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Write $p=(1+\mu {)}^{-1}$. Since $X$ and the geometric random variable $Z$ have the same expected value $\mu$,This is also the entropy deficit relative to the maximum entropy distribution on the nonnegative integers.

The random variables $X$ and $-Z$ are independent, so part a gives $H(X-Z)\ge H(-Z)=H(Z)$. Consequentlywhich is the required bound in terms of the Entropic Ruzsa distance.

Superpose the two independent Poisson processes. The combined process has rate $9$, and each arrival is independently a Number 1 bus with probability $1/9$ and a Number 8 bus with probability $8/9$. Therefore the number of Number 8 buses before the first Number 1 has the geometric distribution on $\{0,1,2,\dots \}$, andThis is number of one Poisson type before another with rates $1$ and $8$.

Using the mean $1/p$ of a geometric distribution and the decomposition in part (ii),

First suppose $L\in \mathbf{RP}\cap \mathbf{co}\text{-}\mathbf{RP}$. Run the RP algorithm for $L$ and the RP algorithm for its complement with fresh random bits. If the first accepts, output one; if the second accepts, output zero; otherwise repeat. Neither output can be wrong, and on every input the appropriate algorithm accepts in each round with probability at least $1/2$. The number of rounds is dominated by a geometric distribution of mean two, so this is an always-correct algorithm with polynomial expected running time.

Conversely, let $T$ be always correct when it halts and have expected running time at most $p(n)$. Run it for $2p(n)$ steps. Markov inequality givesAccept exactly when $T$ halts and outputs one; this is an RP algorithm for $L$. Accepting exactly when it halts and outputs zero is an RP algorithm for the complement. Thereforeis equivalent to zero-error expected polynomial time.

Each completed visit to radius $r_2$ begins a new radial excursion. By part (a), the conditional probability that the following excursion reaches radius $R$ before radius $r_1$ isThe Strong Markov property at the successive stopping times ${\tau }_{2k}$ makes these trials independent with the same success probability. Consequently $N(R)$ has a geometric distribution on $\{1,2,\dots \}$ with parameter $q_R$:As $R\to \infty$, $q_R\to 0$ andSlutsky theorem now gives

For $\mu >0$, let $Q$ be the geometric distributionIf $P$ is the law of $X$, Gibbs inequality givesThereforeFor $\mu =0$, the nonnegative random variable $X$ is zero almost surely and both sides vanish. This proves that the maximum entropy distribution on the nonnegative integers with fixed expected value is geometric.

Let $N$ have the positive-integer geometric distribution with success parameter $p\in (0,1)$, and conditionally toss $N$ independent coins with head probability $\theta \in (0,1)$. Exactly one head has conditional probability $n\theta (1-\theta {)}^{n-1}$. Combining this with the prior by Bayes theorem gives the displayed posterior with $r=(1-p)(1-\theta )$; its normalization uses ${\sum }_{n\ge 1}n{r}^{n-1}=(1-r{)}^{-2}$. For $p=\theta =1/2$, the posterior is $9n/4^{n+1}$. It is also a shifted negative binomial distribution, with $N-1$ equal in law to the number of failures before two successes of probability $1-r$.

For independent repeated time blocks, an event with exceedance probability $p$ has mean waiting time $1/p$ blocks by the geometric distribution. The return period is a long-run average waiting time, not a deterministic recurrence interval. For annual exceedance events, it is measured in years.

When the random intensity is $\alpha +Y$ with $Y$ having an exponential distribution of rate $\beta$, the mixed count is the sum of independent Poisson distribution and support-zero geometric distribution counts. The Poisson parameter is $\alpha$ and the geometric success probability is $\beta /(\beta +1)$. At $\alpha =0$ the law is geometric.