跳到主要内容

京都大学 情報学研究科 知能情報学専攻 2024年8月実施 専門科目 S-1

Author

itsuitsuki

Description

Q.1

Suppose the probability density function f(x)f(x) of a random variable XX is as follows.

f(x)={0(x<0)cx(3x)(0x<3)0(3x)f(x) = \begin{cases} 0 & (x < 0) \\ cx(3-x) & (0 \leq x < 3) \\ 0 & (3 \leq x) \end{cases}

cc is a positive constant (c>0c > 0).

(1) Compute the value of the constant cc.

(2) Compute the mean and variance of the random variable XX.

Q.2

Let XX and YY be independent random variables following the binomial distributions B(m,p)\mathrm{B}(m, p) and B(n,p)\mathrm{B}(n, p), respectively. Derive the distribution of Z=X+YZ = X + Y.

Q.3

Consider normal populations A and B with a common population variance. One sample of size 18 is selected from the normal population A, denoted as (x1,x2,,x18)(x_1, x_2, \dots, x_{18}), and the other one of size 18 from the normal population B, denoted as (y1,y2,,y18)(y_1, y_2, \dots, y_{18}). The statistics derived from the samples are as follows.

xˉ=118i=118xisx2=117i=118(xixˉ)2yˉ=118i=118yisy2=117i=118(yiyˉ)2sxy=117i=118(xixˉ)(yiyˉ) \bar{x} = \frac{1}{18} \sum_{i=1}^{18} x_i \qquad s_x^2 = \frac{1}{17} \sum_{i=1}^{18} (x_i - \bar{x})^2 \\ \bar{y} = \frac{1}{18} \sum_{i=1}^{18} y_i \qquad s_y^2 = \frac{1}{17} \sum_{i=1}^{18} (y_i - \bar{y})^2 \qquad s_{xy} = \frac{1}{17} \sum_{i=1}^{18} (x_i - \bar{x})(y_i - \bar{y})

The values 2.110 and 2.032 may be used for the upper 2.5% point of the tt-distribution with 17 and 34 degrees of freedom, respectively.

(1) Assuming that the samples are paired as (xi,yi)(x_i, y_i), i=1,2,,18i = 1, 2, \dots, 18 and randomly selected, compute the 95% confidence interval for the difference between the two population means using necessary statistics among those mentioned above.

(2) Assuming that the samples are unpaired and randomly selected from each population, compute the 95% confidence interval for the difference between the two population means using necessary statistics among those mentioned above.

Q.4

Consider a linear regression model Yi=α+βxi+ϵi (i=1,2,,16)Y_i = \alpha + \beta x_i + \epsilon_i \ (i = 1, 2, \dots, 16) where the variation in random variables Yi (i=1,2,,16)Y_i \ (i = 1, 2, \dots, 16) are explained by the corresponding constants xi (i=1,2,,16)x_i \ (i = 1, 2, \dots, 16) with the regression coefficients of α\alpha and β\beta. Assume that ϵi (i=1,2,,16)\epsilon_i \ (i = 1, 2, \dots, 16) are independent and follow a normal distribution N(0,σ2)\mathrm{N}(0, \sigma^2) with mean 0 and variance σ2\sigma^2. Let α^\hat{\alpha} and β^\hat{\beta} be the least squares estimators of α\alpha and β\beta, respectively.

(1) Compute the standard deviation of β^\hat{\beta}.

(2) Given a new constant x17x_{17}, show the 95% prediction interval of Y17=α+βx17+ϵ17Y_{17} = \alpha + \beta x_{17} + \epsilon_{17} using α^,β^,σ\hat{\alpha}, \hat{\beta}, \sigma, and x1,x2,,x17x_1, x_2, \dots, x_{17} where ϵ17\epsilon_{17} is independent of ϵi (i=1,2,,16)\epsilon_i \ (i = 1, 2, \dots, 16) and follows a normal distribution N(0,σ2)\mathrm{N}(0, \sigma^2). The value 2.145 for the upper 2.5% point of the tt-distribution with 14 degrees of freedom may be used.

题目描述

  1. 随机变量 XX 的概率密度函数为

    f(x)={0(x<0),cx(3x)(0x<3),0(x3),f(x)= \begin{cases} 0 & (x<0),\\ cx(3-x) & (0\leq x<3),\\ 0 & (x\geq 3), \end{cases}

    其中 c>0c>0。(1)求常数 cc;(2)求 XX 的均值与方差。

  2. XXYY 相互独立,且分别服从二项分布 B(m,p)\mathrm{B}(m,p)B(n,p)\mathrm{B}(n,p)。推导 Z=X+YZ=X+Y 的概率分布。

  3. 从方差相同的两个正态总体 A、B 中分别随机抽取容量为 18 的样本 x1,,x18x_1,\ldots,x_{18}y1,,y18y_1,\ldots,y_{18}。定义

    xˉ=118i=118xi,sx2=117i=118(xixˉ)2,\bar{x}=\frac{1}{18}\sum_{i=1}^{18}x_i,\qquad s_x^2=\frac{1}{17}\sum_{i=1}^{18}(x_i-\bar{x})^2,
    yˉ=118i=118yi,sy2=117i=118(yiyˉ)2,\bar{y}=\frac{1}{18}\sum_{i=1}^{18}y_i,\qquad s_y^2=\frac{1}{17}\sum_{i=1}^{18}(y_i-\bar{y})^2,

    以及

    sxy=117i=118(xixˉ)(yiyˉ).s_{xy}=\frac{1}{17}\sum_{i=1}^{18}(x_i-\bar{x})(y_i-\bar{y}).

    自由度为 17 和 34 的 tt 分布上侧 2.5%2.5\% 分位点可分别取 2.110 和 2.032。(1)若两组样本按 (xi,yi)(x_i,y_i) 配对且为随机抽样,请使用上述必要统计量求两个总体均值之差的 95%95\% 置信区间;(2)若两组样本不配对、分别从两个总体随机抽取,请求该均值差的 95%95\% 置信区间。

  4. 考虑线性回归模型

    Yi=α+βxi+ϵi(i=1,2,,16),Y_i=\alpha+\beta x_i+\epsilon_i\qquad(i=1,2,\ldots,16),

    其中 xix_i 为常数,ϵi\epsilon_i 相互独立且服从 N(0,σ2)\mathrm{N}(0,\sigma^2)α^\hat{\alpha}β^\hat{\beta} 分别为 α\alphaβ\beta 的最小二乘估计量。(1)求 β^\hat{\beta} 的标准差;(2)给定新的常数 x17x_{17},设

    Y17=α+βx17+ϵ17,Y_{17}=\alpha+\beta x_{17}+\epsilon_{17},

    其中 ϵ17\epsilon_{17} 与前 16 个误差项独立且同样服从 N(0,σ2)\mathrm{N}(0,\sigma^2)。使用 α^,β^,σ\hat{\alpha},\hat{\beta},\sigma 以及 x1,,x17x_1,\ldots,x_{17} 写出 Y17Y_{17}95%95\% 预测区间。自由度为 14 的 tt 分布上侧 2.5%2.5\% 分位点可取 2.145。

考点

  • 概率密度归一化与矩:利用积分确定归一化常数,并由一阶、二阶矩计算均值和方差。
  • 独立二项随机变量之和:通过卷积或概率生成函数推导同成功概率下二项分布的可加性。
  • 配对与独立两样本置信区间:区分配对样本与等方差独立样本,选取相应方差估计和 tt 分布自由度。
  • 简单线性回归推断:计算斜率估计量的标准差,并构造新观测值的预测区间。