2. The entropy of ULC distribution

Suppose that $\mathscr{F}$ is the set that contains all pmfs for nonnegative integer-valued random variables. Consider two operators S_p and T_n defined on $\mathscr{F}$ [Reference Yu6]:

1. (1) For all $p\in (0,1)$, S_p maps pmf $f=\{\,f_i, i\in\mathbb{N}\}$ to another pmf $g=\{g_i, i\in\mathbb{N}\}$, where

   \begin{equation*} g_i=p f_{i-1} + (1-p) f_i, \quad i\ge 0, \end{equation*}

   and define $f_i=0$ for all i < 0. Suppose that f is the pmf of X, $Z\sim B(1,p)$, independent of X, then $S_p f$ is the pmf of X + Z.

2. (2) For all n > 1, T_n maps pmf $g=\{g_i, i=0,\ldots, n\}$ to another pmf $\,f=\{f_i, i=0, \ldots, n-1\}$, where

   (2.1)\begin{equation} f_i= \frac {n-i}{n} g_i + \frac {i+1}{n} g_{i+1},\quad i=0,\ldots, n-1. \end{equation}

   Denote g as the pmf of Y. Given Y, consider hypergeometric distribution $(n, Y, n-1)$, that is, suppose there are n balls in an empty box, in which the number of white balls is Y. Now take n − 1 balls out of the box randomly without putting them back, and define random variable X as the number of white balls in the taken balls. Then $T_n g$ is the pmf of X.

Operators S_p and T_n satisfies [Reference Yu6]:

1. (1) If $b_{n,p}=\{b(i,n,p), i=0,\ldots, n\}$ is the pmf of $B(n,p)$, then

   \begin{equation*} S_p b_{n,p}=b_{n+1,p},\quad T_{n+1} b_{n+1,p}=b_{n,p},\quad n\ge 1. \end{equation*}

   So, $T_{n+1}\circ S_p b_{n,p} =b_{n,p}$.

2. (2) ${\rm ULC}(n)$ contains all the pmfs that are Ultra-log-concave of order n. Moreover, we have $S_p \,f\in {\rm ULC}(n+1)$ and $T_n\, f\in {\rm ULC}(n-1)$ for all $f\in {\rm ULC}(n)$.

[Reference Yu6] proved that if X has pmf $f\in {\rm ULC}(n)$ with $\mathbb{E} [X] =np$, $p\in (0,1)$, and $Z\sim B(1,p)$, independent of X, and if Y is a hypergeometric distribution with parameters $(n+1, X+Z, n)$, then $H(X)\le H(Y)$. The proof uses the properties of operators S_p and T_n. By constructing a Markov chain whose limiting distribution is $B(n, p)$ and showing that the entropy never decreases along the iterations of this Markov chain. Here, we give another method to prove the non-decreasing property of the corresponding entropy based on Theorem 1.1.

Proof. Define differential operation $\Delta h_i=h_i-h_{i-1}$ for any sequence $\{h_i\}$, and let $g=\{g_i, i=0,\ldots, n\}$ denote the pmf of Y. By (2.1), we have

(2.2)\begin{equation} g_i =\frac {n+1-i}{n+1} (\,p f_{i-1} + q \,f_i) + \frac {i+1}{n+1} (\,p f_i + q f_{i+1}),\quad q=1-p. \end{equation}

Hence,

\begin{equation*} f_i-g_i =\frac {1}{n+1} \Delta \big [ p(n-i) f_i -q(i+1) f_{i+1}\big ] =\Delta (u_i h_i), \end{equation*}

where

\begin{equation*} u_i=p-q \frac {(i+1) f_{i+1}}{(n-i) f_i},\quad h_i= \frac {(n-i)f_i}{n+1},\quad 0\le i \lt n, \end{equation*}

and $u_i h_i=0$, $i=-1, n$. Therefore, for any convex function sequence $\psi=\{\psi_i\}$, we have

\begin{equation*} \mathbb{E} [\psi(Y)]-\mathbb{E} [\psi(X)] = \sum^n_{i=0} \psi_i (g_i-f_i)= - \sum^n_{i-0}\psi_i \Delta (u_i h_i)=\sum^{n-1}_{i=0} (\psi_{i+1}-\psi_i) u_i h_i. \end{equation*}

