questions stringlengths 80 1.7k | answers stringlengths 74 1.45k |
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True or False: For a fixed number of observations in a data set, introducing more vari ables normally generates a model that has a better fit to the data. What may be the drawback of such a model fitting strategy? | True. However, when an excessive and unnecessary number of variables is used in a lo gistic regression model, peculiarities (e.g., specific attributes) of the underlying data set dis proportionately affect the coefficients in the model, a phenomena commonly referred to as “overfitting”. Therefore, it is important that a logistic regression model does not start training with more variables than is justified for the given number of observations. 2.3. SOLUTIONS The odds of success are defined as the ratio between the probability of success p ∈ [0, 1] |
Define the term “odds of success” both qualitatively and formally. Give a numerical example that stresses the relation between probability and odds of an event occurring. | and the probability of failure 1 − p. Formally: Odds(p) ≡ ! p 1 − p . For instance, assuming the probability of success of an event is p = 0.7. Then, in our example, the odds of success are 7/3, or 2.333 to 1. Naturally, in the case of equal probabilities where p = 0.5, the odds of success is 1 to 1. |
1. Define what is meant by the term "interaction", in the context of a logistic regression predictor variable. LOGISTIC REGRESSION 2. What is the simplest form of an interaction? Write its formulae. 3. What statistical tests can be used to attest the significance of an interaction term? | 1. An interaction is the product of two single predictor variables implying a nonadditive effect. 2. The simplest interaction model includes a predictor variable formed by multiplying two ordinary predictors. Let us assume two variables X and Z. Then, the logistic regression model that employs the simplest form of interaction follows: β0 + β1X + β2Z + β3XZ, where the coefficient for the interaction term XZ is represented by predictor β3. 3. For testing the contribution of an interaction, two principal methods are commonly employed; the Wald chisquared test or a likelihood ratio test between the model with and without the interaction term. Note: How does interaction relates to information theory? What added value does it employ to enhance model performance? |
True or False: In machine learning terminology, unsupervised learning refers to the mapping of input covariates to a target response variable that is attempted at being predicted when the labels are known. | False. This is exactly the definition of supervised learning; when labels are known then supervision guides the learning process. LOGISTIC REGRESSION |
Complete the following sentence: In the case of logistic regression, the response vari able is the log of the odds of being classified in [...]. | In the case of logistic regression, the response variable is the log of the odds of being clas sified in a group of binary or multiclass responses. This definition essentially demonstrates that odds can take the form of a vector. |
Describe how in a logistic regression model, a transformation to the response variable is applied to yield a probability distribution. Why is it considered a more informative repres entation of the response? | When a transformation to the response variable is applied, it yields a probability distribu tion over the output classes, which is bounded between 0 and 1; this transformation can be employed in several ways, e.g., a softmax layer, the sigmoid function or classic normalization. This representation facilitates a softdecision by the logistic regression model, which permits construction of probabilitybased processes over the predictions of the model. Note: What are the pros and cons of each of the three aforementioned transformations? |
