Forecasting elections with tricks and tools from Ch. 2 in BDA3

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1 Forecasting elections with tricks and tools from Ch. 2 in BDA3 1 The data Emil Aas Stoltenberg September 7, 2017 In this example we look at the political party Arbeiderpartiet (Ap) and try to predict their share of the votes in the upcoming parliamentary elections, 11 September The tools used to do this are all found in BDA3 Chapter 2 (that is Gelman et al. (2014)). They can be generalised to a model for more than one party. We use the normal approximation to the normal throughout. A more correct model (for all nine relevant parties) might be a Dirichlet-multinomial model, see Nils Course Notes and Exercises, 13. All the data used in this example can be found at the end of this document. In Figure 1 I have plotted the raw data, these are all the polls conducted from 1 January 2016 until present, which includes the dates the poll were carried out and the sample size. The data are scraped from pollofpolls.no/, a site run by a Bergen based group of lawyers (I think) that like elections. Ap poll day Figure 1: All polls conducted from until present, with point estimates for Arbeiderpartiet as well as approximate 95% confidence intervals. The solid vertical line on the right indicates election day , while the dotted horizontal line is the result Ap got in 2013, 30.8%. 1

2 2 A model This model is built with tools introduced in Chapter 2 of BDA3. For more on models of this type see e.g. Shumway and Stoffer (2006), Jackman (2009) or West and Harrison (1997). We assume that the support for Ap, denoted θ t evolves as a random walk, θ t = θ t 1 + w t, t = 0, 1, 2,..., T, (2.1) where T is the election day and w t iid N(0, τ 2 ). On some days we observe a distorted version of θ t, these are the polls, denoted y t. So instead of observing θ t directly, we see y t given by y t = θ t + v t, (2.2) where v t are independent N(0, σ 2 t ). We also assume that v t, w t are independent of each other. It is the latent support θ t that we want to make inference on. The natural point zero in time is the last election in The dataset does not include data this far back, so we set θ 0 N(m 0 = 0.308, τ 2 0 = (0.025) 2 ), (2.3) and let day zero be the 1 January That is, at the moment we start collecting data, we believe with 95% certainty that the support for Ap is within plus-minus 4.9 percentage points of the last election result. Since the sample sizes in the polls are of usually bigger than 1000, we take σ 2 t as known and equal y t (1 y t )/n, where n is the sample size of the poll conducted on day t. Furthermore, we take τ 2 as constant and known. For example, we might postulate that with 95% certainty, the latent support for a party moves by about half a percentage point from one day to the next. In which case we set τ 2 = (0.0025) 2. Finally, to ease matters, we make one simplifying assumption: there is no more than one poll published on a given day. With tools from Chapter 2 of BDA3 we can now track the latent support of Ap (or any other party in the dataset) from 1 January 2016 up until today. We can also try to predict the outcome on monday (11 September 2017). 2.1 Updating the model Let D t denote the information we have one day t. In our case, D t includes all the polls conducted up to and including day t as well as (2.3). We will work our way forward in time to get a feeling for how this works. For a more formal treatment see Theorem 2.1 in West and Harrison (1997, p. 35). 2

3 Say the first poll is published on day k 1. The posterior of θ k given D k = {y k, m 0, τ 0 } is then p(θ k D k ) p(y k θ k )p(θ k ) = N(θ k, σ 2 k )N(m 0, τ 2 k ) N( m k, ν 2 k), where τ 2 k = τ 0 + kτ 2 and m k = τ k 2 τk 2 + y k + ( 1 τ 2 ) k σ2 k τk 2 + m0, ν σ2 k 2 = σ2 k τ k 2 k τk 2 +. σ2 k See BDA3 p. 42 or my notes 1 for this updating. From now on we define τk 2 and νk 2 to be the prior and posterior variance on day k, respectively. Assume that on day k + 1 there are no polls. Then D k+1 = D k, so θ k+1 D k+1 = d θ k+1 D k, which is the posterior predictive, see BDA3 p. 41. In other words, p(θ k+1 y k ) = p(θ k+1, θ k y k ) dθ k = p(θ k+1 θ k, y k )p(θ k y k ) dθ k = p(θ k+1 θ k )p(θ k y k ) dθ k, where the last line follows because θ k+1 and y k are independent given θ k. Now, p(θ k+1 θ k )p(θ k y k ) exp{ 1 2τ 2 (θ k+1 θ k ) 2 } exp{ 1 1 = exp{ 2(νk 2 + τ 2 ) (θ k+1 m k ) 2 } exp{ 1 2 2ν 2 k (θ k m k ) 2 } τ 2 + νk 2 ( τ 2 νk 2 θk ν2 k θ k+1 + τ 2 m k ) 2}, νk 2 + τ 2 from which we see that In effect, for any j 1, p(θ k+1 y k ) = N(m k, τ 2 k ), τ 2 k = ν2 k + τ 2. p(θ k+j y k ) = N(m k, τ 2 j ), τ 2 j = ν 2 k + jτ 2. (2.4) Now, assume that the second poll appears on day j > k. The posterior on day j is p(θ j D j ) p(y j θ j )p(θ j y k ) N(m j, ν 2 j ), (2.5) 1 3

