability $r^k(1-r{)}^{n-k}r$. The arrangements are disjoint, soThis is the negative binomial stopping argument, with $n+1$ counting all tosses up to and including the $(k+1)$st success. It is not merely the probability of $k+1$ heads somewhere among $n+1$ tosses: the last toss is prescribed to be a head.

Since the $B_i$ form a partition of a set of the sample space, the events $A\cap B_i$ are pairwise disjoint and their set union is $A$. By countable additivity of probability, followed by the definition of conditional probability,Every ratio is legitimate because each $\mathbb{P}(B_i)>0$. This proves the countable law of total probability.

For events $A,B$ with $\mathbb{P}(B)>0$, the conditional probability isThis renormalizes the probability of $A$ within $B$. The elementary ratio definition does not assign a value when $\mathbb{P}(B)=0$.

At $x=0$, $g^{\prime \prime }(0)=36>0$, so the second-derivative test gives a strict local minimum. In fact $g^{\prime }(x)=4x(x-3{)}^2$ is negative for $x<0$ and positive for $x>0$ except at $x=3$, so $g(0)=5$ is the global minimum as well.

At $x=3$, the second-derivative test is inconclusive because $g^{\prime \prime }(3)=0$. But $g^{\prime \prime }(x)=12(x-1)(x-3)$ changes from negative to positive across $3$, while $g^{\prime }(3)=0$. Hence this is a stationary inflection point. Equivalently, the leading nonconstant term $4(x-3{)}^3$ in the Taylor polynomial changes sign across $3$. The paper explicitly uses “saddle point” to mean an inflection point, so in its terminology the classification isHere the “saddle point” is a one-dimensional stationary inflection point, rather than the usual multivariable use of saddle point of a scalar function.^[ref]

The relevant derivatives at $x=3$ areThus the cubic Taylor polynomial isIn fact the exact expansion is $g(x)=32+4(x-3{)}^3+(x-3{)}^4$, so the omitted term has order $O((x-3{)}^4)$ in big O notation.

The derivative factors asThe polynomial is differentiable everywhere, so its critical points occur precisely where this derivative vanishes. HenceTheir values are $g(0)=5$ and $g(3)=32$.

At $x=1$, the left limit is $1$. The right limit is $-1$ because of the period-two extension. Therefore Dirichlet's theorem on Fourier series gives the Fourier series limitEvery Fourier partial sum is also exactly zero there because $\sin ((2j+1)\pi )=0$.

The left and right limits at zero are $-1$ and $1$. By Dirichlet's theorem on Fourier series, the Fourier series converges toIndeed every term $\sin ((2j+1)\pi x)$ is zero at $x=0$.

Use the period-two Fourier series conventionAway from its jumps this square wave is an odd function, so $a_0=a_n=0$. Its Fourier sine coefficients areOnly odd Fourier harmonics remain. ThereforeThe Fourier series equals $f(x)$ away from the jumps; at a jump it gives the midpoint of the one-sided limits by Dirichlet's theorem on Fourier series. The printed definition leaves the values at the integers unspecified. Choosing finite values there consistent with period two changes neither the Fourier coefficients nor these limiting values.

A function $f:\mathbb{R}\to \mathbb{R}$ is periodic with period $\alpha >0$ whenThe period need not be the smallest positive one; for example, every positive integer multiple of a period is also a period.

Put $a=\pi /2$. Substituting $x=-a$ and $x=a$ into part (ii), using $\cos (\pi /4)=\sin (\pi /4)=1/\sqrt{2}$, givesThus $A-B=\sqrt{2}(1+e^{-a})$ and $A+B=\sqrt{2}(1-e^a)$. Solving these two linear equations givesConsequently the solution of the boundary value problem isThe two independent boundary conditions determine $A,B$ uniquely.

For a particular solution, try $y_p=a\sin x+b\cos x$. Substitution givesMatching coefficients requires $a-8b=1$ and $8a+b=8$, so $a=1$ and $b=0$. Adding the homogeneous solution from part (i), the general solution is