Complete the following sentence: Minimizing the negative log likelihood also means maximizing the [...] of selecting the [...] class. 2.2.2 Odds, Logodds | Minimizing the negative log likelihood also means maximizing the likelihood of selecting the correct class. 2.3.2 Odds, Logodds |
Assume the probability of an event occurring is p = 0.1. 1. What are the odds of the event occurring?. 2. What are the logodds of the event occurring?. 3. Construct the probability of the event as a ratio that equals 0.1. 2.2. PROBLEMS | 1. The odds of the event occurring are, by definition: odds = (0.1 0.9) = 0.11. 2. The logodds of the event occurring are simply taken as the log of the odds: logodds = ln(0.1/0.9) = −2.19685. 3. The probability may be constructed by the following representation: probability = odds odds + 1 = 0.11 1.11 = 0.1, or, alternatively: p = exp (ln odds) exp (ln odds) + 1 = 0.11 1.11 = 0.1. Note: What is the intuition behind this representation? 2.3. SOLUTIONS |
ility of success is 0.8. True or False: If the odds of success in a binary response is 4, the corresponding probab | True. By definition of odds, it is easy to notice that p = 0.8 satisfies the following relation: odds = (0.8 0.2) = 4 |
their respective odds. Draw a graph of odds to probabilities, mapping the entire range of probabilities to | The graph of odds to probabilities is depicted in Figure 2.12. Odds odds(p) = p 1−p 10,0 8,0 6,0 4,0 2,0 0,1 0,2 0,3 0,4 0,5 0,6 Probability 0,7 0,8 0,9 FIGURE 2.12: Odds vs. probability values. LOGISTIC REGRESSION |
The logistic regression model is a subset of a broader range of machine learning models known as generalized linear models (GLMs), which also include analysis of variance (AN OVA), vanilla linear regression, etc. There are three components to a GLM; identify these three components for binary logistic regression. | A binary logistic regression GLM consists of there components: 1. Random component: refers to the probability distribution of the response variable (Y ), e.g., binomial distribution for Y in the binary logistic regression, which takes on the values Y = 0 or Y = 1. 2. Systematic component: describes the explanatory variables: (X1, X2, ...) as a combination of linear predictors. The binary case does not constrain these variables to any degree. 3. Link function: specifies the link between random and systematic components. It says how the expected value of the response relates to the linear predictor of explanatory variables. Note: Assume that Y denotes whether a human voice activity was detected (Y = 1) or not (Y = 0) in a give time frame. Propose two systematic components and a link function adjusted for this task. |
logit forms the linear decision boundary: Let us consider the logit transformation, i.e., logodds. Assume a scenario in which the log Pr(Y = 1|X) Pr(Y = 0|X) ! = θ0 + θT X, for a given vector of systematic components X and predictor variables θ. Write the mathem atical expression for the hyperplane that describes the decision boundary. | The hyperplane is simply defined by: θ0 + θT X = 0. Note: Recall the use of the logit function and derive this decision boundary rigorously. |
of each other. True or False: The logit function and the natural logistic (sigmoid) function are inverses LOGISTIC REGRESSION 2.2.3 The Sigmoid The sigmoid (Fig. 2.1) also known as the logistic function, is widely used in binary classification and as a neuron activation function in artificial neural networks. σ(x) = 1 σ(x) = 1+e−4x 1+e−15x y 1,0 0,8 0,6 0,4 0,2 −1,0 −0,8 −0,6 −0,4 −0,2 0,2 0,4 0,6 0,8 x 1,0 FIGURE 2.1: Examples of two sigmoid functions. | True. The logit function is defined as: z(p) = logit(p) = log ! , p 1 − p 2.3. SOLUTIONS for any p ∈ [0, 1]. A simple set of algebraic equations yields the inverse relation: p(z) = exp z 1 + exp z , which exactly describes the relation between the output and input of the logistic function, also known as the sigmoid. 2.3.3 The Sigmoid |
