NHY examples. Bernt Arne Ødegaard. 23 November Estimating dividend growth in Norsk Hydro 8

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1 NHY examples Bernt Arne Ødegaard 23 November 2017 Abstract Finance examples using equity data for Norsk Hydro (NHY) Contents 1 Calculating Beta 4 2 Cost of Capital 7 3 Estimating dividend growth in Norsk Hydro 8 4 Asset Pricing with NHY 9 We use equity data for Norsk Hydro to illustrate empirical finance calculations. 1

2 Figure 1 Stock Price (adjusted) of NHY last five years Stock Price(adj) Time 2

3 Figure 2 Market value equity Market value equity Norsk Hydro (mill)

4 1 Calculating Beta Exercise 1. You want to estimate the beta of the company Norsk Hydro (NHY), traded at the Oslo Stock Exchange. Use the observations of the stock and the market (an Oslo Stock Exchange Index) in the period Use daily returns to estimate the beta. 2. Do you get a different estimate if you use monthly returns? Hint: Data on stock and index prices at the OSE can be downloaded from their web site (Advanced graph and historical prices) Solution to Exercise 1. The first step in this exercise is to gather returns data necessary to do the calculations. At the Oslo Stock Exchange you can download historical data on stock and index prices. If you choose to download five years worth of data, you will get series of daily prices. Note that the stock prices are adjusted for corporate events, such as the SEO in july of The beta calculation is to find cov(rm, rnhy ) β NHY = var(r m) where r m is the return on the index, and r NHY is the return on NHY. We therefore first transform prices to returns using r i,t = Pi,t Pi,t 1 P i,t 1 and then use the excel functions for covariance (COVAR) and variance (VAR) to do the calculation Doing so we find the beta of NHY as β NHY = cov(rnhy, rm) var(r m) = 1.21 This calculation is illustrated in the spreadsheet nhy_beta.xls (available on the course page). To implement the second part of the question, do the calculation from monthly returns, one need to first find monthly returns. While this can be done in a spreadsheet, it is by no means straightforward to pick end-of-month prices. It is much easier to do these calculations as regressions in R, as illustrated next. Note that the data for doing this is, as well as the code for the R examples, is available on the course homepage First, using daily returns, running the regression. Reading the data # read files pulled from OSE homepage library(zoo) OSEBXPric <- read.zoo("../data/osebx.csv", header=true,format="%d.%m.%y",sep=",",skip=1) # pick first column, the closing value OSEBXPric <- OSEBXPric[,1] names(osebxpric)[1] <- "OSEBXClose" NHYPric <- read.zoo("../data/nhy.csv", header=true,format="%d.%m.%y",sep="\t",skip=1) NHYPric <- NHYPric[,1] names(nhypric)[1] <- "NHYClose" Running the regression on daily data: > OSEBXdRet <- (OSEBXPric-lag(OSEBXPric))/lag(OSEBXPric) > NHYdRet <- (NHYPric - lag(nhypric))/lag(nhypric) > beta <- cov(osebxdret,nhydret) / var(osebxdret) > print(beta) [1]

5 > lm(nhydret~osebxdret) Call: lm(formula = NHYdRet ~ OSEBXdRet) Coefficients: (Intercept) OSEBXdRet More detailed output lm(formula = NHYdRet ~ OSEBXdRet) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) OSEBXdRet <2e-16 *** Residual standard error: on 1256 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: 2389 on 1 and 1256 DF, p-value: < 2.2e-16 Then doing the regression on monthly data > # now find monthly returns, first monthly prices > OSEBXMpric <- OSEBXPric[endpoints(OSEBXPric,on="months")] > NHYMPric <- NHYPric[endpoints(NHYPric,on="months")] > OSEBXMrets <- (OSEBXMpric-lag(OSEBXMpric))/lag(OSEBXMpric) > NHYMrets <- (NHYMPric - lag(nhympric))/lag(nhympric) > lm (NHYMrets ~ OSEBXMrets) Call: lm(formula = NHYMrets ~ OSEBXMrets) Coefficients: (Intercept) OSEBXMrets More detailed output lm(formula = NHYMrets ~ OSEBXMrets) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) * OSEBXMrets <2e-16 *** Residual standard error: on 58 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 1 and 58 DF, p-value: < 2.2e-16 5

6 Frequency Beta estimate Daily Monthly

7 2 Cost of Capital Exercise 2. We want to use data for Norsk Hydro to calculate the cost of capital for the firm. We are currently in 2010, and have access to the 2009 accounts of the company. Liabilities and equity Bank Loans and other interest-bearing short-term debt Trade and other payables Provisions Taxes payable Other current financial liabilities 826 Total current liabilities Long-term debt 88 Provisions Pension obligation Other non-current financial liabilities Other liabilities 906 Deferred tax liabililities 849 Total non-current liabilities Total liabilities Total equity The market value of equity in NHY estimated from the stock price times no shares outstanding, yearend 2009: mill. For purposes of calculating cost of capital, what are the weights to put on debt and equity? Solution to Exercise 2. So, from the figure in total liabilities for NHY, we subtract what is obviously not interest-bearing, such as taxes and trade credit. Total liabilities Trade and other payables Taxes payable Deferred tax liabilities 849 = Estimate of debt used for calculating cost of capital Weight on debt = = 0.26 Weight on equity = = =

