# Open data might be a false good opportunity…

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I am always surprised to see many people on Twitter tweeting about #opendata,
e.g. @data4all, @usdatagov, @datapublicatwit, @ProPublica or
@open3 among so many others…
Initially, I was also very enthousiastic, but I have to admit that **Freakonometrics - Tag - R-english**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

*open data*are rarely

*raw data*. Which is what I am usually looking for, as a statistician…

Consider the following example: I was wondering (Valentine’s day is approaching) when will a man born in 1975 (say) get married – if he ever gets married ? More technically, I was looking for a distribution of the age of first marriage (given the year of birth), including the proportion of men that will never get married, for that specific cohort.

The only data I found on the internet is the following, on statistics.gov.uk/

Note that we can also focus on women (e.g. here). Is it possible to use that

*open*data to get an estimation of the distribution of first marriage for some specific cohort ? (and to answer the question I asked).

Here, we have two dimensions: on line , the year (of the marriage), and on column , the age of the man when he gets married. Assume that those were

*raw*data, i.e. that we have the number of marriages of men of age during the year .

We are interested at a longitudinal lecture of the table, i.e. consider some man born year , we want to estimate (or predict) the age he will get married, if he gets married. With raw data, we can do it… The first step is to build up triangles (to have a cohort vs. age lecture of the data), and then to consider a model, e.g.

where is a year effect, and is a cohort effect.

base=read.table("http://freakonometrics.free.fr/mariage-age-uk.csv", sep=";",header=TRUE) m=base[1:16,] m=m[,3:10] m=as.matrix(m) triangle=matrix(NA,nrow(m),ncol(m)) n=ncol(m) for(i in 1:16){ triangle[i,]=diag(m[i-1+(1:n),]) } triangle[nrow(m),1]=m[nrow(m),1] triangle [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 12 104 222 247 198 132 51 34 [2,] 8 89 228 257 202 102 75 49 [3,] 4 80 209 247 168 129 92 50 [4,] 4 73 196 236 181 140 88 45 [5,] 3 78 242 206 161 114 68 47 [6,] 11 150 223 199 157 105 73 39 [7,] 12 117 194 183 136 96 61 36 [8,] 11 118 202 175 122 92 62 40 [9,] 15 147 218 162 127 98 72 48 [10,] 20 185 204 171 138 112 82 NA [11,] 31 197 240 209 172 138 NA NA [12,] 34 196 233 202 169 NA NA NA [13,] 35 166 210 199 NA NA NA NA [14,] 26 139 210 NA NA NA NA NA [15,] 18 104 NA NA NA NA NA NA [16,] 10 NA NA NA NA NA NA NA Y=as.vector(triangle) YEARS=seq(1918,1993,by=5) AGES=seq(22,57,by=5) X1=rep(YEARS,length(AGES)) X2=rep(AGES,each=length(YEARS)) reg=glm(Y~as.factor(X1)+as.factor(X2),family="poisson") summary(reg) Call: glm(formula = Y ~ as.factor(X1) + as.factor(X2), family = "poisson") Deviance Residuals: Min 1Q Median 3Q Max -5.4502 -1.1611 -0.0603 1.0471 4.6214 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 2.8300461 0.0712160 39.739 < 2e-16 *** as.factor(X1)1923 0.0099503 0.0446105 0.223 0.823497 as.factor(X1)1928 -0.0212236 0.0449605 -0.472 0.636891 as.factor(X1)1933 -0.0377019 0.0451489 -0.835 0.403686 as.factor(X1)1938 -0.0844692 0.0456962 -1.848 0.064531 . as.factor(X1)1943 -0.0439519 0.0452209 -0.972 0.331082 as.factor(X1)1948 -0.1803236 0.0468786 -3.847 0.000120 *** as.factor(X1)1953 -0.1960149 0.0470802 -4.163 3.14e-05 *** as.factor(X1)1958 -0.1199103 0.0461237 -2.600 0.009329 ** as.factor(X1)1963 -0.0446620 0.0458508 -0.974 0.330020 as.factor(X1)1968 0.1192561 0.0450437 2.648 0.008107 ** as.factor(X1)1973 0.0985671 0.0472460 2.086 0.036956 * as.factor(X1)1978 0.0356199 0.0520094 0.685 0.493423 as.factor(X1)1983 0.0004365 0.0617191 0.007 0.994357 as.factor(X1)1988 -0.2191428 0.0981189 -2.233 0.025520 * as.factor(X1)1993 -0.5274610 0.3241477 -1.627 0.103689 as.factor(X2)27 2.0748202 0.0679193 30.548 < 2e-16 *** as.factor(X2)32 2.5768802 0.0667480 38.606 < 2e-16 *** as.factor(X2)37 2.5350787 0.0671736 37.739 < 2e-16 *** as.factor(X2)42 2.2883203 0.0683441 33.482 < 2e-16 *** as.factor(X2)47 1.9601540 0.0704276 27.832 < 2e-16 *** as.factor(X2)52 1.5216903 0.0745623 20.408 < 2e-16 *** as.factor(X2)57 1.0060665 0.0822708 12.229 < 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 5299.30 on 99 degrees of freedom Residual deviance: 375.53 on 77 degrees of freedom (28 observations deleted due to missingness) AIC: 1052.1 Number of Fisher Scoring iterations: 5

We can now predict the number of marriages per year, and per cohort

Here, given the cohort , the shape of is the following

Yp=predict(reg,type="response") tYp=matrix(Yp,nrow(m),ncol(m)) tYp[16,] tYp[16,] [1] 10.00000 222.94525 209.32773 159.87855 115.06971 42.59102 [7] 18.70168 148.92360

The errors (Pearson error) look like that

Ep=residuals(reg,type="pearson")

I guess that study here is not valid. The problem is that we deal with

*open*data, and

*numbers*of marriages are not given here: what is given is a he proportion of marriage of men of age during the year , with a yearly normalization. There is a constraint on lines, i.e. we observe

so that

This is mentioned in the title

It is still possible to consider a Poisson regression on the , but unfortunately, I do not think any interpretation is valid (unless demography did not change last century). For instance, the following sum

looks like that

apply(tYp,1,sum) [1] 919.948 838.762 846.301 816.552 943.559 930.280 857.871 896.113 [9] 905.086 948.087 895.862 853.738 826.003 816.192 813.974 927.437i.e. if we look at the graph

But I do not think we can interpret that sum as the probability (if we divide by 1,000) that a man in that cohort gets married.... And more basically, I cannot do anything with that dataset...

So *open* data might be
interesting. The problem is that most of the time, the data are somehow
normalized (or aggregated). And then, it becomes difficult to use them...

*to be continued*.

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