4 with p(θ j y k ) given in (2.5) and m j = τ 2 j τ 2 j + σ2 j τ 2 j y j + ( 1 τj 2 + σ2 j ) mk, νk 2 = σ2 k τ j 2 τj 2 +, σ2 k and τj 2 = νj 2 + (j k)τ 2. The updating generalises to the third poll, the fourth poll and so on, see the theorem referred to above (West and Harrison, 1997, p. 35). 2.2 Implementing the model Figure 2 displays all polls conducted from until present, with point estimates for Arbeiderpartiet. The black line is E[θ t D t ] for each t = 0, 1, 2,..., T. The red shades indicates the 95% posterior intervals. Based on this model our prediction for Ap on monday is with a 95% predictive interval [0.259, 0.284]. Ap poll day Figure 2: All polls conducted from until present, with point estimates for Arbeiderpartiet. The black line is E[θ t D t ] for each t = 0, 1, 2,..., T. The red shades indicates the 95% posterior intervals. Here is a script implementing the model in R. polls txt refers to the data included at the end of this document. 1 data <- read. table (" polls txt ",sep =";",header = TRUE ) 2 head ( data ) 3 # throw out polls with missing date or sample size 4 data <- data [is.na( data $n)== FALSE,] 4

5 5 data <- data [is.na(as. Date ( data $dato, format ="%d -%m -%Y"))== FALSE,] 6 # sort on dates 7 data <- data [ order (as. Date ( data $dato, format ="%d -%m -%Y"), decreasing = FALSE ),] 8 ## ## The election result in # Ap H Frp Sv Sp Krf V Mdg R Andre 11 elec13 <- c (30.8,26.8,16.3,4.1,5.5,5.6,5.2,2.8,1.1,1.8) 12 ## start. date <- " " 14 data $ day <- as. Date ( data $dato, format ="%d -%m -%Y") - min (as. Date ( data $dato, format ="%d -%m -%Y")) elecday <- as. Date (" ",format ="%d -%m -%Y") - as. Date ( start.date, format ="%d -%m -%Y") ## make a plot 19 # postscript (" ap_ dagtildag. eps ") 20 plot (NA, xlim =c(1, elecday ),ylim =c (0.225,0.41),frame. plot = FALSE, xlab =" day ",ylab =" poll ",main ="Ap") 21 points ( data $day, data $Ap/ 100) 22 ## get approximate confidence 23 p. hat <- data $Ap/ lower <- p. hat * sqrt (p. hat *(1 -p. hat )/ data $n) 25 upper <- p. hat * sqrt (p. hat *(1 -p. hat )/ data $n) 26 segments ( data $day, lower, data $day, upper, col =" pink ",lty =1) 27 segments ( elecday,0.2, elecday,0.42, lty =1) 28 abline (0.308,0, lty =2) 29 # dev. off () ## Average of Ap polls appearing the same day 32 nn <- data $n; Ap <- data $ Ap/ 100; 33 poll. dates <- data $ dato ; 34 day. number <- as. Date ( poll. dates, format ="%d -%m -%Y") - as. Date ( start.date, format ="%d -%m -%Y") 35 Ap.avg <- nn.avg <- c() 36 for (j in 1: length ( unique ( poll. dates ))){ 37 dd <- unique ( poll. dates )[j] 38 find <- which ( poll. dates == dd) 39 Ap.avg [j] <- sum (nn[ find ]/ sum (nn[ find ])*Ap[ find ]) 40 nn.avg [j] <- sum (nn[ find ]) 41 } # The model 45 # fix some things 46 t <- 1:( as. Date (" ",format ="%d -%m -%Y") - as. Date ( start.date, format ="%d -%m -%Y")) 47 m0 < ; tau0 < ; tau < # initial values 48 today. hat <- 0* 1:( length (t) + 1) 49 today. var <- 0* 1:( length (t) + 1) 50 today. hat [1] <- m0 ; today. var [1] <- tau0 ^2 51 y <- nn <- 0* 1: length ( t) 5

