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Department of Economics The University of Melbourne ECOM30001/ECOM90001: Basic Econometrics Semester 1, 2022 Solutions: Tutorial 12 This tutorial reviews some concepts using the econometrics software package R. Specifi- cally, the tutorial reviews: – methods for estimating and interpreting the Linear Probability Model (LPM) in R – methods for estimating and interpreting the Probit Model in R – calculating marginal (partial) effects for the Probit Model in R This tutorial requires one (1) data file: – tut12 churn.csv This file can be obtained from the Canvas subject page. In addition, the R script file tut12.R provides the program code necessary to complete the tutorial. This R script file uses the following package(s) which need to be installed prior to running the R script file: stargazer : for easily producing output tables in R ggplot2 : for easily producing graphs in R car : for easily conducting hypothesis tests in R lmtest : for easily conducting the Ramsey RESET test in R sandwich : for easily calculating robust (Huber-White) heteroskedasticty consistent standard errors in R margins : for easily calculating marginal effects in R These can be installed directly in RStudio from the packages tab or by using the com- mand install.packages() and inserting the name of the package in the brackets. 1 Question 1 Suppose you have been engaged as a consultant for a large telecommunications provider. Your current role is to investigate the characteristics of customers who decide to switch to another competitor. Consider the following econometric model for customer churn in a clearly defined narrow geographic segment of the market: churn i = β0 + β1 partneri + β2 dependentsi + β3 tenurei + β4 tenure 2 i + β5 monthlychargesi + β6 contract2i + β7 contract3i + β8 paperlessi + εi (1) where ε|Xi ～ N (0, σ2ε) and: churn = latent variable determining whether a customer switched providers in the last month partner = 1 if customer has a partner, 0 otherwise dependents = 1 if customer has dependents, 0 otherwise tenure = number of months customer has stayed, 0 otherwise monthlycharges = monthly account charges, in dollars contract1 = 1 if customer does not have a contract (‘month-to-month’), 0 otherwise contract2 = 1 if customer has a 12-month contract, 0 otherwise contract3 = 1 if customer has a 24-month contract, 0 otherwise paperless = 1 if customer receives bills electronically, 0 otherwise Note that contract1 is the omitted category. a) Provide a brief interpretation of the latent variable churn Solution: Suppose, each month, customers weigh up the advantages and disadvan- tages of remaining with the provider. They would trade-off the cost of the service, against the quality and reliability of the service. In principle, customers (implicitly) calculate a score that determines their decision to remain with the provider. Let this ‘customer dissatisfaction’ score be denoted by churn i . Customers that perceive the benefits of remaining with the provider exceed the disadvantages will choose to remain with the provider for another month (churn i sufficiently low). Similarly, cus- tomers that perceive the benefits of remaining with the provider do not exceed the disadvantages will choose to switch providers (churn i sufficiently high). There will be some implied threshold level where customers are indifferent to remaining with the provider, or switching providers. For example, there might be some switching rule as follows: churni = { 1 if churn i ≥ H 0 if churn i < H where H represents the threshold ‘cut-off’ level for switching providers, based upon 2 the sufficiently high level of the ‘customer dissatisfaction’ variable churn i . b) The data set tut12 churn contains 7, 043 observations on the population of cus- tomers for this large telecommunications provider in a specific geographic segment. This data contain the following indicator variable: