1005

设离散型随机变量 ${\displaystyle X}$ 和 ${\displaystyle Y}$ 有相同的可能取值 ${\displaystyle a_1,\cdots ,a_n}$, 它们的分布列为 ${\displaystyle P(X=a_k)=x_k,P(Y=a_k)=y_k,x_k>0,y_k>0,k=1,2,\cdots ,n,{\displaystyle \sum _{k=1}^n{x}_k=\sum _{k=1}^n{y}_k=1}}$. 指标 ${\displaystyle D(X||Y)}$ 可用来刻画 ${\displaystyle X}$ 和 ${\displaystyle Y}$ 的相似程度，其定义为 ${\displaystyle D(X||Y)={\displaystyle \sum _{k=1}^n{x}_k\ln \frac{x_k}{y_k}}}$. 设 ${\displaystyle X\sim B(n,p),0<p<1}$.

1. 若 ${\displaystyle Y\sim B(n,q),0<q<1}$, 求 ${\displaystyle D(X||Y)}$;
2. 若 ${\displaystyle n=2,P(Y=k-1)={\displaystyle \frac{1}{3},k=1,2,3}}$, 求 ${\displaystyle D(X||Y)}$ 的最小值;
3. 对任意与 ${\displaystyle X}$ 有相同可能取值的随机变量 ${\displaystyle Y}$, 证明: ${\displaystyle D(X||Y)\ge 0}$, 并指出取等号的充要条件.

1. 解 容易知道 ${\displaystyle x_k=C_n^k{p}^k(1-p{)}^{n-k},y_k=C_n^k{q}_k(1-q{)}^{n-k},0\le k\le n}$. 这样结合二项式定理有$$\begin{array}{rl}D(X||Y)=& \sum _{k=0}^n{C}_n^k{p}^k(1-p{)}^{n-k}\ln {\left(\frac{p}{q}\right)}^k{\left(\frac{1-p}{1-q}\right)}^{n-k}\\ =& \sum _{k=0}^n{p}^k(1-p{)}^{n-k}(k\ln \frac{p(1-q)}{q(1-p)}+n\ln \frac{1-p}{1-q})\\ =& \ln \frac{p(1-q)}{q(1-p)}\sum _{k=1}^{n}k{C}_n^k{p}^k(1-p{)}^{n-k}+n\ln \frac{1-p}{1-q}\sum _{k=0}^n{C}_n^k{p}^k(1-p{)}^{n-k}\\ =& \ln \frac{p(1-q)}{q(1-p)}\sum _{k=1}^{n}pn{C}_{n-1}^{k-1}{p}^{k-1}(1-p{)}^{(n-1)-(k-1)}+n\ln \frac{1-p}{1-q}\\ =& np\ln \frac{p(1-q)}{q(1-p)}+n\ln \frac{1-p}{1-q}.\end{array}$$
2. 解 当 ${\displaystyle n=2}$ 时，${\displaystyle (x_0,x_1,x_2)=((1-p{)}^2,2p(1-p),p^2)}$, 从而$$\begin{array}{r}D(X||Y)=(1-p{)}^2\ln (1-p{)}^2+p^2\ln p^2+2p(1-p)\ln 2p(1-p)-\ln 3.\end{array}$$
   记右边为${\displaystyle f(p)}$, 则经过一番整理得$$\begin{array}{rl}& f^{\prime }(p)=2\ln p-2\ln (p-1)+(2-4p)\ln 2,\\ & f^{″}(p)=\frac{2}{p}+\frac{2}{1-p}-4\ln 2\ge 2\cdot \frac{(1+1{)}^2}{p+1-p}-4\ln 2>0.\end{array}$$
   而${\displaystyle f^{\prime }{\displaystyle \left(\frac{1}{2}\right)=0}}$, 故${\displaystyle f(p)}$在${\displaystyle {\displaystyle (0,\frac{1}{2})}}$上减，在${\displaystyle {\displaystyle (\frac{1}{2},1)}}$上增，从而${\displaystyle f(p{)}_{min}=f{\displaystyle \left(\frac{1}{2}\right)=\ln 3-\frac{3}{2}\ln 2}}$.
3. 证明 利用 ${\displaystyle \ln x\ge 1-{\displaystyle \frac{1}{x}}}$, 有$$\begin{array}{r}D(X||Y)\ge \sum _{k=1}^n{x}_k(1-\frac{y_k}{x_k})=\sum _{k=1}^n{x}_k-\sum _{k=1}^n{y}_k=0.\end{array}$$
   取等当且仅当${\displaystyle {\displaystyle \frac{x_k}{y_k}=1\Rightarrow x_k=y_k,\mathrm{\forall }1\le k\le n}}$.