Since $h_i\ge 0$ and $\sum^n_{i=0} h_i=nq/(n+1)$, h can be modified to a probability function. On the other hand, $f\in {\rm ULC}(n)$ means that u_i is increasing in $i\in \{0, \ldots, n-1\}$; the convexity of ψ means that $\psi_{i+1}-\psi_i$ is increasing in i, and $\sum^{n-1}_{i=0} u_i h_i=0$. Hence, by Chebyshev rearrangement theorem, we have $\sum^{n-1}_{i=0} (\psi_{i+1}-\psi_i) u_i h_i\ge 0$, that is, $\mathbb{E} [\psi(Y)]\ge \mathbb{E} [\psi(X)]$.

Under the assumptions of Proposition 2.1, $Y\in {\rm ULC}(n)$, and the corresponding pmf is log-concave. Hence, by Theorem 1.1, $X\le_{\rm cx} Y$ leads to $H(X)\le H(Y)$.

3. Log-concave ordering between the original distribution and its corresponding compound distribution

Suppose that S is a nonnegative integer-valued random variable, defined by:

\begin{equation*} S=\sum^M_{i=1} X_i, \end{equation*}

where $\{X_n, n\ge 1\}$ is a sequence of independent and identically distributed (iid) nonnegative integer-valued random variables, and M is a counting random variable independent of all X_i’s. Let f and h be the pmfs of X_i and M, respectively. The distribution of S is called a compound distribution with its pmf denoted by $c_h\ (\,f)$. If $M\sim{\rm Poi}(\lambda)$, $B(n,p)$ or ${\rm NB}(\alpha, p)$, then the distribution of S is called compound Poisson distribution, compound Binomial distribution or compound Negative Binomial distribution. For α = 1, ${\rm NB}(1,p)$ reduces to the geometric distribution ${\rm Geo}(p)$.

In the following, the pmfs of Poi$(\lambda)$, ${\rm B}(n,p)$, ${\rm NB}(\alpha, p)$ and ${\rm Geo}(p)$ distributions are denoted by poi$(\lambda)$, ${\rm b}(n,p)$, ${\rm nb}(\alpha, p)$ and ${\rm geo}(p)$, respectively. Similarly, the pmfs of the corresponding compound distributions are denoted by $c_{{\rm poi}(\lambda)}\ (\,f)$, $c_{{\rm b}(n,p)}\ (\,f)$, $c_{{\rm nb}(\alpha,p)}\ (\,f)$ and $c_{{\rm geo}(p)}\ (\,f)$. Denote by µ_f the mean of a distribution with pmf f. Then the means of the distributions with pmf $c_{{\rm poi}(\lambda)}\ (\,f)$, $c_{{\rm b}(n,p)}\ (\,f)$, $c_{{\rm nb}(\alpha,p)}\ (\,f)$ are $\lambda\mu_f$, $np\mu_f$ and $\alpha (1-p)\mu_f/p$, respectively.

By Theorem 1.1 and Proposition 3.1, we have

(3.1)\begin{equation} \lambda^\ast = \lambda \mu_f \Longrightarrow H({\rm poi}(\lambda^\ast))\leq H(c_{{\rm poi}(\lambda)}(\,f)), \end{equation}

(3.2)\begin{equation} p^\ast = p \mu_f \Longrightarrow H({\rm b}(n,p^\ast))\leq H(c_{{\rm b}(n,p)}(\,f)). \end{equation}

Here, $\lambda^\ast=\lambda\mu_f$ ensures that two pmfs ${\rm poi}(\lambda^\ast)$ and $c_{{\rm poi}(\lambda)}(\,f)$ have the same mean. Similarly, $p^\ast=p\mu_f$ ensures that two pmfs ${\rm b}(n, p^\ast)$ and $c_{{\rm b}(n, p)}(\,f)$ have the same mean. Suppose that $M\sim {\rm B}(n,p)$, $M^\ast\sim {\rm B}(n,p^\ast)$, and $\{X_i, i\geq 1\}$ are iid random variables with a common pmf f. Assume that all random variables considered here and below are independent of each other. The explanation of (3.2) is as follows, and the explanation of (3.1) can be given similarly. We consider two following cases:

