Start with $(d[1],d[2],d[3],d[4])=(0,\infty ,\infty ,\infty )$. With in-place edge relaxation, the Bellman-Ford algorithm makes the following changes during its first lexicographic pass:The next pass makes no change; any remaining fixed-count passes also make none. Thus the final vector is $(0,4,2,3)$. These are sequential updates within a pass, rather than the simultaneous values of the Synchronous Bellman-Ford recurrence.

Part (a) gives $\mathbb{P}(X\in \mathbb{Z})=0$. Therefore the ceiling function and floor function satisfy $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely. Both are integrable, since $0\le \lfloor X\rfloor \le X$ and $\lceil X\rceil \le X+1$. Thus $\mathbb{E}[\lceil X\rceil ]=\zeta (t-1)+1$.

Now take expectations in $\lfloor X\rfloor \le X\le \lceil X\rceil$:Writing $s=t-1$, which ranges over every $s>1$, and rearranging yieldsThese are the zeta bounds from rounded Pareto variables: rounding below and above converts one elementary expected value into bounds on an infinite series.

The nonnegative integer-valued tail-sum formula for expectation givesThis is finite because $t-1>1$. Here $\zeta$ is the Riemann zeta function on its convergent real domain. The floor function is nonnegative on the support, so the tail-sum identity applies directly.

For $k\ge 1$, the event $\{\lfloor X\rfloor \ge k\}$ equals $\{X\ge k\}$. Integrating the Pareto distribution's density givesThe mass formula follows by subtracting consecutive tail probabilities, or by integrating over $[k,k+1)$. With the paper's convention $0\in \mathbb{N}$, include the separate endpoint values $\mathbb{P}(\lfloor X\rfloor \ge 0)=1$ and $\mathbb{P}(\lfloor X\rfloor =0)=0$, since $X\ge 1$ almost surely. Substituting $k=0$ into the power expression would be invalid.

The expected value is finite because $t>2$:The condition on $t$ is precisely the first moment threshold for this parameterization of the Pareto distribution.

Normalization of the probability density function requires $1=c{\int }_1^{\infty }{x}^{-t}dx=c/(t-1)$. Hence $c=t-1$. It is positive for the given $t>2$, and the density is nonnegative, so this does define a probability distribution. It is a Pareto distribution with scale $1$ and shape $t-1$.

The events $\{X=k\}$, for $k\in \mathbb{Z}$, are pairwise disjoint. Their union is $\{X\in \mathbb{Z}\}$, and the integers form a countable set. By countable additivity,Countability is the necessary justification: one cannot apply this argument to an arbitrary uncountable union of zero-probability singletons. This is a special case of countable sets have zero probability under a density.

An absolutely continuous probability distribution assigns zero probability to a singleton, because its density has zero integral over that set. Thus $\mathbb{P}(X=x)=0$ for every real $x$, including points where a chosen version of $f$ is nonzero.

The cumulative distribution function is $F_X(x)=\mathbb{P}(X\le x)={\int }_{-\infty }^{x}f(u)du$. This formula uses the assumed probability density function; the density itself need not be a continuous function.

Set $p_{\ast }=p/[p+(1-p)r]$. The preceding probability generating function becomes $p_{\ast }/[1-(1-p_{\ast })s]$, identifyingEquivalently, for $z\in \mathbb{N}$, $\mathbb{P}(Z=z)=p_{\ast }[(1-p_{\ast }){]}^z$. Since $0<p,r<1$, also $0<p_{\ast }<1$. This is the existing geometric count under Bernoulli thinning, with the zero-based convention specified.

The probability generating function of an independent random sum states that, for a nonnegative integer-valued $N$ independent of independent and identically distributed nonnegative integer-valued summands, $G_{{\sum }_{i=1}^N{X}_i}(s)=G_N(G_{X_1}(s))$ wherever the expectations converge. Here $G_{X_1}(s)=1-r+rs$, so, writing $q=1-p$,It is valid at least for $|s|\le 1$, and its power series in fact converges for $|s|<(p+qr)/(qr)$.

Take an independent and identically distributed sequence with $X_i\sim \mathrm{Bernoulli}(r)$, independent of $N$. Conditional on $N=n$, the random sum ${\sum }_{i=1}^n{X}_i$ has binomial distribution with parameters $n,r$, including the empty sum at $n=0$. This construction therefore has exactly the prescribed joint law of $(N,Z)$ and represents $Z$ as the required random sum.

For a literal representation that retains an already given pair $(N,Z)$, enlarge the probability space by auxiliary randomness. Conditional on $N=n,Z=z$, choose uniformly a size-$z$ subset of $\{1,\dots ,n\}$, put ones at those indices and zeros at the others, and append independent Bernoulli variables of parameter $r$. Given $N=n$, a specified length-$n$ pattern with $z$ ones has probability $\left(\genfrac{}{}{0ex}{}{n}{z}\right)r^z(1-r{)}^{n-z}/(\genfrac{}{}{0ex}{}{n}{z})=r^z(1-r{)}^{n-z}$. Thus the entire sequence is independent and identically distributed and independent of $N$, while its first $N$ terms sum to the original $Z$.

Differentiate the probability generating function inside its radius of convergence:The first moment and second factorial moment are $\mathbb{E}[N]=G_N^{\prime }(1)=q/p$ and $\mathbb{E}[N(N-1)]=G_N^{\prime \prime }(1)=2q^2/p^2$. Consequently

Let $q=1-p$. By the definition of a probability generating function and the geometric series,The series converges absolutely when $|qs|<1$, equivalently $|s|<1/(1-p)$.

In independent Bernoulli trials with success probability $p$, $N$ counts failures before the first success. The event $N=n$ requires $n$ failures followed by one success, giving $(1-p{)}^{n}p$. This is the zero-based geometric distribution, rather than the count of trials including the success.

Put $\beta =1-2/n$. The fixed point of this affine recurrence is $n/2$. Subtracting it gives $r_{k+1}-n/2=\beta (r_k-n/2)$. Repeated substitution, or mathematical induction, therefore yieldsFor $n>2$ this expression applies to every $k\ge 0$. For $n=2$, state the initial value $r_0=2$ separately and obtain $r_k=1$ for every $k\ge 1$, avoiding any need to interpret $0^0$. The expected red imbalance decays towards zero, although the urn composition itself remains random.

Apply the law of total expectation to the bounded random variable $R_{k+1}$. Linearity of expectation givesThis also holds for the first step from $r_0=n$.

Compute the conditional expectation from the possible increments:Therefore $\alpha =1$ and $\beta =1-2/n$. This calculation uses the conditional independence of the two selections, not independence of the successive counts $R_k$.

At state $r$, urn 1 has $r$ red and $n-r$ blue balls, while urn 2 has $n-r$ red and $r$ blue balls. Conditional on the state, the two choices are independent, each assigning probability $1/n$ to every ball. A red–blue exchange decreases $r$; a blue–red exchange increases it; equal colours leave it unchanged. Thus the conditional probability mass function isThe three numerators sum to $n^2$. At $r=0$ or $n$, the outward transition has probability zero. These are the transitions of the Bernoulli-Laplace diffusion model. The displayed formula is a transition kernel on every state; a literal conditional probability at a particular time requires the conditioning event to have positive probability.

The first exchange removes one red ball from urn 1 and replaces it with a blue ball. Hence $R_1=n-1$ with probability one.