中科院机器学习题库new.docx

1、中科院机器学习题库new机器学习题库一、 极大似然1、 ML estimation of exponential model (10)A Gaussian distribution is often used to model data on the real line, but is sometimes inappropriate when the data are often close to zero but constrained to be nonnegative. In such cases one can fit an exponential distribution, whos

2、e probability density function is given byGiven N observations xi drawn from such a distribution:(a) Write down the likelihood as a function of the scale parameter b.(b) Write down the derivative of the log likelihood.(c) Give a simple expression for the ML estimate for b.2、换成Poisson分布：3、二、 贝叶斯假设在考试

3、的多项选择中，考生知道正确答案的概率为p，猜测答案的概率为1-p，并且假设考生知道正确答案答对题的概率为1，猜中正确答案的概率为，其中m为多选项的数目。那么已知考生答对题目，求他知道正确答案的概率。1、 Conjugate priorsThe readings for this week include discussion of conjugate priors. Given a likelihood for a class models with parameters , a conjugate prior is a distribution with hyperparameters ,

4、such that the posterior distribution与先验的分布族相同(a) Suppose that the likelihood is given by the exponential distribution with rate parameter :Show that the gamma distribution _is a conjugate prior for the exponential. Derive the parameter update given observations and the prediction distribution .(b) S

5、how that the beta distribution is a conjugate prior for the geometric distributionwhich describes the number of time a coin is tossed until the first heads appears, when the probability of heads on each toss is . Derive the parameter update rule and prediction distribution.(c) Suppose is a conjugate

6、 prior for the likelihood ; show that the mixture prioris also conjugate for the same likelihood, assuming the mixture weights wm sum to 1. (d) Repeat part (c) for the case where the prior is a single distribution and the likelihood is a mixture, and the prior is conjugate for each mixture component

7、 of the likelihood.some priors can be conjugate for several different likelihoods; for example, the beta is conjugate for the Bernoulliand the geometric distributions and the gamma is conjugate for the exponential and for the gamma with fixed (e) (Extra credit, 20) Explore the case where the likelih

8、ood is a mixture with fixed components and unknown weights; i.e., the weights are the parameters to be learned.三、判断题（1）给定n个数据点，如果其中一半用于训练，另一半用于测试，则训练误差和测试误差之间的差别会随着n的增加而减小。（2）极大似然估计是无偏估计且在所有的无偏估计中方差最小，所以极大似然估计的风险最小。（）回归函数A和B，如果A比B更简单，则A几乎一定会比B在测试集上表现更好。（）全局线性回归需要利用全部样本点来预测新输入的对应输出值，而局部线性回归只需利用查询点附近的

9、样本来预测输出值。所以全局线性回归比局部线性回归计算代价更高。（）Boosting和Bagging都是组合多个分类器投票的方法，二者都是根据单个分类器的正确率决定其权重。() In the boosting iterations, the training error of each new decision stump and the training error of the combined classifier vary roughly in concert （F）While the training error of the combined classifier typically

10、 decreases as a function of boosting iterations, the error of the individual decision stumps typically increases since the example weights become concentrated at the most difficult examples.() One advantage of Boosting is that it does not overfit. （F）() Support vector machines are resistant to outli

11、ers, i.e., very noisy examples drawn from a different distribution. （）（9）在回归分析中，最佳子集选择可以做特征选择，当特征数目较多时计算量大；岭回归和Lasso模型计算量小，且Lasso也可以实现特征选择。（10）当训练数据较少时更容易发生过拟合。（11）梯度下降有时会陷于局部极小值，但EM算法不会。（12）在核回归中，最影响回归的过拟合性和欠拟合之间平衡的参数为核函数的宽度。(13) In the AdaBoost algorithm, the weights on all the misclassified point

12、s will go up by the same multiplicative factor. （T）(14) True/False: In a least-squares linear regression problem, adding an L2 regularization penalty cannot decrease the L2 error of the solution w on the training data. （F）(15) True/False: In a least-squares linear regression problem, adding an L2 re

13、gularization penalty always decreases the expected L2 error of the solution w on unseen test data （F）.(16)除了EM算法，梯度下降也可求混合高斯模型的参数。 (T)(20) Any decision boundary that we get from a generative model with class-conditional Gaussian distributions could in principle be reproduced with an SVM and a polyno

14、mial kernel.True! In fact, since class-conditional Gaussians always yield quadratic decision boundaries, they can be reproduced with an SVM with kernel of degree less than or equal to two.(21) AdaBoost will eventually reach zero training error, regardless of the type of weak classifier it uses, prov

15、ided enough weak classifiers have been combined.False! If the data is not separable by a linear combination of the weak classifiers, AdaBoost cant achieve zero training error.(22) The L2 penalty in a ridge regression is equivalent to a Laplace prior on the weights. （F）(23) The log-likelihood of the

16、data will always increase through successive iterations of the expectation maximation algorithm. (F)(24) In training a logistic regression model by maximizing the likelihood of the labels given the inputs we have multiple locally optimal solutions. (F)一、 回归1、考虑回归一个正则化回归问题。在下图中给出了惩罚函数为二次正则函数，当正则化参数C取