churni = { = 1 if churn i ≥ 0 = 0 if churn i < 0 This suggests the following econometric model: churni = β0 + β1 partneri + β2 dependentsi + β3 tenurei + β4 tenure 2 i + β5 monthlychargesi + β6 contract2i + β7 contract3i + β8 paperlessi + εi (2) 3 Estimate model (2) by Ordinary Least Squares (OLS). Solution: The estimation results are provided in Figure 1: i) What is the interpretation of your estimate for β5 What is your interpreta- tion of your estimate of β8 Solution: The econometric model (2) is a linear model. Recall also that for a binary dependent variable: E[churni|Xi] = Pr[churni = 1|Xi] so the interpretation of β5: the effect on Pr[churni = 1|Xi] associated with a one dollar increase in the monthly charges, holding all other variables con- stant. Note that the magnitude of the marginal effect is in terms of probability points—a one dollar increase in monthly charges, holding all other variables constant, changes the probability of churn by 100 β5 probability points. The results in column (1) of Figure 1 provide the estimate b5 = 0.0032 so a one dol- lar increase in monthly charges, holding all other variables constant, changes the probability of churn by 0.32 probability points. The parameter β8 is the coefficient attached to the paperless indicator vari- able. Consequently, it has the ‘usual’ interpretation for a linear model: β8 = Pr[churni = 1|paperlessi = 1,Xi] Pr[churni = 1|paperlessi = 0,Xi] so the interpretation of β8 is the difference in Pr[churni = 1|Xi] for customers who receive electronic bills, relative to customer who do not receive electronic bills, holding all other variables constant. Note that the magnitude of the marginal effect is in terms of probability points—holding all other variables constant, the probability of churn for customers who receive electronic bills is 100 β8 different to that for customers who do not receive electronic bills. The results in column (1) of Figure 1 provide the estimate b8 = 0.0717 so, holding all other variables constant, customers who receive electronic bills, have a probability of churn that is higher by 7.17 probability points, relative to customers who receive paper bills. ii) At the 5% level, test for the presence of heteroskedasticity using White’s test. Since there are numerous indicator variables in model (2), use the ‘no cross products’ form of White’s test. Is there any evidence of heteroskedasticity Solution: The OLS estimation results are presented in column 1 of Figure 1 The null hypothesis is that the errors are homoskedastic against the alternative hypothesis that the errors are heteroskedastic. The test statistic is calculated from the auxiliary regression: e 2i = α0 + α1 partneri + α2 dependentsi + α3 tenurei + α4 tenure 2 i + α5 monthlychargesi + α6 contract2i + α7 contract3i + α8 paperlessi + α9 tenure 4 i + α10 monthlycharges 2 i + υi 4 !"!# $%&'( ))) $%&'( ))) $%$*& $%$( +,! $%$$-. $%$$-. $%$$ $%$ /!#!0! 1$%$*& ))) 1$%$*& ))) $%$( $%$$2 3!! 1$%$- ))) 1$%$- ))) $%$$$2 $%$$$2 3!!4,!0 $%$$$ ))) $%$$$ ))) $%$$$$ $%$$$$ 567,8! $%$$& ))) $%$$& ))) $%$$$ $%$$$ 7!,," 1$%(2- ))) 1$%(2- ))) $%$&. $%$& 7!,," 1$%2. ))) 1$%2. ))) $%$'& $%$&& +,#!6!9:6 $%$.. ))) $%$.. ))) $%$$ $%$$-2 ;,::" -$%'*$ &'&%**& ;#<,6! $ $ ;=0> 2 2 ;0!=0> .$&* .$&* !<,: . $*& . $*& $%*2* $%*2* @0A!0 $%*.' $%*.' !:0,60%B0>C.$&* $%&2&$ $%&2&$ DE ) #F$%$(G )) #F$%$G ))) #F$%$$ Figure 1: OLS Estimates for Model (2) Untitled studentized Breusch‐Pagan test data: lpm BP = 1619.8, df = 10, p‐value < 0.00000000000000022 Page 1 Figure 2: Model 2: White Test for Heteroskedasticity 5 050 100 150 0.200 0.100 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.100 1.200 Predicted Probability churn=1 Co un t Figure 3: Predicted Values for OLS Model 2 where e 2i represent the OLS residuals from model (2). The test statistic is N R2 ～ χ2(K) where N is the sample size. The test statistic is asymptotically distributed as χ2 where the df is the number