1. (i) Suppose $\mu_f\ge 1$. Then $p^\ast\ge p$ since $p^\ast=p\mu_f$ in (3.2). Let $\{I_i, i\geq 1\}$ be a sequence of iid Bernoulli random variables with $\mathbb{P}(I_i=1)= p/p^\ast\in(0,1]$. Since $\sum^{M^\ast}_{i=1} I_i$ has the same distribution as M, it follows that the pmf of $\sum_{i=1}^{M^\ast}I_iX_i$ is $c_{{\rm b}(n,p^\ast)}(\tilde{\,f})=c_{{\rm b}(n,p)}(\,f)$, where $\tilde{f}$ is the pmf of $I_iX_i$, given by:

   \begin{equation*} \tilde{f}=\frac{p}{p^\ast}f + \left (1-\frac{p}{p^\ast}\right )\delta_0, \end{equation*}

   where δ ₀ is the pmf of a degenerate random variable Z = 0. Notice that $\mu_{\tilde{f}}=p\mu_f/p^\ast=1$, and the uncertainty of $\sum_{i=1}^{M^\ast}I_iX_i$ is obviously stronger than that of $M^\ast$. Thus, (3.2) holds.

2. (ii) Suppose $\mu_f\in(0,1)$. In this case, $p^\ast \lt p$. Let $\{I_i, i\geq 1\}$ be a sequence of iid Bernoulli random variables with $\mathbb{P}(I_i=1)=p^\ast/p\in(0,1]$. Then $\sum_{i=1}^{M}I_i\sim {\rm B}(n,p^\ast)$, and the pmf of $\sum_{i=1}^{M}X_i$ is $c_{{\rm b}(n,p)}(\,f)$. Note that $\mathbb{E} [I_i] = \mathbb{E} [X_i]= p^\ast/p$ and $I_i\le_{{\rm cx}}X_i$ for each i. Thus, the uncertainty of $\sum_{i=1}^{M}X_i$ is obviously stronger than $\sum_{i=1}^{M}I_i$, that is (3.2).

To obtain the similar result for a compound Negative Binomial distribution, we need the recursive expression for the corresponding pmf.

Lemma 3.2. Denote the pmf of $c_{{\rm nb}(\alpha,p)}(\,f)$ as g. Then

(3.3)\begin{equation} (n+1) g_{n+1}= \frac{q}{1-qf_0} \sum^n_{j=0} [(\alpha-1) j + n+\alpha] f_{j+1} g_{n-j},\quad n\ge 0. \end{equation}

Proof. In [Reference Panjer3], it is assumed that $f_0=0$. We consider a more general situation, $f_0\ge 0$. The pmf of ${\rm NB}(\alpha,p)$ is denoted by $\{p_n, n\ge 0\}$, that is

\begin{equation*} p_n =\binom{\alpha+n-1}{n} p^\alpha q^n, \quad n\ge 0,\ q=1-p, \end{equation*}

so

\begin{equation*} \frac {p_n}{p_{n-1}}= a+\frac {b}{n}, \quad n\ge 1, \end{equation*}

where a = q, $b=(\alpha-1) q$. Hence, for all $n\ge 0$,

(3.4)\begin{eqnarray} \sum^{n+1}_{j=0} \left (a+\frac {bj}{n+1}\right ) f_{j} g_{n+1-j}=q f_0 g_{n+1}+ \frac {q}{n+1} \sum^n_{j=0} [(\alpha-1)j +n+\alpha] f_{j+1} g_{n-j}. \end{eqnarray}

Denote by $f^{(k)}=\{f^{(k)}_i, i\in\mathbb{N}\}$ the pmf of $\sum^k_{i=1} X_i$, where $X_1, \ldots, X_k$ are iid random variables with a common pmf f. On the other hand, by using $\sum^{n+1}_{j=0} j f_j f^{(k)}_{n+1-j}=\frac {n+1}{k+1} f_{n+1}^{(k+1)}$, we have

