2016.3.12 Question 12

1. Let $X\sim \text{B}<!--\; FUNCTION\; APPLICATION\; -->(100n,0.2)$. We have $\mu =100n\cdot 0.2=20n$, and ${\sigma }^2=100n\cdot 0.2\cdot 0.8=16n$.

   We have that

   $$\begin{array}{llll}\hfill \alpha & =\text{P}<!--\; FUNCTION\; APPLICATION\; -->(16n\le X\le 24n)& \hfill & \\ \hfill & =\text{P}<!--\; FUNCTION\; APPLICATION\; -->(\left|(X-20n)\right|\le 4n)& \hfill & \\ \hfill & =\text{P}<!--\; FUNCTION\; APPLICATION\; -->(\left|(X-\mu )\right|\le \sigma \sqrt{n})& \hfill & \\ \hfill & =1-\text{P}<!--\; FUNCTION\; APPLICATION\; -->(\left|(X-\mu )\right|>\sigma \sqrt{n})& \hfill & \\ \hfill & \ge 1-\frac{1}{{\sqrt{n}}^2}& \hfill & \\ \hfill & =1-\frac{1}{n},& \hfill & \end{array}$$

   as desired, where we applied the Chebyshev Inequality for $k=\sqrt{n}>0$.

2. Let $X\sim \text{Po}<!--\; FUNCTION\; APPLICATION\; -->(n)$. Therefore, $\mu =\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X)=n$, $\sigma =\sqrt{\text{Var}<!--\; FUNCTION\; APPLICATION\; -->(X)}=\sqrt{n}$. To show the desired inequality is equivalent to showing that

   $$\frac{1+n+\frac{n^2}{2!}+\cdot +\frac{n^{2n}}{(2n!)}}{e^n}\ge 1-\frac{1}{n}.$$

   Notice that the left-hand side is simply $\text{P}<!--\; FUNCTION\; APPLICATION\; -->(0\le X\le 2n)$. By the Chebyshev Inequality, we have

   $$\begin{array}{llll}\hfill \text{LHS}& =\text{P}<!--\; FUNCTION\; APPLICATION\; -->(0\le X\le 2n)& \hfill & \\ \hfill & =\text{P}<!--\; FUNCTION\; APPLICATION\; -->(|X-\mu |\le n)& \hfill & \\ \hfill & =\text{P}<!--\; FUNCTION\; APPLICATION\; -->(|X-\mu |\le \sqrt{n}\sigma )& \hfill & \\ \hfill & =1-\text{P}<!--\; FUNCTION\; APPLICATION\; -->(|X-\mu |>\sqrt{n}\sigma )& \hfill & \\ \hfill & \ge 1-\frac{1}{n}& \hfill & \\ \hfill & =\text{RHS},& \hfill & \end{array}$$

   as desired, where we applied the Chebyshev Inequality for $k=\sqrt{n}>0$.