6 52 day. number. uniq <- unique ( day. number ) 53 for (i in 1: length ( day. number. uniq )){ 54 y[as. numeric ( day. number. uniq [i])] <- Ap.avg [i] 55 nn[as. numeric ( day. number. uniq [i])] <- nn.avg [i] 56 } 57 ## compute state for each day 58 for (j in 1: length (t)){ 59 if(y[j ]==0) { 60 today. hat [j +1] <- today. hat [j] 61 today. var [j +1] <- today. var [j] + tau ^2 62 } 63 else {# 64 #if(y[j]!=0){ 65 tauj.sq <- today. var [j] 66 sigmaj.sq <- y[j]*(1 -y[j])/nn[j] 67 today. hat [j +1] <- ( tauj.sq)/( tauj.sq+ sigmaj.sq)*y[j] + (1 -( tauj.sq)/( tauj.sq+ sigmaj.sq))* today. hat [j] 68 today. var [j +1] <- ( tauj.sq* sigmaj.sq)/( tauj.sq+ sigmaj.sq ) 69 } 70 } 71 ## make a plot 72 # postscript (" apstate _ dagtildag. eps ") 73 plot (NA, xlim =c(1, elecday ),ylim =c (0.225,0.41),frame. plot = FALSE, xlab =" day ",ylab =" poll ",main ="Ap") 74 ## get confidence 75 lower <- today.hat * sqrt ( today. var ) 76 upper <- today. hat * sqrt ( today. var ) 77 segments (0: length (t),lower,0: length (t),upper, col =" pink ",lty =1) 78 ## 79 points ( day. number.uniq,ap.avg, cex =.8) 80 lines ( today.hat, pch =1, cex =.5, col =" black ") 81 segments ( elecday,0.2, elecday,0.42, lty =1) 82 abline (0.308,0, lty =2) 83 # dev. off () ## prediction 87 tail ( lower <- today.hat * sqrt ( today. var )) 88 tail ( upper <- today. hat * sqrt ( today. var )) 89 tail ( today. hat ) References Gelman, A., Carlin, J. B., Stern, H. S., Dunson, D. B., Vehtari, A., and Rubin, D. B. (2014). Bayesian Data Analysis. Third Edition. CRC Press. Jackman, S. (2009). Bayesian Analysis for the Social Sciences. John Wiley & Sons. 6

7 Shumway, R. H. and Stoffer, D. S. (2006). Time Series Analysis and Its Applications: with R Examples. Springer. West, M. and Harrison, J. (1997). Bayesian Forecasting and Dynamic Models. Second Edition. Springer. A Data dato inst Ap H Frp Sv Sp Krf V Mdg R Andre n Norfakta TNS Respons Norstat Respons Opinion Ipsos Sentio Sentio InFact TNS Norstat Norfakta Ipsos InFact Norstat TNS Norfakta Norstat Ipsos Sentio TNS Norstat Norfakta InFact Norstat Respons Opinion Ipsos Sentio TNS Norstat Norfakta Norstat Respons Opinion InFact Sentio Ipsos

8 TNS Norstat Norfakta Respons Opinion Sentio Ipsos Norstat TNS Norfakta Norfakta TNS Opinion Norstat Sentio Ipsos Norstat TNS Norstat Norfakta InFact Opinion Respons Sentio Ipsos Norstat Norstat Norfakta Opinion InFact Respons Sentio Ipsos Norstat Norstat TNS Norfakta Opinion Respons Sentio Ipsos InFact Norstat Norstat InFact TNS Opinion Norfakta Respons

9 Sentio Ipsos Norstat Norfakta TNS Norstat InFact Respons Opinion Sentio Ipsos Norstat InFact Norstat Norfakta Opinion Sentio Ipsos Norstat Norstat Norfakta InFact Respons Opinion InFact Sentio Norstat Ipsos InFact Opinion Sentio Ipsos Norstat InFact Norfakta Norstat TNS Respons Opinion Sentio Ipsos Norstat InFact Norstat Norfakta InFact Opinion Respons TNS

10 Ipsos Norstat Sentio InFact Norfakta Norfakta InFact Opinion Respons TNS Norstat InFact TNS Sentio Ipsos TNS InFact Respons TNS Norstat Norstat TNS Respons TNS InFact TNS TNS Norstat InFact TNS TNS Respons TNS