(3.5)\begin{align} \sum^{n+1}_{j=0} \left (a+\frac {bj}{n+1}\right ) f_{j} g_{n+1-j} &= \sum^{n+1}_{j=0} \left (a+\frac {bj}{n+1}\right ) f_j \sum^\infty_{k=0} p_k f^{(k)}_{n+1-j} \nonumber \\ &= \sum^\infty_{k=0} p_k \sum^{n+1}_{j=0} \left (a+\frac {bj}{n+1}\right ) f_j f^{(k)}_{n+1-j} \nonumber \\ &= \sum^\infty_{k=0} p_k \left (a f_{n+1}^{(k+1)}+ \frac {b}{n+1} \sum^{n+1}_{j=0} j f_j f^{(k)}_{n+1-j}\right ) \nonumber \\ &= \sum^\infty_{k=0} p_k \left (a+\frac {b}{k+1}\right ) f_{n+1}^{(k+1)} \nonumber \\ &= \sum^\infty_{k=1} p_k f_{n+1}^{(k)} \nonumber \\ &= p_0 f^{(0)}_{n+1}+ \sum^\infty_{k=1} p_k f_{n+1}^{(k)}\qquad [f^{(0)}_\ell=0,\ \ell\ge 1] \nonumber \\ &= g_{n+1}. \end{align}

By (3.4) and (3.5), we conclude (3.3).

Proof. Denote $g=c_{{\rm nb}(\alpha,p)}(\,f)$. Since g is non-degenerate and log-concave, we have $g_n \gt 0$ for $n\in \mathbb{N}$ and

(3.6)\begin{equation} \frac {g_{n-j}}{g_{n-j-1}} \ge \frac {g_n}{g_{n-1}},\quad 0 \lt j \lt n. \end{equation}

Since $\alpha\in (0,1)$ and $\alpha^\ast\ge \alpha$, it follows that:

(3.7)\begin{equation} \frac{(\alpha-1)j+n+\alpha}{(\alpha-1)j+n-1+\alpha}\geq\frac{n+\alpha}{n+\alpha-1}\ge \frac{n+\alpha^\ast}{n+\alpha^\ast-1},\quad j\ge 1. \end{equation}

In view of (3.6) and (3.7), we have

\begin{eqnarray*} (n+1) g_{n+1} & \ge & \frac {q}{1-q f_0} \sum^n_{j=0} [(\alpha-1) j +n+\alpha] f_{j+1} g_{n-j-1} \cdot \frac {g_n}{g_{n-1}}\\ & \ge & \frac {g_n}{g_{n-1}}\cdot \frac {q}{1-q f_0} \sum^{n-1}_{j=0} [(\alpha-1) j +n+\alpha] f_{j+1} g_{n-j-1} \\ & \ge & \frac {g_n}{g_{n-1}}\cdot \frac {q}{1-q f_0} \sum^{n-1}_{j=0} [(\alpha-1) j +n-1+\alpha] \frac {\alpha^\ast+n}{\alpha^\ast +n-1} f_{j+1} g_{n-j-1} \\ & \ge & \frac {g_n (\alpha^\ast +n)}{g_{n-1}(\alpha^\ast +n-1)} \cdot \frac {q}{1-q f_0} \sum^{n-1}_{j=0} [(\alpha-1) j +(n-1)+\alpha] f_{j+1} g_{n-j-1}\\ & = & \frac {g_n (\alpha^\ast+n)}{g_{n-1}(\alpha^\ast +n-1)} \cdot n g_n,\quad n\ge 1. \end{eqnarray*}

The above inequality can be simplified to:

\begin{equation*} \left [\frac {g_n}{\binom{\alpha^\ast+n-1}{n}}\right ]^2 \le \frac {g_{n-1}}{\binom{\alpha^\ast+n-2}{n-1}} \cdot \frac {g_{n+1}}{\binom{\alpha^\ast +n}{n+1}},\quad n\ge 1, \end{equation*}

that is, $g_n/\binom{\alpha^\ast +n-1}{n}$ is log-convex, so ${\rm nb}(\alpha^\ast,p^\ast) \le_{\rm lc} g$.

Corollary 3.4. Suppose that f is a pmf defined on $\mathbb{N}$ with the mean µ > 0. For $\alpha\in(0,1]$, if there exists $\alpha^\ast \gt \alpha$ and $p^\ast\in(0,1)$ such that:

\begin{equation*} \frac {\alpha^\ast(1-p^\ast)}{p^\ast} = \frac {\alpha \mu (1-p)}{p}, \end{equation*}

and $c_{{\rm nb}(\alpha,p)}(\,f)$ is non-degenerate and log-concave, then $H\left ({\rm nb}(\alpha^\ast,p^\ast)\right ) \le H\left ( c_{{\rm nb}(\alpha,p)}(\,f)\right )$.