8 3 Estimating dividend growth in Norsk Hydro The usual method for estimating the growth of dividends is to calculate the implicit growth (g) from stock prices (P ) and dividends (D), using the relationship P 0 = D 1 r g Behind this pricing formula is an assumption that the company s dividend will grow at a rate of g each year, with r as the equity cost of capital, P 0 is today s stock price, and D 1 the future dividend (at time 1). The future dividend is unknown, but it can be estimated as D 1 = D 0 (1 + g), where D 0 is the current dividend. This gives the relationship P 0 = D 0(1 + g). r g Solving this wrt g, we find that g = r D0 P D0 P 0 This expression mainly contains information observable at time t = 0. The only exception is the cost of equity capital r. Let us set this to r = 14.3%. Using this r, the following table show the calculation using data for Norsk Hydro in the period Stock Price Estimated Date Dividend Week Day Day Week growth before before after after (g) 27 apr may may may may may may average The average of the growth estimates, ĝ = 10.86%, can be used as an estimate of the future growth of dividends in Norsk Hydro. 1 The dividend payment in 2006 has been adjusted for a stock split at the dividend date. 8

9 4 Asset Pricing with NHY Exercise 3. You are investigating the market model r it = a + br mt + e it in the Norwegian Market, and apply it to the company Norsk Hydro (NHY). Collect monthly returns for NHY for the period 1980-, and monthly returns for a value weighted market index for the same period. Estimate the model and evaluate the results. You worry about the possibility of the variance of the errors varying, i.e. heteroskedasticity. To investigate this you run a regression with the squared residuals as dependent variable, and as explanatory variables a constant, the market return, and the squared market return, Do you find signs of heteroskedasticity? What can be done to remedy any problems? Solution to Exercise 3. Reading the data ê 2 t = a + b 1 r mt + b 2 r 2 mt + ε > library(zoo) > rets <- read.zoo("../data/monthly_rets.csv", + format="%y%m%d",skip=2,header=true,sep=",") > Rm <- read.table("../data/market_portfolio_returns_monthly.txt", + header=true,sep=";"); > rnhy <- rets$norsk.hydro > vw <- Rm$VW > data <- merge(rnhy,vw,all=false) > rnhy <- data$rnhy > rm <- data$vw Let us first do the standard estimation > reg <- lm(rnhy ~ rm) > summary(reg) Call: lm(formula = rnhy ~ rm) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) ** rm < 2e-16 *** --- Signif. codes: 0 *** ** 0.01 * Residual standard error: on 370 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 1 and 370 DF, p-value: < 2.2e-16 Now, doing the regression that will test for heteroskedasticity 9

10 > rm2 <- rm^2 > e2 <- residuals(reg)^2 > het <- lm(e2~rm + rm2) > summary(het) Call: lm(formula = e2 ~ rm + rm2) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-12 *** rm ** rm * --- Signif. codes: 0 *** ** 0.01 * Residual standard error: on 405 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 2 and 405 DF, p-value: Now, the coefficient estimates are not significant. Still, one can calculate so called Heteroskedasticity consistent standard errors, or White corrected standard errors. These typically will be slightly larger than the usual OLS errors. Doing so in this case, let us compare the two. Consider HC or HAC consistent estimation from package sandwich. > library(sandwich) > sandwich(reg) (Intercept) rm (Intercept) e e-05 rm e e-03 > sqrt(diag(vcov(reg))) (Intercept) rm > sqrt(diag(vcovhc(reg))) (Intercept) rm Compare the two cases OLS HC Exercise 4. Collect monthly stock return data for the stock Norsk Hydro for the period For the same period collect data for a market index of Norwegian stocks. 1. Estimate the market model 2. Test the hypothesis that β = Estimate a similar model in excess return form where r ft is an estimate of a risk free return. r it = α + βr mt + e t (r it r ft ) = α + β(r mt r ft ) + e t 10

11 4. Test the hypothesis that α = Add two additional factors HM L and SM B (norwegian versions of the FF factors). Estimate the model with these two factors and test whether these two factors have any explanatory power. Solution to Exercise 4. Reading the data > rnhy <- read.zoo("../data/nhy_mret.txt",header=false,format="%y%m%d") > names(rnhy)[1] <- "rnhy" > Rm <- read.zoo("../data/market_portfolios_monthly.txt", + header=true,sep=";",format="%y%m%d"); > ew <- Rm$EW > names(ew)[1] <- "ew" > rnhy <- window(rnhy, start=as.date(" "),end=as.date(" ")) > data <- merge(rnhy,ew,all=false) > rnhy <- data$rnhy > rm <- data$ew > summary(cbind(rnhy,rm)) Index rnhy rm Min. : Min. : Min. : st Qu.: st Qu.: st Qu.: Median : Median : Median : Mean : Mean : Mean : rd Qu.: rd Qu.: rd Qu.: Max. : Max. : Max. : Doing the market model regression > mm <- lm(rnhy ~ rm) > summary(mm) Call: lm(formula = rnhy ~ rm) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) rm <2e-16 *** --- Signif. codes: 0 *** ** 0.01 * Residual standard error: on 262 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 1 and 262 DF, p-value: < 2.2e-16 11

