问题及解答

随机变量 $X$ 被称为服从带参数 $\alpha,\beta$(都是正的), 以及参数 $A$ 和 $B$ 的 $\beta$-分布(beta distribution), 如果其 pdf (概率密度函数)定义为

\[
f(x;\alpha,\beta,A,B)=\begin{cases}
\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\biggl(\frac{x-A}{B-A}\biggr)^{\alpha-1}\biggl(\frac{B-x}{B-A}\biggr)^{\beta-1}, & x\in[A,B]\\
0, & \text{其他}
\end{cases}
\]

特别的, 当 $A=0$, $B=1$ 时, 该分布称为标准 $\beta$-分布 (standard beta distribution).

首先, 可以证明, 这样定义的 pdf 是合理的, 即

\[
\int_{A}^{B}f(x;\alpha,\beta,A,B)dx=1.
\]

其次, 可以计算服从 $\beta$-分布的随机变量 $X$, 其均值和方差为:

\[
\mu=A+(B-A)\cdot\frac{\alpha}{\alpha+\beta},\qquad\sigma^2=\frac{(B-A)^2\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}
\]

(1)

\[
\begin{split}
\int_{A}^{B}f(x;\alpha,\beta,A,B)dx&=\int_{A}^{B}\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\biggl(\frac{x-A}{B-A}\biggr)^{\alpha-1}\biggl(\frac{B-x}{B-A}\biggr)^{\beta-1}dx\\
&=\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{A}^{B}\biggl(\frac{x-A}{B-A}\biggr)^{\alpha-1}\biggl(\frac{B-x}{B-A}\biggr)^{\beta-1}dx
\end{split}
\]

令 $t=\frac{x-A}{B-A}$, 则 $\frac{B-x}{B-A}=1-t$. 且由于 $x=A+(B-A)t$ 得 $dx=(B-A)dt$. 故

\[
\begin{split}
\mu&=\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}t^{\alpha-1}\cdot(1-t)^{\beta-1}\cdot(B-A)dt\\
&=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}t^{\alpha-1}\cdot(1-t)^{\beta-1}dt\\
&=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot B(\alpha,\beta)\\
&=1
\end{split}
\]

这里倒数第二个等号是根据 B-函数的定义, 最后一个等号是因为 B-函数和 $\Gamma$-函数的性质.

因此, 该 pdf 的定义是合理的.

(2) 求均值(或期望) $\mu=E(X)$.

\[
\begin{split}
\mu&=\int_{A}^{B}xf(x;\alpha,\beta,A,B)dx\\
&=\int_{A}^{B}x\cdot\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\biggl(\frac{x-A}{B-A}\biggr)^{\alpha-1}\biggl(\frac{B-x}{B-A}\biggr)^{\beta-1}dx\\
&=\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}[A+(B-A)t]\cdot t^{\alpha-1}\cdot(1-t)^{\beta-1}\cdot(B-A)dt\\
&=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}[A+(B-A)t]\cdot t^{\alpha-1}\cdot(1-t)^{\beta-1}dt\\
&=A\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}t^{\alpha-1}\cdot(1-t)^{\beta-1}dt+(B-A)\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}t^{\alpha}\cdot(1-t)^{\beta-1}dt\\
&=A\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot B(\alpha,\beta)+(B-A)\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}t^{\alpha}\cdot(1-t)^{\beta-1}dt\\
&=A\cdot 1+(B-A)\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}t^{\alpha}\cdot(1-t)^{\beta-1}dt\\
\end{split}
\]

这里第三个等号是令 $t=\frac{x-A}{B-A}$ 所得. 倒数第二个等号是根据 B-函数的定义, 最后一个等号是根据 B-函数与 $\Gamma$-函数的性质.

下面计算积分

\[
\begin{split}
\int_{0}^{1}t^{\alpha}\cdot(1-t)^{\beta-1}dt&=\int_{0}^{1}t^{1+\alpha-1}\cdot(1-t)^{\beta-1}dt\\
&=B(1+\alpha,\beta)\\
&=\frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(\alpha+1+\beta)}\\
&=\frac{\alpha\Gamma(\alpha)\Gamma(\beta)}{(\alpha+\beta)\Gamma(\alpha+\beta)}
\end{split}
\]

代入上面的式子, 得

\[
\begin{split}
\mu&=A+(B-A)\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\frac{\alpha\Gamma(\alpha)\Gamma(\beta)}{(\alpha+\beta)\Gamma(\alpha+\beta)}\\
&=A+(B-A)\cdot\frac{\alpha}{\alpha+\beta}.
\end{split}
\]