Proof. The notations are the same as in the proof of Proposition 3.3. Obviously, $g_n \gt 0$ for $n\ge 0$ and by the log-concavity of g, we have

\begin{equation*} \frac{g_{n-j}}{g_{n+1-j}}\leq\frac{g_n}{g_{n+1}},\quad j=0,\ldots, n. \end{equation*}

Hence, for $n\ge 0$,

\begin{align*} (n+1) g_{n+1} & \le \frac {q}{1-q f_0} \sum^n_{j=0} [(\alpha-1) j +n+\alpha] f_{j+1} g_{n+1-j} \cdot \frac {g_n}{g_{n+1}}\\ & \le \frac {g_n}{g_{n+1}}\cdot \frac {q}{1-q f_0} \sum^{n+1}_{j=0} [(\alpha-1) j +n+1+\alpha] f_{j+1} g_{n+1-j} \\ & = \frac {g_n}{g_{n+1}} \cdot (n+2) g_{n+2}, \end{align*}

that is, $n!g_n$ is log-convex in $n\in\mathbb{N}$. Thus, ${\rm poi}(\lambda)\leq_{{\rm lc}}c_{{\rm nb}(\alpha,p)}(\,f)$. The rest can be derived directly from Theorem 1.1.

4. Log-concavity of a compound distribution

[Reference Yu8] proved that if f is log-concave, then $c_{{\rm poi}(\lambda)}(\,f)$ is log-concave if and only if $\lambda f_1^2\ge 2 f_2$. In order to show that Propositions 3.3, 3.5 and Corollary 3.4 are meaningful, we need to investigate the log-concavity of compound Negative Binomial distribution. Firstly, we study the necessary and sufficient condition for a compound Geometric distribution to be log-concave.

Proof. Denote the pmf of ${\rm Geo}(p)$ as $\{p_n, n\ge 0\}$. Then the pmf of $c_{{\rm geo}(p)}(\,f)$ is g. By (3.3), we have:

(4.1)\begin{equation} g_{n+1}=\eta\sum_{j=0}^nf_{j+1}g_{n-j}, \quad n\ge 0, \end{equation}

where $\eta= q/(1-qf_0) \gt 0$.

First of all, observe that

1. (1) $g_0 = p_0+\sum_{n=1}^\infty p_nf_0^n = p+\sum_{n=1}^\infty pq^nf_0^n= p/(1-qf_0) = p\eta/q \gt 0$;

2. (2) $f_1\neq 0 \Longrightarrow g_n \gt 0$ for $n\ge 0$.

$(\Longrightarrow)$ Prove that $f_n=0$ for all $n\ge 2$ by induction. For n = 2, by (4.1), we have $g_1=\eta f_1 g_0, g_2=\eta[f_1g_1+f_2g_0]$. Substituting in $g_1^2\ge g_0 g_2$, we have $f_2g_0=0$, that is, $f_2=0$ and $g_1^2 = g_0g_2$.

Now assume that $f_n=0$ for $n=2,\cdots,k$ and $g_{k-1}^2=g_{k-2}g_{k}$. Notice that

\begin{equation*} g_{k-1}=\eta f_1g_{k-2}, ~g_k=\eta f_1g_{k-1}, ~g_{k+1}=\eta[f_1g_k+f_{k+1}g_0]. \end{equation*}

Substituting in $g_k^2\geq g_{k-1}g_{k+1}$, we have

\begin{equation*} f_1^2g_{k-1}^2\geq f_1^2g_kg_{k-2}+f_1f_{k+1}g_0g_{k-2}. \end{equation*}

So, $f_1f_{k+1}g_0g_{k-2}=0$. Thus, $f_{k+1}=0$ and in the meantime, $g_k^2= g_{k-1}g_{k+1}$. The necessity is proved by induction.

