2.1
(1) 規格化条件より, \begin{align*} \int_{-\infty}^\infty f(x)dx&=\int_0^1 cx^4(1-x)^3dx\\ &=c\frac{4!\cdot3!}{(4+3+1)!}\\ &=\frac{c}{280} \end{align*} となるので, これより$c=280$である. ここでベータ関数の公式
\begin{align*} \mathrm{B}(p, q)&=\int_0^1x^{p-1}(1-x)^{q-1}dx\\ &=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} \end{align*} を用いた. また, \begin{align*} E(X)&=\int_{-\infty}^\infty xf(x)dx\\
&=280\int_0^1 x^5(1-x)^3dx\\ &=\frac{280\cdot5!\cdot3!}{(5+3+1)!}\\ &=\frac{5}{9}\\\\ V(X)&=E(X^2)-[E(X)]^2\\ &=\int_{-\infty}^\infty x^2f(x)dx-\left(\frac{5}{9}\right)^2\\
&=280\int_0^1 x^6(1-x)^3dx-\frac{25}{81}\\ &=\frac{280\cdot6!\cdot3!}{(6+3+1)!}-\frac{25}{81}\\ &=\frac{1}{3}-\frac{25}{81}=\frac{2}{81} \end{align*} となる.
(2)
(2)
2.2
(1) \begin{align*} E[X(X-1)]&=E(X^2)-E(X)\\ &=E(X^2)-{[E(X)]}^2+{[E(X)]}^2-E(X)\\ &=\sigma^2+\mu^2-\mu \end{align*}
(2) \begin{align*} E[X(X+5)]&=E(X^2)+5E(X)\\ &=E(X^2)-{[E(X)]}^2+{[E(X)]}^2+5E(X)\\ &=\sigma^2+\mu^2+5\mu \end{align*}
(2) \begin{align*} E[X(X+5)]&=E(X^2)+5E(X)\\ &=E(X^2)-{[E(X)]}^2+{[E(X)]}^2+5E(X)\\ &=\sigma^2+\mu^2+5\mu \end{align*}
2.3
連続型で考える. \begin{align*} M_1(a)&=E(|x-a|)\\ &=\int_{-\infty}^\infty |x-a|f(x)dx\\ &=\int_{-\infty}^a -(x-a)f(x)dx+\int_{a}^\infty (x-a)f(x)dx\\ &=\int_{-\infty}^a
-xf(x)dx+aF(a)+\int_{a}^\infty xf(x)dx-a[1-F(a)]\\ &=2aF(a)-a-\int_{-\infty}^a xf(x)dx+\int_{a}^\infty xf(x)dx \end{align*} が成り立つ. 期待値が存在することから, $xf(x)は\pm\infty$において$0$をとるので, \begin{align*}
\frac{\partial M_1(a)}{\partial a}&=2[F(a)+aF'(a)]-1-af(a)-af(a)\\ &=2F(a)-1\ \ \ \left(\because f(x)=\dfrac{dF(x)}{dx}\right) \end{align*} となる. これより, \begin{align*} \frac{\partial
M_1(a)}{\partial a}=0 &\Leftrightarrow F(a)=\frac{1}{2}\\ &\Leftrightarrow a=m \end{align*} が成り立つ. ゆえに$M_1(a)$は$a=m$で, 最小値$M_1(m)=E[|X-m|]$をとる.
2.4
$\min M_2(a)=\sigma^2$を示す. \begin{align*} \frac{\partial M_2(a)}{\partial a}&=\frac{\partial}{\partial a}E[(x-a^2)]\\ &=\frac{\partial}{\partial a}\left[E(X^2)-2aE(X)+a^2\right]\\
&=2a-2E(X)=2(a-\mu) \end{align*} が成り立つので, \begin{align*} \frac{\partial M_2(a)}{\partial a}=0 &\Leftrightarrow a=\mu \end{align*} となる. ゆえに$M_2(a)$は$a=\mu$で,
最小値$M_2(\mu)=E[(X-\mu)^2]=\sigma^2$をとる.
2.5
規格化条件より, \begin{align*} \int_{-\infty}^\infty f(x)dx&=\int_{-\infty}^\infty \frac{1}{n}\sum_{i=1}^n f_i(x)dx\\ &=\frac{1}{n}\sum_{i=1}^n 1\\ &=\frac{1}{n}\cdot {n}=1 \end{align*} となる.
よって平均と分散は, \begin{align*} E(X)&=\int_{-\infty}^\infty xf(x)dx\\ &=\frac{1}{n}\sum_{i=1}^n \int_{-\infty}^\infty xf_i(x)dx\\ &=\frac{1}{n}\sum_{i=1}^n E(X_i)=\frac{1}{n}\sum_{i=1}^n
\mu_i\\\\ V(X)&=E(X^2)-[E(X)]^2\\ &=\frac{1}{n}\sum_{i=1}^n \int_{-\infty}^\infty x^2f_i(x)dx-\mu^2\\ &=\frac{1}{n}\sum_{i=1}^n ({\sigma_i}^2+{\mu_i}^2)-2\mu^2+\mu^2\ \ \ \left(\because
{\sigma_i}^2=\displaystyle\int_{-\infty}^\infty x^2f_i(x)dx-{\mu_i}^2\right)\\ &=\frac{1}{n}\sum_{i=1}^n {\sigma_i}^2+\frac{1}{n}\sum_{i=1}^n {\mu_i}^2-2\mu\cdot\frac{1}{n}\sum_{i=1}^n
{\mu_i}+\frac{1}{n}\cdot n \mu^2\\ &=\frac{1}{n}\sum_{i=1}^n {\sigma_i}^2+\frac{1}{n}\sum_{i=1}^n (\mu_i-\mu)^2 \end{align*}
2.6
2.7
2.8
(1) 確率母関数は, \begin{align*} P(t)=E(t^Y)&=\sum_{y=0}^\infty \frac{t^y \theta^y f(y)}{A(\theta)}\\ &=\sum_{y=0}^\infty \frac{{(t\theta)}^y f(y)}{A(\theta)}=\frac{A(t\theta)}{A(\theta)}
\end{align*} となる.
