# Solution to Introduction of Mathematical Statistics.

We are given that $Y = K(X)$ is such that $E(Y) = \theta$, and we need to show that in this case, $Y \sim N(\theta, \,1)$.

Remember that the expected value of a random variable can be expressed in terms of its moment-generating function in the following way:

$E(Y) = E(K(X)) = M'(0)\,$

where M is the moment-generating function of Y=K(X).

Using the result of the previous exercise, we have that

$M(t) = \exp(q(\theta) - q(\theta + t))\,$,

which means that the derivative (with respect to t) is

$M'(t) = \underbrace{\exp(q(\theta) - q(\theta + t))}_{=\, M(t)}\cdot(-q'(\theta + t))\,.$

Since M(0) = 1, for any moment-generating function, then

$\theta = M'(0) = 1\cdot (-q'(\theta + 0)) = -q'(\theta) \iff q'(\theta) = - \theta\, .
θ=M′(0)=1⋅(−q′(θ+0))=−q′(θ)⟺q′(θ)=−θ$.

Integrating both sides with respect to \thetaθ (or simply guessing the solution) yields the solution to this differential equation:

$q(\theta) = -\frac{\theta^2}{2} + C\, , \, \, \, \, \, C \in \mathbb R.$

Therefore, the moment-generating function of Y=K(X) is

$M(t) = \exp(-\frac{\theta^2}{2} + \frac{(\theta+t)^2}{2}) = \exp(\theta\cdot t + \frac{t^2}{2})\,$

which is exactly the moment-generating function of a $N(\theta, \, 1)$$N(θ,1)$ random variable.

Since this moment-generating function MM exists for values of tt contained in an open interval containing 0 (it's defined on the entire $\mathbb R$), then it uniquely determines the distribution of Y = K(X), i.e. we can conclude that

$Y = K(X) \sim N(\theta, \, 1)\, ,
Y=K(X)∼N(θ,1)$,

as was to be shown.

3.3.21. Show that the graph of the β pdf is symmetric about the vertical line through x = 1/2 if α = β.

The probability density function of beta distribution with parameters $\alpha$ and $\beta$ is given by

\begin{align*} f(x) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \Gamma(\beta)} x^{\alpha-1} (1-x)^{\beta - 1}, \hspace{3mm} 0<x<1, \end{align*}
and zero otherwise. Assume now that $\alpha$ = $\beta$, it becomes

\begin{align*} f(x) & = \frac{\Gamma(\alpha + \alpha)}{\Gamma(\alpha) \Gamma(\alpha)} x^{\alpha-1} (1-x)^{\alpha - 1} \\ & = \frac{\Gamma(2\alpha)}{\Gamma(\alpha) \Gamma(\alpha)} \left[ x (1-x)\right]^{\alpha - 1} , \hspace{3mm} 0<x<1. \end{align*}

Take a point z such that $z \in (0,1/2)$. Then it follows that

\begin{align*} \left( \frac{1}{2} + z \right) \end{align*}
is a point to the right of $x=1/2$ on the same distance as point

\begin{align*} \left( \frac{1}{2} - z \right) \end{align*}

is on the same distance from x=1/2. Also, it follows that

\begin{align*} f\left( \frac{1}{2} + z \right) & = \frac{\Gamma(2\alpha)}{\Gamma(\alpha) \Gamma(\alpha)} \left[ \left(\frac{1}{2} + z\right) \left(1-\frac{1}{2} - z\right)\right]^{\alpha - 1} \\ & = \frac{\Gamma(2\alpha)}{\Gamma(\alpha) \Gamma(\alpha)} \left[ \left(\frac{1}{2} + z\right) \left( \frac{1}{2}- z\right)\right]^{\alpha - 1}, \end{align*}

For $z \in (0,1/2)$, it follows that

\begin{align*} f\left( \frac{1}{2} - z \right) & = \frac{\Gamma(2\alpha)}{\Gamma(\alpha) \Gamma(\alpha)} \left[ \left(\frac{1}{2} - z\right) \left(1-\frac{1}{2} + z\right)\right]^{\alpha - 1} \\ & = \frac{\Gamma(2\alpha)}{\Gamma(\alpha) \Gamma(\alpha)} \left[ \left(\frac{1}{2} - z\right) \left( \frac{1}{2}+ z\right)\right]^{\alpha - 1} \\ \end{align*}

which implies that

\begin{align*} f\left( \frac{1}{2} + z \right) = f\left( \frac{1}{2} - z \right) \end{align*}

for every $z \in (0,1/2)$ or equally the pdf is symmetric.