$(\Longleftarrow)$ Suppose that $f_k=0$ for all $k\ge 2$. By (4.1), we have

(4.2)\begin{equation} g_{n+1}=\eta f_1g_n=(\eta f_1)^{n+1}g_0,\quad n\ge 0. \end{equation}

It is obvious that $g_n^2=g_{n-1}g_{n+1}$ for $n\ge 1$ and, hence, $g\in {\rm LC}$.

Proposition 4.3. Suppose that $f\in {\rm LC}$, and denote $q=1-p$ and $\eta=q/(1-qf_0)$. Then $g:=c_{{\rm nb}(\alpha,p)}(\,f) \in {\rm LC}$ if and only if

(4.3)\begin{equation} (\alpha-1)\eta f_1^2\geq 2f_2. \end{equation}

The proof of Proposition 4.3 is postponed to Appendix A.1.

Next, the necessary and sufficient condition for a compound Binomial distribution to be log-concave is discussed. To this end, we first give the recursive expression of its corresponding pmf.

Lemma 4.5. Denote $g:= c_{{\rm b}(n,p)}(\,f)$. Then the pmf $\{g_k, k\in\mathbb{N}\}$ has recursive expression as follows:

(4.4)\begin{equation} (k+1)g_{k+1}=\delta\sum_{j=0}^k[(n+1)j+n-k]f_{j+1}g_{k-j},\quad k\ge 0, \end{equation}

where $\delta = p/(pf_0+q)$ and $q=1-p$.

Proof. The proof of (4.4) is similar to (3.3) since

\begin{equation*} \frac{p_k}{p_{k-1}}=-\frac{p}{q}+\frac{(n+1)p}{qk}=a+\frac{b}{k},\quad \forall\ k\ge 1. \end{equation*}

Here, (3.5) still holds,

(4.5)\begin{equation} \sum_{j=0}^{k+1}\left(a+\frac{bj}{k+1}\right)f_jg_{k+1-j} = g_{k+1}, ~k\geq 0, \end{equation}

and (3.4) is replaced by the following formula

(4.6)\begin{align} & \sum_{j=0}^{k+1}\left(a+\frac{bj}{k+1}\right)f_jg_{k+1-j} \nonumber\\ & \qquad= -\frac{p}{q}f_0g_{n+1}+\frac{p}{q(n+1)}\sum_{j=0}^k[(n+1)j+n-k]f_{j+1}g_{k-j},\ \ k\ge 0. \end{align}

Thus, (4.4) follows from (4.5) and (4.6).

Proposition 4.6. Denote $g:=c_{{\rm b}(n,p)}(\,f)$, and let $f\in{\rm LC}$. If $g\in{\rm LC}$, then

(4.7)\begin{equation} (n+1)\delta f_1^2\geq 2f_2, \end{equation}

where $\delta=p/(pf_0+q)$ and $q=1-p$. In particular, for n = 1, (4.7) is also the sufficient condition for $c_{{\rm b}(1,p)}(\,f)\in{\rm LC}$. But for $n\geq 2$, (4.7) is not the sufficient condition for $g\in{\rm LC}$.

Proof. (1)  By lemma 4.5, we have $g_1 = n\delta f_1g_0$ and $g_2= [(n-1)\delta f_1g_1+2n\delta f_2g_0]/2$. In view of $g_1^2\geq g_0g_2$, we obtain (4.7).

(2)  For n = 1, (4.7) holds, that is $\delta f_1^2\geq f_2$. Based on the random expression of the random variable corresponding to g, we have $g_0=q+pf_0$ and $g_k=pf_k$ for $k\ge 1$. To prove $g\in{\rm LC}$, we only need to prove $g_1^2\geq g_0g_2$, that is, (4.7).

(3)  Take a counterexample to illustrate: assume n = 2, and $c_{{\rm b}(2,p)}(\,f) = c_{{\rm b}(1,p)}(\,f)\ast c_{{\rm b}(1,p)}(\,f)$. Denote $h = c_{{\rm b}(1,p)}(\,f)$, by (2), we have $h_0=q+pf_0, ~h_k=pf_k, ~k\geq 1$. Especially, take $p=20/23$,

\begin{equation*} f = \left(\frac{1}{20}, \frac{1}{5}, \frac{3}{10}, \frac{9}{20}, 0, 0, \cdots\right). \end{equation*}