of parameters (excluding the constant) to be estimated in the auxiliary regression. Since there are so many dummy variables in the model, the White test with no cross product terms is conducted, (K) = 10. The R output from the bptest command is presented in Figure 2. This provides a value for the test statistic as N R2 = 1619.8 with a p-value of 0.0000. Since p < 0.05—reject H0. The sample is not consistent with the hypothesis of homoskedastic errors. This should come as no surprise since we already know that the linear probability model will exhibit heteroskedasticity and we know the exact form of this heteroskedasticity. iii) Calculate the predicted values for churn in model (2). How many observations have predicted values less than zero How many observations have predicted values greater than one Is this a problem Why Hint: Create two indicator variables. The sample means of these dummy vari- ables will tell you the proportion of observations with negative predicted values (lpm preds neg) or predicted values greater than one (lpm preds ones): tut12$lpm preds ones <- as.numeric(tut12$lpm preds>1) print(summary(tut12$lpm preds ones)) tut11$lpm preds neg <- as.numeric(tut12$lpm preds<0) print(summary(tut12$lpm preds neg)) Solution: The predicted values of churn for model (2) have a sample mean of 0.26537 with a maximum value of 0.71371 and a minimum value of -0.19688. Creating a dummy variable lpm preds ones, we can see that the estimated model produces a predicted probability greater than one for none of the obser- vations in the sample. Similarly, creating a dummy variable lpm preds neg, we can see that the estimated model produces a predicted probability less than zero for 946 observations (out of 7,043 total observations or 13.43% of obser- vations). This is quite large, for over 10% of observations, the OLS model produces a predicted probability that is less than zero. Why is this a prob- lem The predicted values for model (2) represent predicted probabilities. A negative predicted probability makes no sense in light of this interpretation. Moreover, these negative probabilities would imply negative estimated error variances for these observations. In this case, a FGLS transformation would 6 Untitled # Predicted Values for OLS model churn$lpm_preds <‐lpm$fitted.values print(summary(churn$lpm_preds)) Min. 1st Qu. Median Mean 3rd Qu. Max. ‐0.19688 0.09831 0.26642 0.26537 0.44438 0.71371 #‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ # create dummy variables # dummy variable if predicted values > 1 churn$lpm_preds_ones <‐ as.numeric(churn$lpm_preds>1) print(summary(churn$lpm_preds_ones)) Min. 1st Qu. Median Mean 3rd Qu. Max. 0 0 0 0 0 0 lpm_preds_ones_tab <‐ table(churn$lpm_preds_ones) lpm_preds_ones_tab 0 7043 # dummy variable if predicted values < 0 churn$lpm_preds_neg <‐ as.numeric(churn$lpm_preds<0) print(summary(churn$lpm_preds_neg)) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0000 0.0000 0.0000 0.1343 0.0000 1.0000 lpm_preds_ones_neg_tab <‐ table(churn$lpm_preds_neg) lpm_preds_ones_neg_tab 0 1 6097 946 Page 1 Figure 4: Model 2: Summary Statistics for Predicted Values not be defined for these observations. iv) Report the estimates for model (2) that use the Huber-White (robust) standard errors. Compare and contrast the OLS standard errors with these Robust standard errors. Solution: The estimation results, using the robust variance estimator are provided in column 2 in Figure 1. Overall, the White standard errors are similar to the conventional standard errors (that assume homoskedasticity). In particular, there is no qualitative difference in the p-values for the two- sided test about zero for the estimated coefficients. The same substantive conclusions regarding statistical significance would be obtained if we used the OLS standard errors or the White standard errors. However, since we know there is heteroskedasticity in the linear probability model, the