Compute the derivative of the natural sigmoid function: σ(x) = 1 + e−x ∈ (0, 1). | or derivation via the softmax function. There are various approaches to solve this problem, here we provide two; direct derivation d 1. Direct derivation: dx σ(x) = d dx((1 + e 2. Softmax derivation: −x)−1) = −((1 + e −x)(−2)) d dx(1 + e −x) = e −x (1+e−x)2 . In a classification problem with mutually exclusive classes, where all of the values are positive and sum to one, a softmax activation function may be used. By definition, the softmax activation function consists of n terms, such that ∀i ∈ [1, n]: f(θi) = eθiP = 1 + e−θi k evk P . k6=i eθk To compute the partial derivative of 2.18, we treat all θk where k 6= i as constants and then differentiate θi using regular differentiation rules. For a given θi, let us define: X k6=i β = eθk, and f(θi) = It can now be shown that the derivative with respect to θi holds: = (1 + βe −θi)−1. 1 + βe−θi cid:16 (cid:17)−2 0(θi) = f 1 + βe −θi −θi, βe LOGISTIC REGRESSION which can take on the informative form of: 0(θi) = f(θi)(1 − f(θi)). f It should be noted that 2.21 holds for any constant β, and for β = 1 it clearly reduces to the sigmoid activation function. Note: Characterize the sigmoid function when its argument approaches 0,∞ and −∞. What undesired properties of the sigmoid function do this values entail when considered as an activation function? |
Remember that in logistic regression, the hypothesis function for some parameter vector β and measurement vector x is defined as: hβ(x) = g(βT x) = = P(y = 1|x; β), 1 + e−βT x 2.2. PROBLEMS Suppose the coefficients of a logistic regression model with independent variables are as where y holds the hypothesis value. follows: β0 = −1.5, β1 = 3, β2 = −0.5. Assume additionally, that we have an observation with the following values for the dependent variables: x1 = 1, x2 = 5. As a result, the logit equation becomes: logit = β0 + β1x1 + β2x2. 1. What is the value of the logit for this observation? 2. What is the value of the odds for this observation? 3. What is the value of P(y = 1) for this observation? 2.2.4 Truly Understanding Logistic Regression | 1. The logit value is simply obtained by substituting the values of the dependent variables and model coefficients into the linear logistic regression model, as follows: logit = β0 + β1x1 + β2x2 = −1.5 + 3 · 1 + −0.5 · 5 = −1. 2. According to the natural relation between the logit and the odds, the following holds: odds = elogit = eβ0+β1x1+β2x2 = e −1 = 0.3678794. 3. The odds ratio is, by definition: odds = P(y = 1) P(y = 0) , so the logistic response function is: P(y = 1) = 1 + e−logit = 1 1 + e1 = 0.2689414. 2.3.4 Truly Understanding Logistic Regression 2.3. SOLUTIONS |
Proton therapy (PT) [2] is a widely adopted form of treatment for many types of cancer including breast and lung cancer (Fig. 2.2). FIGURE 2.2: Pulmonary nodules (left) and breast cancer (right). A PT device which was not properly calibrated is used to simulate the treatment of cancer. As a result, the PT beam does not behave normally. A data scientist collects inform ation relating to this simulation. The covariates presented in Table 2.1 are collected during LOGISTIC REGRESSION the experiment. The columns Yes and No indicate if the tumour was eradicated or not, re spectively. Tumour eradication Cancer Type Yes Breast Lung No TABLE 2.1: Tumour eradication statistics. Referring to Table 2.1: 1. What is the explanatory variable and what is the response variable? 2. Explain the use of relative risk and odds ratio for measuring association. 3. Are the two variables positively or negatively associated? Find the direction and strength of the association using both relative risk and odds ratio. 4. Compute a 95% confidence interval (CI) for the measure of association. 5. Interpret the results and explain their significance. | 1. Tumour eradication (Y ) is the response variable and cancer type (X) is the explanatory variable. 