Then

\begin{equation*} h = \left(\frac{4}{23}, \frac{4}{23}, \frac{6}{23}, \frac{9}{23}, 0, 0, \cdots\right). \end{equation*}

Hence,

\begin{equation*} g = h\ast h = \left(g_0, g_1, g_2, \frac{120}{23^2}, \frac{108}{23^2}, \frac{108}{23^2}, g_6, \cdots\right). \end{equation*}

Obviously, $g_4^2 \lt g_3g_5$, which means that g is not log-concave. On the other hand, $\delta = p/(pf_0+q)=5$, $f\in{\rm LC}$, and $3\delta f_1^2= 3/5 =2f_2$, that is (4.7) holds. Therefore, (4.7) is not the sufficient condition for $g\in{\rm LC}$.

5. The relative log-concavity

Lemma 2 in [Reference Yu8] states that, for pmfs $f, g, f^\ast$ and $g^\ast$ defined on $\mathbb{Z}_+$, if $f\le_{\rm cx} f^\ast$ and $g\le_{\rm cx}g^\ast$, then $c_g(\,f)\le_{\rm cx}c_{g^\ast}(\,f^\ast)$. On the other hand, $g\le_{\rm lc}{\rm poi}(\lambda)$ for any $g\in{\rm ULC}$ and λ > 0. Connected with Theorem 1.1, it is easy to establish the following proposition.

Notice that $g\in{\rm ULC}(n)\Longleftrightarrow g\le_{\rm lc} b(n,p)$ for all $p\in(0,1)$, and that $g\in{\rm LC}\Longleftrightarrow g\le_{\rm lc} {\rm geo}(\lambda)$ for all λ > 0. The following two propositions are easily established by Theorem 1.1.

Naturally, the following problems will arise:

Question

1. (1) If $g\in{\rm ULC}$ and $\mu_g = \lambda$, is it correct $c_g(\,f)\leq_{\rm lc}c_{{\rm poi}(\lambda)}(\,f)$?

2. (2) If $g\in{\rm ULC}(n)$ and $\mu_g = np$, is it correct $c_g(\,f)\leq_{\rm lc}c_{{\rm b}(n,p)}(\,f)$?

3. (3) If $g\in{\rm LC}$ and $\mu_g = (1-\lambda)/\lambda$, is it correct that $c_g(\,f)\leq_{\rm lc}c_{{\rm geo}(\lambda)}(\,f)$?

The following three counterexamples show that the assertions in Question 1 are negative.

Example 5.1. $g\in{\rm ULC}$, $\mu_g =\lambda$ and $f\in{\rm LC} \not\!\Longrightarrow c_g(\,f)\le_{\rm lc}c_{{\rm poi}(\lambda)}(\,f)$.

Suppose that $g=B(2,\lambda/2)$ and $0 \lt \lambda \lt 2$, f satisfies $f_j=0$ for $j\ge 3$, and denote $s=c_g(\,f)$ and $ t=c_{{\rm poi}(\lambda)}(\,f)$. Then

\begin{align*} s_0 & = \left ( 1-\frac{\lambda}{2}\right )^2+\lambda\left(1-\frac{\lambda}{2}\right)f_0+\frac{\lambda^2}{4}f_0^2,\qquad s_1 = \lambda\left (1-\frac{\lambda}{2}\right )f_1+\frac{\lambda^2}{2}f_0f_1,\\ s_2 & = \lambda\left (1-\frac{\lambda}{2}\right )f_2+\frac{\lambda^2}{2}f_0f_2+\frac{\lambda^2}{4}f_1^2,\qquad s_3 = \frac{\lambda^2}{2}f_1f_2, \qquad s_4=\frac{\lambda^2}{4}f_2^2, \end{align*}

and

\begin{align*} t_0 & = e^{\lambda(\,f_0-1)}, \qquad t_1=\lambda f_1e^{\lambda(\,f_0-1)},\qquad t_2 =\frac{\lambda}{2}(\lambda f_1^2+2f_2)e^{\lambda(\,f_0-1)},\\ t_3 & = \lambda^2f_1 \left (\frac{1}{6}\lambda f_1^2+f_2\right ) e^{\lambda(\,f_0-1)},\qquad t_4=\frac{\lambda^2}{4} \left (\frac{1}{6}\lambda^2 f_1^4+2\lambda f_1^2f_2+2f_2^2\right )e^{\lambda(\,f_0-1)}. \end{align*}

