Kyoto-University
京都大学 情報学研究科 知能情報学専攻 2022年8月実施 専門科目 S-2
Author Isidore
Description
Kai 設問1
\[
E[X] = \int^{1}_{-1}x(-\frac{1}{2}x+\frac{1}{2})\mathrm{d}x = -\frac{1}{3}
\]
\[
E[X^2] = \int^{1}_{-1}x^2(-\frac{1}{2}x+\frac{1}{2})\mathrm{d}x = \frac{1}{3} \\
\]
\[
Var[X] = E[X^2] - E^2[X] = \frac{2}{9}
\]
設問2
By the convolution Rule, we have
\[
\begin{align}
f_{Z}(z) &= P(Z=z) = \sum^{z}_{i=0}f_{X}(i)f_{Y}(z-i) \\
& = \sum^{z}_{i=0} \frac{\lambda_1^{i}e^{-\lambda_1}}{i!} \frac{\lambda_2^{z - i}e^{-\lambda_2}}{(z - i)!} \\
& = \frac{e^{-(\lambda_1 + \lambda_2)}}{z!}\sum^{z}_{i=0}\frac{z!}{(z-i)!i!}\lambda_1^{i}\lambda_2^{z - i}
\end{align}
\]
By the Binomial Theorem, we can insert \(\sum^{z}_{i=0}\frac{z!}{(z-i)!i!}\lambda_1^{i}\lambda_2^{z - i} = (\lambda_1 + \lambda_2)^z\)
\[
f_{Z}(z) = \frac{e^{-(\lambda_1 + \lambda_2)}}{z!}(\lambda_1 + \lambda_2)^z
\]
So is the PMF for a Poisson Distribution with the parameter \((\lambda_1 + \lambda_2)\)
PS : A easier solution is to use Moments Generating Function.
設問3
(1)
By the Least Square Method, we have the sum of square residuals,
\[
S(\hat{\beta}) = \sum^{n}_{i=1}(y_i - \hat{\beta}x_i)^2
\]
Calculate its derivative with the root
\[
\begin{align}
S'(\hat{\beta}) &= \sum^{n}_{i=1}(-2x_i)(y_i-\hat{\beta}x_i) \\
&= 2(\sum^{n}_{i=1}\hat{\beta}x_i^2 - \sum^{n}_{i=1}x_iy_i) \\
\hat{\beta} &= \frac{\sum^{n}_{i=1}x_iy_i}{\sum^{n}_{i=1}x_i^2}
\end{align}
\]
(2)
By the equation (\(5\) ) and (\(6\) ), we immediately have
\[
\begin{align}
0 &= (\sum^{n}_{i=1}\hat{\beta}x_i^2 - \sum^{n}_{i=1}x_iy_i) \\
&= \sum^{n}_{i=1}x_i(y_i-\hat{\beta}x_i) \\
&= \sum^{n}_{i=1}x_i\hat{\epsilon_i}
\end{align}
\]
設問4
(1)
Denote the events below:
The hypothesis is true: \(T\)
The hypothesis is false: \(F\)
The test results in significance: \(S\)
Then, what is asked can be represented by the probability as \(Pr[T|S]\) .
Given the ratio \(R:1\) , we have
\[
Pr[T] = \frac{R}{R+1} ;\; Pr[F] = \frac{1}{R+1}
\]
By the definition of Significance Level and Statistic Power, we have
\[
Pr[S|F] = \alpha ;\; Pr[S|T] = 1 - \beta
\]
Therefore, with the Bayes' theorem, we have
\[
\begin{align}
Pr[T|S] &= \frac{Pr[S\cap T]}{Pr[S]} \\
&= \frac{Pr[S|T]Pr[T]}{Pr[S|T]Pr[T]+Pr[S|F]Pr[F]} \\
&=\frac{(1-\beta)R}{\alpha+(1-\beta)R}
\end{align}
\]
Insert the values, the answer is
\[
Pr[T|S] = \frac{8}{13}
\]
(2)
By perform the experiments \(k\) times independently, we only need to multiply the probabilities with event \(S\) \(k\) times in equation (\(11\) ), which means,
\[
\begin{align}
Pr[T|S^k] &= \frac{Pr[S^k|T]Pr[T]}{Pr[S^k|T]Pr[T]+Pr[S^k|F]Pr[F]} \\
&= \frac{Pr^k[S|T]Pr[T]}{Pr^k[S|T]Pr[T]+Pr^k[S|F]Pr[F]} \\
&= \frac{(1-\beta)^kR}{\alpha^k+(1-\beta)^kR}
\end{align}
\]
Insert the values, the answer is
\[
Pr[T|S^2] = \frac{128}{133}
\]