2. Relative risk (RR) is the ratio of risk of an event in one group (e.g., exposed group) versus the risk of the event in the other group (e.g., nonexposed group). The odds ratio (OR) is the ratio of odds of an event in one group versus the odds of the event in the other group. 3. If we calculate odds ratio as a measure of association: ˆθ = 560 × 36 69 × 260 = 1.23745. And the logodds ratio is (log(1.23745)) = 0.213052: The odds ratio is larger than one, indicating that the odds for a breast cancer is more than the odds for a lung cancer to be eradicated. Notice however, that this result is too close to one, which prevents conclusive decision regarding the odds relation. Additionally, if we calculate relative risk as a measure of association: RR = 560+260 69+36 = 1.0392. 4. The 95% confidence interval for the oddsratio, θ is computed from the sample confid ence interval for log odds ratio: s cid:16 cid:17 log(ˆθ) ˆσ = 560 + 1 260 + 1 69 + 1 36 = 0.21886. Therefore, the 95% CI for log (θ) is: 0.213052 ± 1.95 × 0.21886 = (0.6398298,−0.2137241). LOGISTIC REGRESSION Therefore, the 95% CI for θ is: −0.210, e0.647) = (0.810, 1.909). (e 5. The CI (0.810, 1.909) contains 1, which indicates that the true odds ratio is not signi ficantly different from 1 and there is not enough evidence that tumour eradication is dependent on cancer type. |
Consider a system for radiation therapy planning (Fig. 2.3). Given a patient with a ma lignant tumour, the problem is to select the optimal radiation exposure time for that patient. A key element in this problem is estimating the probability that a given tumour will be erad icated given certain covariates. A data scientist collects information relating to this radiation therapy system. 2.2. PROBLEMS FIGURE 2.3: A multidetector positron scanner used to locate tumours. The following covariates are collected; X1 denotes time in milliseconds that a patient is irradiated with, X2 = holds the size of the tumour in centimeters, and Y notates a binary re sponse variable indicating if the tumour was eradicated. Assume that each response’ variable Yi is a Bernoulli random variable with success parameter pi, which holds: pi = eβ0+β1x1+β2x2 1 + eβ0+β1x1+β2x2 . The data scientist fits a logistic regression model to the dependent measurements and pro duces these estimated coefficients: ˆβ0 = −6, ˆβ1 = 0.05, ˆβ2 = 1. 1. Estimate the probability that, given a patient who undergoes the treatment for 40 milliseconds and who is presented with a tumour sized 3.5 centimetres, the system eradicates the tumour. 2. How many milliseconds the patient in part (a) would need to be radiated with to have exactly a 50% chance of eradicating the tumour? LOGISTIC REGRESSION | 1. By using the defined values for X1 and X2, and the known logistic regression model, substitution yields: ˆp(X) = −6+0.05X1+X2 e (1 + e−6+0.05X1+X2) = 0.3775. 2. The equation for the predicted probability tells us that: −6+0.05X1+3.5 e (1 + e−6+0.05X1+3.5) = 0.5, which is equivalent to constraining: −6+0.05X1+3.5 = 1. e By taking the logarithm of both sides, we get that the number of milliseconds needed is: X1 = 2.5 0.05 = 50. |
Recent research [3] suggests that heating mercury containing dental amalgams may cause the release of toxic mercury fumes into the human airways. It is also presumed that drinking hot coffee, stimulates the release of mercury vapour from amalgam fillings (Fig. 2.4). FIGURE 2.4: A dental amalgam. To study factors that affect migraines, and in particular, patients who have at least four dental amalgams in their mouth, a data scientist collects data from 200K users with and without dental amalgams. The data scientist then fits a logistic regression model with an indicator of a second migraine within a time frame of one hour after the onset of the first mi graine, as the binary response variable (e.g., migraine=1, no migraine=0). The data scientist believes that the frequency of migraines may be related to the release of toxic mercury fumes. There are two independent variables: 1. X1 = 1 if the patient has at least four amalgams; 0 otherwise. 