conventional OLS standard errors are invalid for statistical inference. The White standard errors should be used for conducting hypothesis tests. v) Based on your estimation results that use the robust variance estimator, test whether the estimated model should include a quadratic term in tenure, using a 5% level of significance. Solution: A test of whether the model should also include a quadratic term in education is a test of the null hypothesis H0 : β4 = 0 against the alternative hypothesis β4 6= 0. in model (2). The estimation results are reported in Figure 7 Untitled Call: glm(formula = churn ~ factor(partner) + factor(dependents) + tenure + I(tenure^2) + monthlycharges + factor(contract2) + factor(contract3) + factor(paperless), family = binomial(link = "probit"), data = churn) Deviance Residuals: Min 1Q Median 3Q Max ‐1.8851 ‐0.7426 ‐0.3107 0.7729 3.7753 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) ‐0.82878677 0.05797610 ‐14.295 < 0.0000000000000002 *** factor(partner)1 0.04305360 0.04413219 0.976 0.329283 factor(dependents)1 ‐0.17163919 0.04914763 ‐3.492 0.000479 *** tenure ‐0.04422843 0.00321537 ‐13.755 < 0.0000000000000002 *** I(tenure^2) 0.00037937 0.00004731 8.019 0.00000000000000106 *** monthlycharges 0.01516475 0.00081032 18.715 < 0.0000000000000002 *** factor(contract2)1 ‐0.57125411 0.05713456 ‐9.998 < 0.0000000000000002 *** factor(contract3)1 ‐1.17528843 0.08833733 ‐13.305 < 0.0000000000000002 *** factor(paperless)1 0.27845126 0.04175776 6.668 0.00000000002588672 *** ‐‐‐ Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 8150.1 on 7042 degrees of freedom Residual deviance: 6032.8 on 7034 degrees of freedom AIC: 6050.8 Number of Fisher Scoring iterations: 6 Page 1 Figure 5: Probit Estimates for Model (2) 1. The sample test statistic will follow a t-distribution with (N K 1) = 7, 034 degrees of freedom, with a critical value of tc ≈ 1.96. The decision rule is to reject H0 is t > tc or t < tc. The test statistic is calculated as: t = b4 0 se(b4) = 0.000099378 0.000010143 = 9.7976 The p-value for the t-test is p = 0.0000. At the 5% le

vel, the null hypothesis should be rejected. The data are consistent with a quadratic relationship between tenure and the probability of churn. c) Estimate model (2) as a Probit model. Solution: The Probit estimation results are provided in Figure 5: i) Estimate p i, the predicted probability that a customer switches providers, for each individual in the sample. Consider the following decision rule: if p i ≥ 0.5, predict that hurni = 1, otherwise hurni = 0. What percentage of outcomes are successfully predicted How does your answer change if the decision rule cutoff is set at p i ≥ 0.6. 8 true predicted frequency correctly predicted 0 0 4,678 yes 1 0 970 no 0 1 496 no 1 1 899 yes Table 1: Predicted Probability Threshold p i ≥ 0.5 true predicted frequency correctly predicted 0 0 4,950 yes 1 0 1,273 no 0 1 224 no 1 1 596 yes Table 2: Predicted Probability Threshold p i ≥ 0.6 0.5 Threshold The R output for a threshold of p i ≥ 0.5 is presented in Table 1. There are 5,174 observations with a ‘true’ value of churni = 0. Of these 5,174 observa- tions, 4,678 observations are (correctly) predicted with a predicted probability P r(churni = 0), according to our rule. Consequently, when the threshold is set at p i ≥ 0.5, the estimated model predicts that (4, 678/5, 174) = 90.41% of the observations with churni = 0 are predicted correctly (based upon the values for their explanatory variables). The model seems to be doing a decent job at predicting churni = 0, the probability of not switching providers. There are 1,869 observations with a ‘true’ value of churni = 1. Of these 1,869 observations, 899 observations are (correctly) predicted with a predicted prob- ability P r(churni = 1). Consequently, when the threshold is set at p i ≥ 0.5, the estimated model predicts that (899/1, 869) = 48.10% of the observations with churni = 1 are predicted correctly (based upon the values for their explanatory variables). The model is not doing as good job at predicting churni = 1, the probability of switching providers. 