## 算幾不等式
今天就跟大家談談算幾不等式。 假設$a,b>0$，那麼$\sqrt{a}$與$\sqrt{b}$均存在。那麼 $$(\sqrt{a}-\sqrt{b})^{2}=a-2\sqrt{ab}+b\geq 0.$$ 所以我們推得$$\frac{a+b}{2}\geq\sqrt{ab}$$ 等號乘立時，$a=b$。由此，我們可以推廣到更一般的情況:
算幾不等式
設$a_{1},\cdots,a_{n}$均為正實數，那麼他們的關係必定滿足下列不等式: $$\displaystyle \frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\geq \sqrt[n]{a_{1}a_{2}\cdots a_{n}}$$ 此不等式就稱為算幾不等式。等號乘立若且唯若$a_{1}=\cdots=a_{n}.$
附註:在不失一般的情況下，我們假設$a_{i}>0$。如果有某數$a_{i}=0,$不等式顯然成立。

證明
因為我們會證明$n=2$的，我們可以證明算幾不等式對$n=2^{k}$均成立。舉例來說，假設$x,y,z,u> 0$。則 $$\displaystyle\frac{x+y+z+u}{4}=\frac{\frac{x+y}{2}+\frac{z+u}{2}}{2}\geq \sqrt{\frac{x+y}{2}\cdot\frac{z+u}{2}}.$$ 利用算幾不等式$x+y\geq 2\sqrt{xy}$且$z+u\geq 2\sqrt{zu}$我們得到 $$\displaystyle\frac{x+y}{2}\cdot\frac{z+u}{2}\geq\sqrt{xyzu}.$$ 所以我們得到$$\frac{x+y+z+u}{4}\geq \sqrt[4]{xyzu}$$ 利用歸納法，我們可以證明對所有的$k\geq 1$，算幾不等式對$n=2^{k}$均成立。 接著我們來呈現一下，如何從$n=2^{k}$算幾不等式成立推到一般$n$的算幾不等式成立。我們就用$n=4$來推$n=3$ 成立。我們取$u=(x+y+z)/3$，利用$n=4$的算幾不等式可以推得 $$\displaystyle\frac{1}{4}\left(x+y+z+\frac{x+y+z}{3}\right)\geq \sqrt[4]{xyz\left(\frac{x+y+z}{3}\right)}.$$ 注意到: $$\displaystyle\frac{1}{4}\left(x+y+z+\frac{x+y+z}{3}\right)=\frac{x+y+z}{3}.$$ 上面不等式兩邊取四次方後得到 $$\displaystyle\left(\frac{x+y+z}{3}\right)^{4}\geq xyz\left(\frac{x+y+z}{3}\right).$$ 假設$x,y,z$其中之一不為零，兩邊同除掉一個$(x+y+z)/3$就可以得到$n=3$的算幾不等式。 一般的情況是類似的。我們利用$n=2^{k}$的算幾不等式之後， 再利用上述的想法可以推得算幾不等式對$2^{k}-1$也成立。 所以我們假設算幾不等式對自然數$n$成立，我們希望推得算幾不等式對$n-1$也要成立。 取正實數$a_{1},\cdots,a_{n-1}$，定義一個新的正實數 $$\displaystyle a_{n}=\frac{a_{1}+\cdots+a_{n-1}}{n-1}.$$ 於是觀察發現 $$\displaystyle \frac{a_{1}+\cdots+a_{n-1}+a_{n}}{n}=\frac{a_{1}+\cdots+a_{n-1}}{n-1}=a_{n}.$$ 因為我們假設算幾不等式對$n$個正實數成立，所以我們推得 $$\displaystyle a_{n}=\frac{a_{1}+\cdots+a_{n}}{n}\geq\sqrt[n]{a_{1}\cdots a_{n-1}a_{n}}$$ 兩邊同取$n$次方之後我們得到 $$ a_{n}^{n}\geq a_{1}\cdots a_{n-1}a_{n}.$$ 假設$a_{n}\neq 0$，則$a_{n}^{n-1}\geq a_{1}\cdots a_{n-1}$。換句話說 $$a_{n}\geq \sqrt[n-1]{a_{1}\cdots a_{n}}.$$ 由於$a_{n}=(a_{1}+\cdots+a_{n-1})/(n-1)$，帶入上式之後，我們證明了算幾不等式對$n-1$也成立。 利用歸納法我們得到了算幾不等式對$n=2^{k}-l$成立，其中$0\leq l<2^{k}$。我們就證明了對任意 $n\geq 2$的算幾不等式。