2. X2 = coffee consumption (0 to 100 hot cups per month). The output from training a logistic regression classifier is as follows: Analysis of LR Parameter Estimates Parameter Estimate Std.Err Zval Pr>|Z| 0.11904 0.05497 Intercept $X_1$ $X_2$ 2.2. PROBLEMS 1. Using X1 and X2, express the odds of a patient having a migraine for a second time. 2. Calculate the probability of a second migraine for a patient that has at least four amalgams and drank 100 cups per month? 3. For users that have at least four amalgams, is high coffee intake associated with an increased probability of a second migraine? 4. Is there statistical evidence that having more than four amalgams is directly associ ated with a reduction in the probability of a second migraine? | 2.3. SOLUTIONS For the purpose of this exercise, it is instructive to predefine z as: z (X1, X2) = −6.36 − 1.02 × X1 + 0.12 × X2. 1. By employing the classic logistic regression model: odds = exp(z (X1, X2)). 2. By substituting the given values of X1, X2 into z (X1, X2), the probability holds: p = exp(z (1, 100))/(1 + exp(z (1, 100))) = 0.99. 3. Yes. The coefficient for coffee consumption is positive (0.119) and the pvalue is less than 0.05 (0.0304). Note: Can you describe the relation between these numerical relations and the positive conclusion? 4. No. The pvalue for this predictor is 0.3818 > 0.05. Note: Can you explain why this inequality implicates a lack of statistical evidence? |
To study factors that affect Alzheimer’s disease using logistic regression, a researcher considers the link between gum (periodontal) disease and Alzheimer as a plausible risk factor [1]. The predictor variable is a count of gum bacteria (Fig. 2.5) in the mouth. FIGURE 2.5: A chain of spherical bacteria. The response variable, Y , measures whether the patient shows any remission (e.g. yes=1). The output from training a logistic regression classifier is as follows: Parameter DF Estimate Intercept gum bacteria Std 4.8792 1.2197 0.0258 0.0194 1. Estimate the probability of improvement when the count of gum bacteria of a patient is 33. LOGISTIC REGRESSION 2. Find out the gum bacteria count at which the estimated probability of improvement is 0.5. 3. Find out the estimated odds ratio of improvement for an increase of 1 in the total gum bacteria count. 4. Obtain a 99% confidence interval for the true odds ratio of improvement increase of 1 in the total gum bacteria count. Remember that the most common confidence levels are 90%, 95%, 99%, and 99.9%. Table 9.1 lists the z values for these levels. Confidence Level 90% 95% 99% 99.9% z TABLE 2.2: Common confidence levels. | 1. The estimated probability of improvement is: ˆπ(gum bacteria) = exp(−4.8792 + 0.0258 × gum bacteria) 1 + exp(−4.8792 + 0.0258 × gum bacteria). Hence, ˆπ(33) = 0.01748. LOGISTIC REGRESSION 2. For ˆπ(gum bacteria) = 0.5 we know that: ˆπ(gum) = exp(ˆα + ˆβx) 1 + exp(ˆα + ˆβx) = 0.5 gum bacteria = −ˆα/ ˆβ = 4.8792/0.0258 = 189.116. 3. The estimated odds ratio are given by: exp( ˆβ) = exp(0.0258) = 1.0504. 4. A 99% confidence interval for β is calculated as follows: ˆβ ± z0.005 × ASE( ˆβ) = 0.0258 ± 2.576 × 0.0194 = (−0.00077, 0.9917). Therefore, a 99% confidence interval for the true odds ratio exp(β) is given by: (exp(−0.00077), exp(0.9917)) = (0.99923, 2.6958). |
Recent research [4] suggests that cannabis (Fig. 2.6) and cannabinoids administration in particular, may reduce the size of malignant tumours in rats. FIGURE 2.6: Cannabis. To study factors affecting tumour shrinkage, a deep learning researcher collects data from two groups; one group is administered with placebo (a substance that is not medicine) and the other with cannabinoids. His main research revolves around studying the relationship (Table 2.3) between the anticancer properties of cannabinoids and tumour shrinkage: 2.2. PROBLEMS Tumour Shrinkage In Rats Group Cannabinoids Placebo Sum Yes No Sum TABLE 2.3: Tumour shrinkage in rats. For the true odds ratio: 1. Find the sample odds ratio. 