Generate random correlated data between a binary and a continuous variable

Question

I want to generate two variables. One is binary outcome variable (say success / failure) and the other is age in years. I want age to be positively correlated with success. For example there should be more successes in the higher age segments than in lower. Ideally I should be in position to control degree of correlation. How do I do that?

Thanks

Answer 1

@ocram's approach will certainly work. In terms of the dependence properties it's somewhat restrictive though.

Another method is to use a copula to derive a joint distribution. You can specify marginal distributions for success and age (if you have existing data this is especially simple) and a copula family. Varying the parameters of the copula will yield different degrees of dependence, and different copula families will give you various dependence relationships (e.g. strong upper tail dependence).

A recent overview of doing this in R via the copula package is available here. See also the discussion in that paper for additional packages.

You don't necessarily need an entire package though; here's a simple example using a Gaussian copula, marginal success probability 0.6, and gamma distributed ages. Vary r to control the dependence.

r = 0.8 # correlation coefficient
sigma = matrix(c(1,r,r,1), ncol=2)
s = chol(sigma)
n = 10000
z = s%*%matrix(rnorm(n*2), nrow=2)
u = pnorm(z)

age = qgamma(u[1,], 15, 0.5)
age_bracket = cut(age, breaks = seq(0,max(age), by=5))
success = u[2,]>0.4

round(prop.table(table(age_bracket, success)),2)

plot(density(age[!success]), main="Age by Success", xlab="age")
lines(density(age[success]), lty=2)
legend('topright', c("Failure", "Success"), lty=c(1,2))

Output:

Table:

           success
age_bracket FALSE TRUE
    (0,5]    0.00 0.00
    (5,10]   0.00 0.00
    (10,15]  0.03 0.00
    (15,20]  0.07 0.03
    (20,25]  0.10 0.09
    (25,30]  0.07 0.13
    (30,35]  0.04 0.14
    (35,40]  0.02 0.11
    (40,45]  0.01 0.07
    (45,50]  0.00 0.04
    (50,55]  0.00 0.02
    (55,60]  0.00 0.01
    (60,65]  0.00 0.00
    (65,70]  0.00 0.00
    (70,75]  0.00 0.00
    (75,80]  0.00 0.00

Answer 2

You can simulate the logistic regression model.

More precisely, you can first generate values for the age variable (for example using a uniform distribution) and then compute probabilities of success using

$$\pi ( x ) = \frac{\exp(\beta_0 + \beta_1 x)}{1 + \exp(\beta_0 + \beta_1 x)}$$

where $\beta_0$ and $\beta_1$ are constant regression coefficients to be specified. In particular, $\beta_1$ controls the magnitude of association between success and age.

Having values of $\pi$, you can now generate values for the binary outcome variable using the Bernoulli distribution.

Illustrative example in R:

n <- 10
beta0 <- -1.6
beta1 <- 0.03
x <- runif(n=n, min=18, max=60)
pi_x <- exp(beta0 + beta1 * x) / (1 + exp(beta0 + beta1 * x))
y <- rbinom(n=length(x), size=1, prob=pi_x)
data <- data.frame(x, pi_x, y)
names(data) <- c("age", "pi", "y")
print(data)

         age        pi y
 1  44.99389 0.4377784 1
 2  38.06071 0.3874180 0
 3  48.84682 0.4664019 1
 4  24.60762 0.2969694 0
 5  39.21008 0.3956323 1
 6  24.89943 0.2988003 0
 7  51.21295 0.4841025 1
 8  43.63633 0.4277811 0
 9  33.05582 0.3524413 0
 10 30.20088 0.3331497 1

Answer 3

You can first generate the success/failure variable ($X$), and then generate the age ($Y$) with a different distribution depending on the value of $X$. That will give you correlation.

To quantify the correlation, the simplest way is to shift $Y$ according to the value of $X$. The amount by which you shift will be a measure of the correlation.