京都大学 情報学研究科 システム科学専攻 2018年8月実施 専門科目 確率統計
Author
uogxtc
Description
問題1
確率変数
(1) 同時確率
(2)
(3) 制約条件
(4) 設問 (3) の
問題2
袋の中に
(1)
(2) 確率変数
袋の中に白いボールが多数入っている。
その個数が分からないので未知パラメータ
(3)
(4) 設問 (3) の
(5)
Kai
問題1
(1)
The posterior is given by
and we easily obtain that
which is exactly
(2)
The likelihood is
and the log-likelihood is
Let
(1-\beta)\sum_{i=1}^nX_iY_i-\beta\sum_{i=1}^nX_i(1-Y_i)=0,
\hat{\beta}n=\frac{\sum{i=1}^nX_iY_i}{\sum_{i=1}^nX_i}.
\log L=\sum_{i=1}^{n} \Big{ X_{i}Y_{i}\log(\alpha(1-\alpha))+X_{i}(1-Y_{i})\log(\alpha^{2})\+(X_{i}-1)Y_{i}\log 0+(1-X_{i})(1-Y_{i})\log(1-\alpha) \Big}.
\hat{\alpha}_n=\frac{2\sum X_i-\sum X_iY_i}{n+\sum X_i-\sum Y_i-\sum X_iY_i}.
\begin{aligned} &\sum X_{i}\to n\mathbb{E}[X_i=1]=\alpha n\ &\sum Y_{i}\to n\mathbb{E}[Y_i=1]=\alpha\beta n\ &\sum X_{i}Y_{i}\to n\mathbb{E}[X_i=1,Y_i=1]=\alpha \beta n \end{aligned}
\lim\limits_{n\to\infty}\hat{\alpha}_n=\frac{2\alpha n-\alpha\beta n}{n+\alpha n-2\alpha\beta n}=\frac{2\alpha-\alpha\beta}{1+\alpha-2\alpha\beta}.
\Pr(X=k)=\frac{\binom{m}{k}\binom{N-m}{n-k}}{\binom{N}{n}}.
\mathbb{E}[X]=\sum_{i=1}^n\Pr(X=k)\cdot k
\begin{aligned} k\binom{m}{k}&=\frac{m!}{(k-1)!(m-k)!}\ &=\frac{(m-1)!m}{(k-1)!(m-k)!}\ &=m\binom{m-1}{k-1}. \end{aligned}
k \cdot \Pr(X=k)=\frac{m\binom{m-1}{k-1}\binom{N-m}{n-k}}{\binom{N}{n}}=\frac{m\binom{m-1}{k-1}\binom{(N-1)-(m-1)}{(n-1)-(k-1)}}{\frac{N}{n}\binom{N-1}{n-1}}.
\begin{aligned} \mathbb{E}[X]& =\sum_{k=1}^n\frac{mn}{N}\bigg[\frac{\binom{m-1}{k-1}\binom{(N-1)-(m-1)}{(n-1)-(k-1)}}{\binom{N-1}{n-1}}\bigg] \ &=\frac{mn}N\underbrace{\sum_{k=1}^n\left[\frac{\binom{m-1}{k-1}\binom{(N-1)-(m-1)}{(n-1)-(k-1)}}{\binom{N-1}{n-1}}\right]}_{=1,\text{ as all probabilities sum to 1.}} \ &=\frac{mn}{N}. \end{aligned}
L(N)=\underbrace{\Pr(X=k)}_{\text{ function of } n,k \text{ and parameterized by }N}=\frac{\binom{m}{k}\binom{N-m}{n-k}}{\binom{N}{n}}.
\begin{aligned} \frac{L(N)}{L(N-1)}& =\frac{\binom{N-m}{n-k}}{\binom{N-m-1}{n-k}}\cdot\frac{\binom{N-1}{n}}{\binom{N}{n}} \ &=\frac{N-m}{N-m-n+k}\cdot\frac{N-n}{N}. \end{aligned}
\begin{aligned} &\frac{N-m}{N-m-n+k} \cdot \frac{N-n}{N}\leq1,\ &\Rightarrow\quad N \geq \frac{mn}{k}, \end{aligned}
N \leq \frac{mn}{k},