2. Find the sample logodds ratio. 3. Compute a 95% confidence interval (z0.95 = 1.645; z0.975 = 1.96) for the true log odds ratio and true odds ratio. 2.2.5 The Logit Function and Entropy | 1. The sample odds ratio is: ˆθ = 130 × 6833 60 × 6778 = 2.1842. 2. The estimated standard error for log cid:16 cid:17 log ˆθ ˆσ = 3. According to previous sections, the 95% CI for the true log odds ratio is: 0.7812 ± 1.96 × 0.1570 = (0.4734, 1.0889). Correspondingly, the 95% CI for the true odds ratio is: (e0.4734, e1.0889) = (1.6060, 2.9710). 2.3. SOLUTIONS s cid:16 cid:17 ˆθ 6833 + 1 60 + 1 is: 130 + 1 6778 = 0.1570. 2.3.5 The Logit Function and Entropy |
defined as: The entropy (see Chapter 4) of a single binary outcome with probability p to receive 1 is H(p) ≡ −p log p − (1 − p) log(1 − p). 1. At what p does H(p) attain its maximum value? 2. What is the relationship between the entropy H(p) and the logit function, given p? LOGISTIC REGRESSION 2.2.6 Python/PyTorch/CPP | 1. The entropy (Fig. 2.13) has a maximum value of log2(2) for probability p = 1/2, which is the most chaotic distribution. A lower entropy is a more predictable outcome, with zero providing full certainty. 2. The derivative of the entropy with respect to p yields the negative of the logit func tion: dH(p) dp = −logit(p). Note: The curious reader is encouraged to rigorously prove this claim. 2.3.6 Python, PyTorch, CPP |
The following C++ code (Fig. 2.7) is part of a (very basic) logistic regression implement ation module. For a theoretical discussion underlying this question, refer to problem 2.17. 1 #include ... 2 std::vector<double> theta {6,0.05,1.0}; 3 double sigmoid(double x) { 4 double tmp =1.0 / (1.0 + exp(x)); 5 std::cout << "prob=" << tmp<<std::endl; 6 return tmp; 7 } 8 double hypothesis(std::vector<double> x){ 9 double z; 10 z=std::inner_product(std::begin(x), std::end(x), ,→ std::begin(theta), 0.0); 11 std::cout << "inner_product=" << z<<std::endl; 12 return sigmoid(z); 13 } 14 int classify(std::vector<double> x){ 15 int hypo=hypothesis(x) > 0.5f; 16 std::cout << "hypo=" << hypo<<std::endl; 17 return hypo; 18 } 19 int main() { 20 std::vector<double> x1 {1,40,3.5}; 21 classify(x1); 22 } FIGURE 2.7: Logistic regression in CPP 1. Explain the purpose of line 10, i.e., inner_product. 2. Explain the purpose of line 15, i.e., hypo(x) > 0.5f. 2.2. PROBLEMS 3. What does θ (theta) stand for in line 2? 4. Compile and run the code, you can use: https://repl.it/languages/cpp11 to evaluate the code. What is the output? | LOGISTIC REGRESSION FIGURE 2.13: Binary entropy. 1. During inference, the purpose of inner_product is to multiply the vector of logistic re gression coefficients with the vector of the input which we like to evaluate, e.g., calculate the probability and binary class. 2. The line hypo(x) > 0.5f is commonly used for the evaluation of binary classification wherein probability values above 0.5 (i.e., a threshold) are regarded as TRUE whereas values below 0.5 are regarded as FALSE. 3. The term θ (theta) stands for the logistic regression coefficients which were evaluated during training. 4. The output is as follows: 1 > inner_product=0.5 2 > prob=0.377541 3 > hypo=0 FIGURE 2.14: Logistic regression in C++ |
matrix. The following Python code (Fig. 2.8) runs a very simple linear model on a twodimensional 1 import torch 2 import torch.nn as nn 4 lin = nn.Linear(5, 7) 5 data = (torch.randn(3, 5)) 7 print(lin(data).shape) 8 >? FIGURE 2.8: A linear model in PyTorch Without actually running the code, determine what is the size of the matrix printed as a result of applying the linear model on the matrix. | Because the second dimension of lin is 7, and the first dimension of data is 3, the result ing matrix has a shape of torch.Size([3, 7]) . 2.3. SOLUTIONS |