0.6 Threshold The R output for a threshold of p i ≥ 0.6 is presented in Table 2. There are 5,174 observations with a ‘true’ value of churni = 0. Of these 5,174 observa- tions, 4,950 observations are (correctly) predicted with a predicted probability P r(churni = 0), according to our rule. Consequently, when the threshold is set at p i ≥ 0.6, the estimated model predicts that (4, 950/5, 174) = 95.67% of the observations with churni = 0 are predicted correctly (based upon the values 9 for their explanatory variables). The model seems to be doing a decent job at predicting churni = 0, the probability of not switching providers. There are 1,869 observations with a ‘true’ value of churni = 1. Of these 1,869 observations, 596 observations are (correctly) predicted with a predicted prob- ability P r(churni = 1). Consequently, when the threshold is set at p i ≥ 0.6, the estimated model predicts that (596/1, 869) = 31.89% of the observations with churni = 1 are predicted correctly (based upon the values for their explana- tory variables). The model is not doing as good job at predicting churni = 1, the probability of switching providers. Moreover, (as expected) raising the threshold considerably lowers the ability of the model to predict churni = 1. ii) Provide an expression for the marginal effect for tenure with the provider. Using the margins package, calculate the average marginal effect for tenure with the provider. Solution: Using the latent variable model (1), and in particular the assump- tion that εi|Xi ～ N (0, σ2), the response probability is given by: Pr(churni = 1|Xi) = Φ ( β0 σ + β1 σ partneri + β2 σ dependentsi + β3 σ tenurei + β4 σ tenure2i + β5 σ monthlychargesi + β6 σ contract2i + β7 σ contract3i + β8 σ paperlessi ) where Φ(·) represents the standard normal distribution function. The variable tenure is a continuous variable. First, note that the observed value of churn can only take two values (zero or one) so: E[churni|Xi] = {Pr(churni = 0|Xi) 0}+ {Pr(churni = 1|Xi) 1} = Pr(churni = 1|Xit) This implies that the marginal effect is given by: E[churni|Xi] tenurei = Pr(churni = 1|Xi) tenurei so: E[churni|Xi] tenurei = Pr(churni = 1|Xi) tenurei = ( β3 + 2 β4 tenurei σ ) φ ( β0 σ + β1 σ partneri + β2 σ dependentsi + β3 σ tenurei + β4 σ tenure2i + β5 σ monthlychargesi + β6 σ contract2i + β7 σ contract3i + β8 σ paperlessi ) where φ(·) is the probability density function for the standard normal distri- 10 Untitled margins_summary(probit) factor AME SE z p lower upper contract21 ‐0.1330 0.0123 ‐10.8071 0.0000 ‐0.1571 ‐0.1088 contract31 ‐0.2369 0.0128 ‐18.5098 0.0000 ‐0.2620 ‐0.2118 dependents1 ‐0.0412 0.0116 ‐3.5378 0.0004 ‐0.0640 ‐0.0184 monthlycharges 0.0037 0.0002 20.2977 0.0000 0.0033 0.0040 paperless1 0.0673 0.0100 6.7526 0.0000 0.0477 0.0868 partner1 0.0104 0.0107 0.9774 0.3283 ‐0.0105 0.0313 tenure ‐0.0064 0.0003 ‐19.9063 0.0000 ‐0.0070 ‐0.0057 # ame for specific variable: check interactions summary(margins(probit,variables=”tenure”)) factor AME SE z p lower upper tenure ‐0.0064 0.0003 ‐19.9063 0.0000 ‐0.0070 ‐0.0057 Page 1 Figure 6: Average Marginal Effects (Probit) for Model (2) Untitled factor tenure AME SE z p lower upper tenure 0.0000 ‐0.0138 0.0010 ‐14.0965 0.0000 ‐0.0158 ‐0.0119 tenure 5.0000 ‐0.0125 0.0009 ‐14.2834 0.0000 ‐0.0142 ‐0.0108 tenure 10.0000 ‐0.0109 0.0007 ‐15.1520 0.0000 ‐0.0123 ‐0.0095 tenure 15.0000 ‐0.0093 0.0006 ‐16.5727 0.0000 ‐0.0104 ‐0.0082 tenure 20.0000 ‐0.0077 0.0004 ‐18.1760 0.0000 ‐0.0085 ‐0.0068 tenure 25.0000 ‐0.0062 0.0003 ‐18.8662 0.0000 ‐0.0069 ‐0.0056 tenure 30.0000 ‐0.0049 0.0003 ‐17.1425 0.0000 ‐0.0055 ‐0.0043 tenure 35.0000 ‐0.0038 0.0003 ‐13.2963 0.0000 ‐0.0043 ‐0.0032 tenure 40.0000 ‐0.0028 0.0003 ‐9.1463 0.0000 ‐0.0034 ‐0.0022 tenure 45.0000 ‐0.0019 0.0003 ‐5.6731 