(2) 積率母関数は, \begin{align*} M(t)=P(e^t)&=\frac{A(e^t \theta)}{A(\theta)}\\ &=\sum_{y=0}^\infty \frac{{(e^t \theta)}^y f(y)}{A(\theta)}\\ &=\sum_{y=0}^\infty \frac{e^{ty} \theta^y f(y)}{A(\theta)} \end{align*} となる. ここで, \begin{align*} M'(t)&=\frac{\partial M(t)}{\partial t}=\sum_{y=0}^\infty \frac{y e^{ty} \theta^y f(y)}{A(\theta)}\\ M''(t)&=\frac{\partial^2 M(t)}{\partial t^2}=\sum_{y=0}^\infty \frac{y^2 e^{ty} \theta^y f(y)}{A(\theta)} \end{align*} が成り立つ. ゆえに平均と分散は, \begin{align*} E(Y)&=M'(0)\\ &=\sum_{y=0}^\infty \frac{y\theta^y f(y)}{A(\theta)}\\ &=\frac{\theta}{A(\theta)}\sum_{y=0}^\infty y{\theta}^{y-1} f(y)\\ &=\frac{\theta A'(\theta)}{A(\theta)}\\ &=\mu\\\\ V(Y)&=M''(0)-[M'(0)]^2\\ &=\left[\frac{\theta^2}{A(\theta)}\sum_{y=0}^\infty y^2{\theta}^{y-2} f(y)\right]-[M'(0)]^2\\ &=\left[\frac{\theta^2}{A(\theta)}\sum_{y=0}^\infty y(y-1){\theta}^{y-2} f(y)\right]+\left[\frac{\theta}{A(\theta)}\sum_{y=0}^\infty y{\theta}^{y-1} f(y)\right]-[M'(0)]^2\\ &=\frac{\theta^2 A''(\theta)}{A(\theta)}+\frac{\theta A'(\theta)}{A(\theta)}-\left[\frac{\theta A'(\theta)}{A(\theta)}\right]^2\\ &=\sigma^2 \end{align*} となる.
(3) 与えられた式を計算すると, \begin{align*} \theta\frac{d\mu}{d\theta}&=\theta\cdot\frac{\left[A'(\theta)+\theta A''(\theta)\right]A(\theta)-\theta[A'(\theta)]^2}{[A(\theta)]^2}\\ &=\frac{\theta^2 A''(\theta)}{A(\theta)}+\frac{\theta A'(\theta)}{A(\theta)}-\left[\frac{\theta A'(\theta)}{A(\theta)}\right]^2\\ &=\sigma^2 \end{align*} が成り立つ.
(2) 積率母関数は, \begin{align*} M(t)=P(e^t)&=\frac{A(e^t \theta)}{A(\theta)}\\ &=\sum_{y=0}^\infty \frac{{(e^t \theta)}^y f(y)}{A(\theta)}\\ &=\sum_{y=0}^\infty \frac{e^{ty} \theta^y f(y)}{A(\theta)} \end{align*} となる. ここで, \begin{align*} M'(t)&=\frac{\partial M(t)}{\partial t}=\sum_{y=0}^\infty \frac{y e^{ty} \theta^y f(y)}{A(\theta)}\\ M''(t)&=\frac{\partial^2 M(t)}{\partial t^2}=\sum_{y=0}^\infty \frac{y^2 e^{ty} \theta^y f(y)}{A(\theta)} \end{align*} が成り立つ. ゆえに平均と分散は, \begin{align*} E(Y)&=M'(0)\\ &=\sum_{y=0}^\infty \frac{y\theta^y f(y)}{A(\theta)}\\ &=\frac{\theta}{A(\theta)}\sum_{y=0}^\infty y{\theta}^{y-1} f(y)\\ &=\frac{\theta A'(\theta)}{A(\theta)}\\ &=\mu\\\\ V(Y)&=M''(0)-[M'(0)]^2\\ &=\left[\frac{\theta^2}{A(\theta)}\sum_{y=0}^\infty y^2{\theta}^{y-2} f(y)\right]-[M'(0)]^2\\ &=\left[\frac{\theta^2}{A(\theta)}\sum_{y=0}^\infty y(y-1){\theta}^{y-2} f(y)\right]+\left[\frac{\theta}{A(\theta)}\sum_{y=0}^\infty y{\theta}^{y-1} f(y)\right]-[M'(0)]^2\\ &=\frac{\theta^2 A''(\theta)}{A(\theta)}+\frac{\theta A'(\theta)}{A(\theta)}-\left[\frac{\theta A'(\theta)}{A(\theta)}\right]^2\\ &=\sigma^2 \end{align*} となる.
(3) 与えられた式を計算すると, \begin{align*} \theta\frac{d\mu}{d\theta}&=\theta\cdot\frac{\left[A'(\theta)+\theta A''(\theta)\right]A(\theta)-\theta[A'(\theta)]^2}{[A(\theta)]^2}\\ &=\frac{\theta^2 A''(\theta)}{A(\theta)}+\frac{\theta A'(\theta)}{A(\theta)}-\left[\frac{\theta A'(\theta)}{A(\theta)}\right]^2\\ &=\sigma^2 \end{align*} が成り立つ.
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