12 Model 1 (Intercept) 0.00 (0.00) rm 1.03 (0.07) R Adj. R Num. obs. 264 *** p < 0.001, ** p < 0.01, * p < 0.05, p < 0.1 The testing of whether the coefficent on the market is equal to one: Can be done in several ways, for example by looking at the confidence intervals: > confint(mm) 2.5 % 97.5 % (Intercept) rm or doing a linear hypothesis test > R=cbind(0,1) > r=cbind(1) > linearhypothesis(mm,r,r) Linear hypothesis test Hypothesis: rm = 1 Model 1: restricted model Model 2: rnhy ~ rm Res.Df RSS Df Sum of Sq F Pr(>F) In neither case can we say that the coefficient is statistically different from one. Let us next look at the regression in excess return form > FF <- read.zoo("../data/pricing_factors_monthly.txt", + header=true,sep=";",format="%y%m%d"); > RF <- read.zoo("../data/nibor_monthly.txt", + header=true,sep=";",format="%y%m%d"); > ernhy <- rnhy - RF > erm <- rm-rf > data <- merge(ernhy,erm,all=false) > ernhy <- data$ernhy > erm <- data$erm > reg <- lm(ernhy ~ erm) Call: lm(formula = ernhy ~ erm) Residuals: Min 1Q Median 3Q Max

13 Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) erm <2e-16 *** --- Signif. codes: 0 *** ** 0.01 * Residual standard error: on 262 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 1 and 262 DF, p-value: < 2.2e-16 Model 1 (Intercept) 0.00 (0.00) erm 1.03 (0.07) R Adj. R Num. obs. 264 The intercept is not statistically different from zero. Then we add the two Fama French factors: Looking at the data *** p < 0.001, ** p < 0.01, * p < 0.05, p < 0.1 > SMB <- FF$SMB > HML <- FF$HML > data <- merge(ernhy,erm,smb,hml,all=false) > ernhy <- data$ernhy > erm <- data$erm > SMB <- data$smb > HML <- data$hml > summary(cbind(ernhy,erm,smb,hml)) Index ernhy erm Min. : Min. : Min. : st Qu.: st Qu.: st Qu.: Median : Median : Median : Mean : Mean : Mean : rd Qu.: rd Qu.: rd Qu.: Max. : Max. : Max. : SMB HML Min. : Min. : st Qu.: st Qu.: Median : Median : Mean : Mean : rd Qu.: rd Qu.: Max. : Max. : and the regression > ff <- lm(ernhy~erm+smb+hml) Call: 13

14 lm(formula = ernhy ~ erm + SMB + HML) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) erm <2e-16 *** SMB <2e-16 *** HML Signif. codes: 0 *** ** 0.01 * Residual standard error: on 260 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 3 and 260 DF, p-value: < 2.2e-16 with results Model 1 (Intercept) 0.00 (0.00) erm 0.91 (0.06) SMB 0.74 (0.08) HML 0.09 (0.07) R Adj. R Num. obs. 264 *** p < 0.001, ** p < 0.01, * p < 0.05, p < 0.1 Here we find that SMB is a significant determinant. To formally investigate the joint hypothesis that both HML and SMB have zero coefficients: > linearhypothesis(ff,r,r) Linear hypothesis test Hypothesis: SMB = 0 HML = 0 Model 1: restricted model Model 2: ernhy ~ erm + SMB + HML Res.Df RSS Df Sum of Sq F Pr(>F) < 2.2e-16 *** --- Signif. codes: 0 *** ** 0.01 * Let us summarize all our estimations in one table. Such a table could with advantage have been used to present all the results, and in an academic paper it will typically be done this way, summarize several regressions in one table. 14

15 Model 1 Model 2 Model 3 (Intercept) (0.00) (0.00) (0.00) rm 1.03 (0.07) erm (0.07) (0.06) SMB 0.74 (0.08) HML 0.09 (0.07) R Adj. R Num. obs *** p < 0.001, ** p < 0.01, * p < 0.05, p < 0.1 In case you used the vw index instead, here are a summary of the same regressions Model 1 Model 2 Model 3 (Intercept) (0.00) (0.00) (0.00) rm 1.10 (0.05) erm (0.05) (0.05) SMB 0.30 (0.08) HML 0.04 (0.06) R Adj. R Num. obs *** p < 0.001, ** p < 0.01, * p < 0.05, p <

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