0.0000 ‐0.0026 ‐0.0013 tenure 50.0000 ‐0.0012 0.0004 ‐2.9949 0.0027 ‐0.0019 ‐0.0004 tenure 55.0000 ‐0.0005 0.0005 ‐1.0005 0.3171 ‐0.0013 0.0004 tenure 60.0000 0.0002 0.0005 0.4349 0.6636 ‐0.0008 0.0013 tenure 65.0000 0.0009 0.0007 1.4263 0.1538 ‐0.0003 0.0022 tenure 70.0000 0.0017 0.0008 2.0821 0.0373 0.0001 0.0033 Page 1 Figure 7: Average Marginal Effects (Probit) for Model (2) bution. For a standard normal random variable (Z), the probability density function is given by: φ(Z) = 1√ 2 pi exp ( Z2 2 ) The average marginal effects, calculated using the margins package are pre- sented in Figure 6. The average marginal effect for tenure is -0.0064—a one- month increase in tenure with the provider reduces the probability of switching providers by 0.64 probability points. iii) Using the margins package and ggplot2, calculate and plot the average marginal effect for tenure for each value for tenure from 0 to 70 months. Provide a brief discussion of the results. Solution: As shown above for part (c)(ii), the marginal effect for tenure with the provider depends upon the level of tenure. The overall average marginal effect for the full sample is give by ame = -0.0064. An examination of Figure 11 0.020 0.015 0.010 0.005 0.000 0.005 0.010 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 Tenure with Provider Av e ra ge M ar gi na l E ffe ct Average Marginal Effect for Tenure, with 95% Confidence Intervals Figure 8: Average Marginal Effects (Probit) for Model (2) 7 and Figure 8 reveal that the average marginal effect (ame) for tenure is generally increasing in the level of tenure. For a value of tenure = 0, the ame = -0.0138, for a value of tenure = 10, the ame = -0.0109, for a value of tenure = 30, the ame = -0.0049, and for a value of tenure = 70, the ame = 0.0017. For values of tenure less than 55 months, the ame is significantly negative. This can also be seen in Figure 8 which provides the 95% confidence intervals around the estimated marginal effects. iv) Calculate the marginal effect for tenure with the provider for a customer with a partner, but no dependents, with a tenure of 30 months, with monthly charges of $70 per month, on a month-to-month contract, and receiving bills electronically. Solution: The marginal effect for tenure is given by: E[churni|Xi] tenurei = Pr(churni = 1|Xi) tenurei = ( β3 + 2 β4 tenurei σ ) φ ( β0 σ + β1 σ partneri + β2 σ dependentsi + β3 σ tenurei + β4 σ tenure2i + β5 σ monthlychargesi + β6 σ contract2i + β7 σ contract3i + β8 σ paperlessi ) where φ(·) is the probability density function for the standard normal distri- bution. For a standard normal random variable (Z), the probability density function is given by: φ(Z) = 1√ 2 pi exp ( Z2 2 ) Now evaluate this at partner=1, dependents=0, tenure = 30, monthlycharges 12 Untitled # short way > summary(margins(probit,variables=”tenure”, at=list(partner =1, dependents = 0, + tenure = 30, monthlycharges = 70, contract2 = 0, + contract3 = 0, paperless = 1))) factor partner dependents tenure monthlycharges contract2 contract3 paperless 1.0000 0.0000 30.0000 70.0000 0.0000 0.0000 1.0000 AME SE z p lower upper tenure ‐0.0078 0.0004 ‐17.8354 0.0000 ‐0.0087 ‐0.0069 Page 1 Figure 9: Model 2: Marginal Effect for tenure Evaluated at Selected Characteristics = 70, contract2 = 0, contract3=0, and paperless = 1. E[churni|Xi] tenurei = Pr(churni = 1|Xi) tenurei = ( β3 + 2 30 β4 σ ) φ ( β0 σ + β1 σ + 30 β3 σ + 900 β4 σ + 70 β5 σ + β8 σ ) Recall the Probit parameter estimates are provided in ratio form (scaled by the error standard deviation variance σ): b0 = ( β0 σ ) b1 = ( β1 σ ) b2 = ( β2 σ ) b3 = ( β3 σ ) b4 = ( β4 σ ) b5 = ( β5 σ ) b6 = ( β6 σ ) b7 = ( β7 σ ) b8 = ( β8 σ ) Substituting for the Probit estimates from Figure 5: E[churni|Xi] tenurei = P r(churni = 1|Xi) tenurei = (b3 + 60 b4)φ (b0 + b1 + [30 b3] + [900 b4] + [70 b5] + b8) = 0.02146607φ ( 0.431167) We can calculate φ( 0.431167) as: φ( 0.431167) = 1√ 2pi