(3) 计算方差 $\sigma^2=V(X)=E(X^2)-(E(X))^2$. 其中 $E(X)=\mu$ 已经计算, 因此只需计算 $E(X^2)$.

\[
\begin{split}
E(X^2)&=\int_{A}^{B}x^2\cdot f(x;\alpha,\beta,A,B)dx\\
&=\int_{A}^{B}x^2\cdot\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\biggl(\frac{x-A}{B-A}\biggr)^{\alpha-1}\biggl(\frac{B-x}{B-A}\biggr)^{\beta-1}dx
\end{split}
\]

令 $t=\frac{x-A}{B-A}$, 则积分化为

\[
\begin{split}
E(X^2)&=\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}\bigl[A+(B-A)t\bigr]^2\cdot t^{\alpha-1}\cdot(1-t)^{\beta-1}\cdot(B-A)dt\\
&=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\int_{0}^{1}\bigl[A^2+2A(B-A)t+(B-A)^2t^2\bigr]t^{\alpha-1}\cdot(1-t)^{\beta-1}dt\\
&=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\biggl[A^2\int_{0}^{1}t^{\alpha-1}(1-t)^{\beta-1}dt+2A(B-A)\int_{0}^{1}t^{\alpha}(1-t)^{\beta-1}dt+(B-A)^2\int_{0}^{1}t^{\alpha+1}(1-t)^{\beta-1}dt\biggr]\\
&=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\biggl[A^2\cdot B(\alpha,\beta)+2A(B-A)\cdot\int_{0}^{1}t^{1+\alpha-1}(1-t)^{\beta-1}dt+(B-A)^2\int_{0}^{1}t^{2+\alpha-1}(1-t)^{\beta-1}dt\biggr]\\
&=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\biggl[A^2\cdot B(\alpha,\beta)+2A(B-A)\cdot B(\alpha+1,\beta)+(B-A)^2\cdot B(\alpha+2,\beta)\biggr]\\
&=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\biggl[A^2\cdot\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}+2A(B-A)\cdot\frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(1+\alpha+\beta)}+(B-A)^2\cdot \frac{\Gamma(\alpha+2)\Gamma(\beta)}{\Gamma(2+\alpha+\beta)}\biggr]\\
&=A^2+2A(B-A)\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\frac{\alpha\Gamma(\alpha)\Gamma(\beta)}{(\alpha+\beta)\Gamma(\alpha+\beta)}+(B-A)^2\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\cdot\frac{(\alpha+1)\alpha\Gamma(\alpha)\Gamma(\beta)}{(\alpha+\beta+1)(\alpha+\beta)\Gamma(\alpha+\beta)}\\
&=A^2+\frac{2\alpha A(B-A)}{\alpha+\beta}+\frac{\alpha(\alpha+1)(B-A)^2}{(\alpha+\beta)(\alpha+\beta+1)}
\end{split}
\]

于是

\[
\begin{split}
\sigma^2&=A^2+\frac{2\alpha A(B-A)}{\alpha+\beta}+\frac{\alpha(\alpha+1)(B-A)^2}{(\alpha+\beta)(\alpha+\beta+1)}-\biggl(A+(B-A)\frac{\alpha}{\alpha+\beta}\biggr)^2\\
&=A^2+\frac{2\alpha A(B-A)}{\alpha+\beta}+\frac{\alpha(\alpha+1)(B-A)^2}{(\alpha+\beta)(\alpha+\beta+1)}-A^2-\frac{2\alpha A(B-A)}{\alpha+\beta}-(B-A)^2\cdot\frac{\alpha^2}{(\alpha+\beta)^2}\\
&=\frac{\alpha(\alpha+1)(B-A)^2}{(\alpha+\beta)(\alpha+\beta+1)}-\frac{\alpha^2(B-A)^2}{(\alpha+\beta)^2}\\
&=\frac{\alpha(B-A)^2}{\alpha+\beta}\cdot\biggl[\frac{\alpha+1}{\alpha+\beta+1}-\frac{\alpha}{\alpha+\beta}\biggr]\\
&=\frac{\alpha(B-A)^2}{\alpha+\beta}\cdot\frac{(\alpha+1)(\alpha+\beta)-\alpha(\alpha+\beta+1)}{(\alpha+\beta)(\alpha+\beta+1)}\\
&=\frac{\alpha(B-A)^2}{\alpha+\beta}\cdot\frac{\beta}{(\alpha+\beta)(\alpha+\beta+1)}\\
&=\frac{\alpha\beta(B-A)^2}{(\alpha+\beta)^2(\alpha+\beta+1)}
\end{split}
\]