Especially, we take λ = 1 and $f=(\,f_0, f_1, f_2)=(1/3, 1/3, 1/3)$, that is, f is a discrete uniform distribution on $\{0, 1, 2\}$. Obviously, $f\in{\rm ULC}$. By calculation,

\begin{equation*} \left(\frac{s_0}{t_0}, \frac{s_1}{t_1}, \frac{s_2}{t_2}, \frac{s_3}{t_3}, \frac{s_4}{t_4}\right) = \left(\frac{4}{9}, \frac{2}{3}, \frac{9}{14}, \frac{9}{19}, \frac{54}{145}\right)e^{2/3}. \end{equation*}

It can be verified that

\begin{equation*} \left(\frac{s_1}{t_1}\right)^2-\frac{s_0}{t_0}\frac{s_2}{t_2} \gt 0, \qquad \left(\frac{s_3}{t_3}\right)^2-\frac{s_2}{t_2}\frac{s_4}{t_4} = -0.015 \lt 0. \end{equation*}

Hence, $s_k/t_k$ is not log-concave, so $s\nleq_{\rm lc}t$.

Example 5.2. $g\in{\rm LC}$, $\mu_g = (1-\lambda)/\lambda$ and $f\in{\rm LC} \not\!\Longrightarrow c_g(\,f)\leq_{\rm lc}c_{{\rm geo}(\lambda)}(\,f)$.

Suppose that f and g satisfy $f_j = g_j = 0$ for $j\ge 3$, and denote $s=c_g(\,f)$ and $t=c_{\rm geo}(\,f)$. Then

\begin{align*} s_0 & = g_0+g_1f_0+g_2f_0^2, \qquad s_1=g_1f_1+2g_2f_0f_1, \\ s_2 & = g_1f_2+g_2(2f_0f_2+f_1^2), \qquad s_3=2g_2f_1f_2, \qquad s_4=g_2f_2^2, \end{align*}

and

\begin{align*} t_1 = \eta f_1t_0, \quad t_2 = \eta(\,f_1t_1+f_2t_0), \quad t_3 = \eta(\,f_1t_2+f_2t_1),\quad t_4 = \eta(\,f_1t_3+f_2t_2), \end{align*}

with $\eta = (1-\lambda)/[1-(1-\lambda)f_0]$. If we take $f=(\,f_0, f_1, f_2)=(1/3, 1/3, 1/3)\in{\rm LC}$, then

\begin{equation*} g = (g_0,g_1,g_2)=\left (\frac{7\lambda-3}{4\lambda}, \frac{1-\lambda}{2\lambda}, \frac{1-\lambda}{4\lambda}\right). \end{equation*}

Given $\lambda= 40/77$, g is log-concave, that is, $g\le_{\rm lc}{\rm Geo}(\lambda)$. It is easy to verify that

\begin{equation*} \left(\frac{s_3}{t_3}\right)^2-\frac{s_2}{t_2}\frac{s_4}{t_4} = -0.082 \lt 0, \end{equation*}

hence, $s_k/t_k$ is not log-concave, so $s\nleq_{\rm lc}t$.

Example 5.3. $g\in{\rm ULC}(n), ~\mu_g = np, ~p\in(0,1)$ and $f\in{\rm LC} \not\!\Longrightarrow c_g(\,f)\le_{\rm lc} c_{{\rm b}(n,p)}(\,f)$.

Suppose that $f=(\,f_0, f_1, f_2)=(1/3, 1/3, 1/3)\in {\rm LC}$ and $g = (g_0,g_1,g_2)= (1-3p/2, p, p/2)$. If $p\in(1/2,2/3)$, then $g\in {\rm ULC}(2)$. Denote $s=c_g(\,f)$ and $t=c_{{\rm b}(2,p)}(\,f)$. Then s_j and t_j can be calculated as in Example 5.2. Choose $p=7/12$. Then

\begin{equation*} \left(\frac{s_3}{t_3}\right)^2-\frac{s_2}{t_2}\frac{s_4}{t_4} = -0.1728 \lt 0, \end{equation*}

hence, $s_k/t_k$ is not log-concave, so $s\nleq_{\rm lc}t$.