exp ( ( 0.431167)2 2 ) = 0.3635309 The marginal effect for this representative customer is given by ( 0.02146607 0.3635309) = 0.0078. An additional one month increase in tenure with the provider reduces the probability of switching by approximately 0.78 probability points, for an individual with the selected characteristics (an individual with a partner but no dependents, a tenure of 30 months with the provider, a monthly 13 account of $70, and currently receiving paperless bills). v) Provide an expression for the marginal effect for paperless bills. Using the margins package, calculate the average marginal effect for paperless bills. Aside: When Xj is an indicator variable (say Di) it is not appropriate to use a derivative to measure the marginal effect derived above since the derivative measures the effect of an infinitesimal change in the variable Xj on the condi- tional mean. For an indicator variable the only changes possible are from (0 to 1) or from (1 to 0) which represent very large changes in Xj. In this case it is possible to approximate the derivative as a discrete change: E[yi|Xi] Di = Pr(yi = 1|Di) Di ≈ Pr(yi = 1|Xi) Di = Pr(yi = 1|Xi) = Pr(yi = 1|Di = 1,Xi) Pr(yi = 1|Di = 0,Xi) where the last equality follows because Di is an indicator variable so the change ( Di) is always one unit. 14 Solution: We have: Pr(churni = 1|Xi) paperlessi ≈ Pr(churni = 1|Xi) = {Pr(churni = 1|Xi)}paperlessi=1 {Pr(churni = 1|Xi)}paperlessi=0 = Φ ( β0 + β1 partneri + β2 dependenentsi + β3 tenurei σ + β4 tenure 2 i + β5 monthlychargesi + β6 contract2i σ + β7 contract3i + β8 paperlessi σ ) paperlessi=1 Φ ( β0 + β1 partneri + β2 dependenentsi + β3 tenurei σ + β4 tenure 2 i + β5 monthlychargesi + β6 contract2i σ + β7 contract3i + β8 paperlessi σ ) paperlessi=0 = Φ ( β0 + β1 partneri + β2 dependenentsi + β3 tenurei σ + β4 tenure 2 i + β5 monthlychargesi + β6 contract2i σ + β7 contract3i + β8 σ ) Φ ( β0 + β1 partneri + β2 dependenentsi + β3 tenurei σ + β4 tenure 2 i + β5 monthlychargesi + β6 contract2i σ + β7 contract3i σ ) where the last equality substitutes for paperlessi = 1 or paperlessi = 0. The average marginal effect, calculated using the margins package is presented in Figure 6. The average marginal effect for paperless is 0.0673—relative to customers who only receive paper bills, customers who receive paperless bills have a probability of switching providers (churn) that is higher by 6.73 probability points. This is somewhat close to the estimated marginal effect for the linear probability model of 0.0717. vi) Calculate the marginal effect for paperless bills for a customer with a partner, but no dependents, with a tenure of 30 months, with monthly charges of $70 per month, and on a month-to-month contract. 15 Untitled # short way > summary(margins(probit,variables=”paperless”, + at=list(partner =1, dependents = 0, + tenure = 30, monthlycharges = 70, contract2 = 0, + contract3 = 0))) factor partner dependents tenure monthlycharges contract2 contract3 1.0000 0.0000 30.0000 70.0000 0.0000 0.0000 AME SE z p lower upper paperles 0.0942 0.0138 6.8396 0.0000 0.0672 0.1212 Page 1 Now evaluate the marginal effect provided in part (c)(iv) at partner=1, dependents=0, tenure = 30, monthlycharges = 70, contract2 = 0, contract3=0, and paperless = 1. Pr(churni = 1|Xi) paperlessi = Φ ( β0 σ + β1 σ + { 30 β3 σ } + { 900 β4 σ } + { 70 β5 σ } + β8 σ ) Φ ( β0 σ + β1 σ + { 30 β3 σ } + { 900 β4 σ } + { 70 β4 σ }) Recall the Probit parameter estimates are provided in ratio form (scaled by the error standard deviation variance σ): b0 = ( β0 σ ) b1 = ( β1 σ ) b2 = ( β2 σ ) b3 = ( β3 σ ) b4 = ( β4 σ ) b5 = ( β5 σ ) b6 = ( β6 σ ) b7 = ( β7 σ ) b8 = ( β8 σ ) Substituting for the Probit estimates, the estimated marginal effect is P r(churni = 1|Xi) paperlessi = Φ (b0 + b1 + [30 b3] + [900 b4] + [70 b4] + b8) Φ (b0 + b1 + [30 b3] + [900 b4] + [70 b4]) = Φ( 0.431167) Φ( 0.7096182) = 0.3331735 0.2389705 = 0.09420303 16