This morning, in the ACT2040 class (on non-life insurance), we’ve discussed the difference between observable and non-observable heterogeneity in ratemaking (from an economic perspective). To illustrate that point (we will spend more time, later on, discussing observable and non-observable risk factors), we looked at the following simple example. Let denote the height of a person. Consider the following dataset

> Davis=read.table( + "http://socserv.socsci.mcmaster.ca/jfox/Books/Applied-Regression-2E/datasets/Davis.txt")

There is a small typo in the dataset, so let us make manual changes here

`> Davis[12,c(2,3)]=Davis[12,c(3,2)] `

Here, the variable of interest is the height of a given person,

`> X=Davis$height `

If we look at the histogram, we have

> hist(X,col="light green", border="white",proba=TRUE,xlab="",main="")

Can we assume that we have a Gaussian distribution ?

Maybe not… Here, if we fit a Gaussian distribution, plot it, and add a kernel based estimator, we get

> (param <- fitdistr(X,"normal")$estimate) > f1 <- function(x) dnorm(x,param[1],param[2]) > x=seq(100,210,by=.2) > lines(x,f1(x),lty=2,col="red") > lines(density(X))

If you look at that black line, you might think of a mixture, i.e. something like

(using standard mixture notations). Mixture are obtained when we have a **non-observable heterogeneity factor**: with probability , we have a random variable (call it type [1]), and with probability , a random variable (call it type [2]). So far, nothing new. And we can fit such a mixture distribution, using e.g.

```
> library(mixtools)
> mix <- normalmixEM(X)
number of iterations= 335
> (param12 <- c(mix$lambda[1],mix$mu,mix$sigma))
[1] 0.4002202 178.4997298 165.2703616 6.3561363 5.9460023
```

If we plot that mixture of two Gaussian distributions, we get

> f2 <- function(x){ param12[1]*dnorm(x,param12[2],param12[4]) + (1-param12[1])*dnorm(x,param12[3],param12[5]) } > lines(x,f2(x),lwd=2, col="red") lines(density(X))

Not bad. Actually, we can try to maximize the likelihood with our own codes,

> logdf <- function(x,parameter){ + p <- parameter[1] + m1 <- parameter[2] + s1 <- parameter[4] + m2 <- parameter[3] + s2 <- parameter[5] + return(log(p*dnorm(x,m1,s1)+(1-p)*dnorm(x,m2,s2))) + } > logL <- function(parameter) -sum(logdf(X,parameter)) > Amat <- matrix(c(1,-1,0,0,0,0, + 0,0,0,0,1,0,0,0,0,0,0,0,0,1), 4, 5) > bvec <- c(0,-1,0,0) > constrOptim(c(.5,160,180,10,10), logL, NULL, ui = Amat, ci = bvec)$par [1] 0.5996263 165.2690084 178.4991624 5.9447675 6.3564746

Here, we include some constraints, to insurance that the probability belongs to the unit interval, and that the variance parameters remain positive. Note that we have something close to the previous output.

Let us try something a little bit more complex now. What if we assume that the underlying distributions have the same variance, namely

In that case, we have to use the previous code, and make small changes,

> logdf <- function(x,parameter){ + p <- parameter[1] + m1 <- parameter[2] + s1 <- parameter[4] + m2 <- parameter[3] + s2 <- parameter[4] + return(log(p*dnorm(x,m1,s1)+(1-p)*dnorm(x,m2,s2))) + } > logL <- function(parameter) -sum(logdf(X,parameter)) > Amat <- matrix(c(1,-1,0,0,0,0,0,0,0,0,0,1), 3, 4) > bvec <- c(0,-1,0) > (param12c= constrOptim(c(.5,160,180,10), logL, NULL, ui = Amat, ci = bvec)$par) [1] 0.6319105 165.6142824 179.0623954 6.1072614

This is what we can do if we cannot observe the heterogeneity factor. But wait… we actually have some information in the dataset. For instance, we have the sex of the person. Now, if we look at histograms of height per sex, and kernel based density estimator of the height, per sex, we have

So, it looks like the height for male, and the height for female are different. Maybe we can use that variable, that was actually observed, to explain the heterogeneity in our sample. Formally, here, the idea is to consider a mixture, with an **observable heterogeneity factor**: the sex,

We now have interpretation of what we used to call class [1] and [2] previously: male and female. And here, estimating parameters is quite simple,

> (pM <- mean(sex=="M")) [1] 0.44 > (paramF <- fitdistr(X[sex=="F"],"normal")$estimate) mean sd 164.714286 5.633808 > (paramM <- fitdistr(X[sex=="M"],"normal")$estimate) mean sd 178.011364 6.404001

And if we plot the density, we have

> f4 <- function(x) pM*dnorm(x,paramM[1],paramM[2])+(1-pM)*dnorm(x,paramF[1],paramF[2]) > lines(x,f4(x),lwd=3,col="blue")

What if, once again, we assume identical variance? Namely, the model becomes

Then a natural idea to derive an estimator for the variance, based on previous computations, is to use

The code is here

> s=sqrt((sum((height[sex=="M"]-paramM[1])^2)+sum((height[sex=="F"]-paramF[1])^2))/(nrow(Davis)-2)) > s [1] 6.015068

and again, it is possible to plot the associated density,

> f5 <- function(x) pM*dnorm(x,paramM[1],s)+(1-pM)*dnorm(x,paramF[1],s) > lines(x,f5(x),lwd=3,col="blue")

Now, if we think a little about what we’ve just done, it is simply a linear regression on a factor, the sex of the person,

where . And indeed, if we run the code to estimate this linear model,

> summary(lm(height~sex,data=Davis)) Call: lm(formula = height ~ sex, data = Davis) Residuals: Min 1Q Median 3Q Max -16.7143 -3.7143 -0.0114 4.2857 18.9886 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 164.7143 0.5684 289.80 <2e-16 *** sexM 13.2971 0.8569 15.52 <2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 6.015 on 198 degrees of freedom Multiple R-squared: 0.5488, Adjusted R-squared: 0.5465 F-statistic: 240.8 on 1 and 198 DF, p-value: < 2.2e-16

we get the same estimators for the means and the variance as the ones obtained previously. So, as mentioned this morning in class, if you have a non-observable heterogeneity factor, we can use a mixture model to fit a distribution, but if you can get a proxy of that factor, that is observable, then you can run a regression. But most of the time, that observable variable is just a proxy of a non-observable one…

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Working through your code from scratch (twice!), I find my mixture distribution is radically different from yours (and each of mine differs one from the other!).

For example, the first time I ran the code, lambda was about 0.15; the second time, about 0.6.

Mixture estimation is a difficult topic… and indeed, numerical computations are rather unstable.

I’m barely literate in Mathematica so I’m just barely able to follow your line of reasoning in R. Still, I have a question (suggestion). Wouldn’t it be better to handle biological distributions with the classical binomial? Seems to me when I see columns I see ‘sources of error’ and I have a better chance of assigning a gene to each ‘source of error’ than when I’m working with smooth approximators like the normal curve.

Great explanation, very helpful. Thanks for posting