tion module in Python. The following Python code snippet (Fig. 2.9) is part of a logistic regression implementa LOGISTIC REGRESSION 1 from scipy.special import expit 2 import numpy as np 3 import math 5 def Func001(x): e_x = np.exp(x np.max(x)) return e_x / e_x.sum() 9 def Func002(x): return 1 / (1 + math.exp(x)) 12 def Func003(x): return x * (1x) FIGURE 2.9: Logistic regression methods in Python. Analyse the methods Func001 , Func002 and Func003 presented in Fig. 2.9, find their purposes and name them. | the interviewer. Ideally, you should be able to recognize these functions immediately upon a request from 1. A softmax function. 2. A sigmoid function. 3. A derivative of a sigmoid function. |
Python. The following Python code snippet (Fig. 2.10) is part of a machine learning module in 2.2. PROBLEMS 1 ^^I^^I 2 from scipy.special import expit 3 import numpy as np 4 import math 5 ^^I^^I 6 def Func006(y_hat, y): if y == 1: return np.log(y_hat) else: return np.log(1 y_hat)^^I FIGURE 2.10: Logistic regression methods in Python. Analyse the method Func006 presented in Fig. 2.10. What important concept in machine learning does it implement? | The function implemented in Fig. 2.10 is the binary crossentropy function. |
same function. The following Python code snippet (Fig. 2.11) presents several different variations of the LOGISTIC REGRESSION 1 ^^I^^I 2 from scipy.special import expit 3 import numpy as np 4 import math 6 def Ver001(x): return 1 / (1 + math.exp(x)) 9 def Ver002(x): return 1 / (1 + (np.exp(x))) 12 WHO_AM_I = 709 14 def Ver003(x): return 1 / (1 + np.exp((np.clip(x, WHO_AM_I, None)))) FIGURE 2.11: Logistic regression methods in Python. 1. Which mathematical function do these methods implement? 2. What is significant about the number 709 in line 11? 3. Given a choice, which method would you use? 2.3 Solutions 2.3.1 General Concepts | 1. All the methods are variations of the sigmoid function. 2. In Python, approximately 1.797e + 308 holds the largest possible valve for a floating point variable. The logarithm of which is evaluated at 709.78. If you try to execute the following expression in Python, it will result in inf : np.log(1.8e + 308). 3. I would use Ver003 because of its stability. Note: Can you entail why is this method more stable than the others? |
DataScience-Interview-Preparation-Bot Dataset
This dataset contains questions and answers related to data science interview preparation. It is designed to help individuals prepare for data science interviews by providing a variety of questions and detailed answers.
Dataset Structure
The dataset is provided in CSV format with the following columns:
question: The question or problem statement.answer: The solution or answer to the problem.
Example
Here is an example of the dataset:
questions,answers
"True or False: For a fixed number of observations in a data set, introducing more variables normally generates a model that has a better fit to the data. What may be the drawback of such a model fitting strategy?","True. However, when an excessive and unnecessary number of variables is used in a logistic regression model, peculiarities (e.g., specific attributes) of the underlying data set disproportionately affect the coefficients in the model, a phenomena commonly referred to as “overfitting”. Therefore, it is important that a logistic regression model does not start training with more variables than is justified for the given number of observations."
"Define the term “odds of success” both qualitatively and formally. Give a numerical example that stresses the relation between probability and odds of an event occurring.","The odds of success are defined as the ratio between the probability of success p ∈ [0, 1] and the probability of failure 1 − p. Formally: Odds(p) ≡ p / (1 − p). For instance, assuming the probability of success of an event is p = 0.7. Then, in our example, the odds of success are 7/3, or 2.333 to 1. Naturally, in the case of equal probabilities where p = 0.5, the odds of success is 1 to 1."
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