17、不同值时，在训练集和测试集上的log似然（mean log-probability）。（10分）（1）说法“随着C的增加，图2中训练集上的log似然永远不会增加”是否正确，并说明理由。（2）解释当C取较大值时，图2中测试集上的log似然下降的原因。2、考虑线性回归模型：，训练数据如下图所示。（10分）（1）用极大似然估计参数，并在图（a）中画出模型。（3分）（2）用正则化的极大似然估计参数，即在log似然目标函数中加入正则惩罚函数，并在图（b）中画出当参数C取很大值时的模型。（3分）（3）在正则化后，高斯分布的方差是变大了、变小了还是不变？（4分） 图(a) 图(b)3. 考虑二维输入空间点上

18、的回归问题，其中在单位正方形。训练样本和测试样本在单位正方形中均匀分布，输出模型为，我们用1-10阶多项式特征，采用线性回归模型来学习x与y之间的关系（高阶特征模型包含所有低阶特征），损失函数取平方误差损失。(1) 现在个样本上，训练1阶、2阶、8阶和10阶特征的模型，然后在一个大规模的独立的测试集上测试，则在下3列中选择合适的模型（可能有多个选项），并解释第3列中你选择的模型为什么测试误差小。（10分）训练误差最小训练误差最大测试误差最小1阶特征的线性模型X2阶特征的线性模型X8阶特征的线性模型X10阶特征的线性模型X(2) 现在个样本上，训练1阶、2阶、8阶和10阶特征的模型，然后在一个大

19、规模的独立的测试集上测试，则在下3列中选择合适的模型（可能有多个选项），并解释第3列中你选择的模型为什么测试误差小。（10分）训练误差最小训练误差最大测试误差最小1阶特征的线性模型X2阶特征的线性模型8阶特征的线性模型XX10阶特征的线性模型X(3) The approximation error of a polynomial regression model depends on the number of training points. (T)(4) The structural error of a polynomial regression model depends on the

20、 number of training points. (F)4、We are trying to learn regression parameters for a dataset which we know was generated from a polynomial of a certain degree, but we do not know what this degree is. Assume the data was actually generated from a polynomial of degree 5 with some added Gaussian noise (

21、that is .For training we have 100 x,y pairs and for testing we are using an additional set of 100 x,y pairs. Since we do not know the degree of the polynomial we learn two models from the data. Model A learns parameters for a polynomial of degree 4 and model B learns parameters for a polynomial of d

22、egree 6. Which of these two models is likely to fit the test data better?Answer: Degree 6 polynomial. Since the model is a degree 5 polynomial and we have enough training data, the model we learn for a six degree polynomial will likely fit a very small coefficient for x6 . Thus, even though it is a

23、six degree polynomial it will actually behave in a very similar way to a fifth degree polynomial which is the correct model leading to better fit to the data.5、Input-dependent noise in regressionOrdinary least-squares regression is equivalent to assuming that each data point is generated according t

24、o a linear function of the input plus zero-mean, constant-variance Gaussian noise. In many systems, however, the noise variance is itself a positive linear function of the input (which is assumed to be non-negative, i.e., x = 0).a) Which of the following families of probability models correctly desc

25、ribes this situation in the univariate case? (Hint: only one of them does.)(iii) is correct. In a Gaussian distribution over y, the variance is determined by the coefficient of y2; so by replacing by , we get a variance that increases linearly with x. (Note also the change to the normalization “cons

26、tant.”) (i) has quadratic dependence on x; (ii) does not change the variance at all, it just renames w1.b) Circle the plots in Figure 1 that could plausibly have been generated by some instance of the model family(ies) you chose.(ii) and (iii). (Note that (iii) works for .) (i) exhibits a large vari

27、ance at x = 0, and the variance appears independent of x.c) True/False: Regression with input-dependent noise gives the same solution as ordinary regression for an infinite data set generated according to the corresponding model.True. In both cases the algorithm will recover the true underlying mode

28、l.d) For the model you chose in part (a), write down the derivative of the negative log likelihood with respect to w1.二、 分类1. 产生式模型 vs. 判别式模型(a) points Your billionaire friend needs your help. She needs to classify job applications into good/bad categories, and also to detect job applicants who lie

29、in their applications using density estimation to detect outliers. To meet these needs, do you recommend using a discriminative or generative classifier? Why? final_sol_s07产生式模型因为要估计密度(b) points Your billionaire friend also wants to classify software applications to detect bug-prone applications usi

30、ng features of the source code. This pilot project only has a few applications to be used as training data, though. To create the most accurate classifier, do you recommend using a discriminative or generative classifier? Why?判别式模型样本数较少，通常用判别式模型直接分类效果会好些(d) points Finally, your billionaire friend al

31、so wants to classify companies to decide which one to acquire. This project has lots of training data based on several decades of research. To create the most accurate classifier, do you recommend using a discriminative or generative classifier? Why?产生式模型样本数很多时，可以学习到正确的产生式模型2、logstic回归 Figure 2: Log-probability of labels as a function of regularization parameter CHere we use a logistic regression model to solve a classification problem. In Figure 2, we have plotted the mean log-probability of labels in the training and test sets after having trained the classifier with quad