接著我提供另一個簡單的證明(凹函數的概念)。如果我們對算幾不等式兩邊同取$\log$我們得到了 $$\displaystyle \frac{1}{n}\left(\log a_{1}+\cdots+\log a_{n}\right)\leq \log\left(\frac{a_{1}+\cdots+a_{n}}{n}\right).$$ 如果令$\lambda_{1}=\cdots=\lambda_{n}=1/n$則$ \sum_{i=1}^{n}\lambda_{i}=1$， 且上述不等式等於 $$\displaystyle\sum_{i=1}^{n}\lambda_{i}\log a_{i}\leq\log\left(\sum_{i=1}^{n}\lambda_{i}a_{i}\right).$$ 其中$f(x)=\log x$是所謂的凹函數。

定義
假設$f:(a,b)\to\mathbb{R}$是一個函數， 並且對任意$0\leq \lambda\leq 1,$ $x_{1},x_{2}\in (a,b)$恆有 $$f(\lambda x_{1}+(1-\lambda)x_{2})\geq \lambda f(x_{1})+(1-\lambda)f(x_{2}),$$ 則我們稱函數$f$是一凹函數。(如果上述不等式中不等號是相反的，我們把他稱為凸函數。)
定理
假設$f:(a,b)\to\mathbb{R}$是二次可微，並且$f''(x)<0$，$x\in (a,b)$則$f$是凹函數。
Jesen不等式
假設$f:(a,b)\to\mathbb{R}$是凹函數。任給 $x_{1},\cdots,x_{n}\in (a,b)$且$\lambda_{1},\cdots,\lambda_{n}\geq 0$ 使得$\sum_{i=1}^{n}\lambda_{i}=1$。則 $$\displaystyle f\left(\sum_{i=1}^{n}\lambda_{i}x_{i}\right)\geq\sum_{i=1}^{n}\lambda_{i}f(x_{i}).$$
證明
:我們利用歸納法。首先我們來看$n=2$怎麼推到$n=3$成立。
假設$x_{1},x_{2},x_{3}\in (a,b)$且$\lambda_{1},\lambda_{2},\lambda_{3}\geq 0$滿足$\lambda_{1}+\lambda_{2}+\lambda_{3}=1$。令$ \lambda=\lambda_{2}+\lambda_{3}$則 $\lambda_{1}=1-\lambda$。於是 $$\displaystyle\lambda_{1}x_{1}+\lambda_{2}x_{2}+\lambda_{3}x_{3}=(1-\lambda)x_{1}+\lambda\left(\frac{\lambda_{2}}{\lambda}x_{2}+\frac{\lambda_{3}}{\lambda}x_{3}\right)$$ 所以利用不等式可知 $$\displaystyle f(\lambda_{1}x_{1}+\lambda_{2}x_{2}+\lambda_{3}x_{3})\geq (1-\lambda)f(x_{1})+\lambda f\left(\frac{\lambda_{2}}{\lambda}x_{2}+\frac{\lambda_{3}}{\lambda}x_{3}\right)$$ 再用一次凹函數的性質，我們可以推得 $$ \displaystyle f\left(\frac{\lambda_{2}}{\lambda}x_{2}+\frac{\lambda_{3}}{\lambda}x_{3}\right)\geq \frac{\lambda_{2}}{\lambda}f(x_{2})+\frac{\lambda_{3}}{\lambda}f(x_{3}).$$ 統整後我們證明了$n=3$的情況。我們假設不等式在$n=k$時成立。利用類似的想法我們做以下的更動:取$\lambda=\lambda_{2}+\cdots+\lambda_{n}$則$\lambda_{1}=1-\lambda$且 $$\displaystyle\sum_{i=1}^{n}\lambda_{i}x_{i}=(1-\lambda)x_{1}+\lambda\sum_{i=2}^{n}\frac{\lambda_{i}}{\lambda}x_{i}.$$ 如果令$\displaystyle y=\sum_{i=2}^{n}\frac{\lambda_{i}}{\lambda}x_{i}$。利用凹函數的性質，我們推得 $$f((1-\lambda) x_{1}+\lambda y)\geq (1-\lambda)f(x_{1})+\lambda f(y).$$ 在一次利用凹函數的性質我們知道 $$\displaystyle f(y)\geq\sum_{i=2}^{n}\frac{\lambda_{i}}{\lambda}f(x_{i}).$$ 所以整理的結果我們推得 $$ \displaystyle f\left(\sum_{i=1}^{n}\lambda_{i}x_{i}\right)\geq \sum_{i=1}^{n}\lambda_{i}f(x_{i}).$$ 於是我們證明了Jensen不等式。接著我們來證明算幾不等式。

證明
接著我們利用Jensen不等式提出算幾不等式的第二個證明。 令$f(x)=\log x$其中$x>0$。則$f$是光滑函數且$f''(x)<0$所以$f$為凹函數。再利用Jesen不等式，我們證明了算幾不等式。