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?



3 Answers 3


@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))



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

enter image description here

  • $\begingroup$ Great answer! Copulas are a beautiful if under-appreciated tool. The probit model (with Gaussian marginal on the continuous variable) is a special case of the Gaussian copula model. But this is a much more general solution. $\endgroup$
    – jpillow
    Commented Jul 11, 2011 at 6:50
  • 1
    $\begingroup$ @JMS: +1 Yes, Copulas are very appealing. I should try to study them in more details! $\endgroup$
    – ocram
    Commented Jul 11, 2011 at 7:04
  • $\begingroup$ @jpillow Indeed; Gaussian copula models subsume multivariate probit-type models of any sort. Via scale mixing they also extend to t/logistic copulae and logit/robit models too. Tres cool :) $\endgroup$
    – JMS
    Commented Jul 11, 2011 at 15:34
  • $\begingroup$ @ocram Do! There are a lot of open questions in mixed data contexts (when using them as models and not just drawing from them) that people like me would love to see solved... $\endgroup$
    – JMS
    Commented Jul 11, 2011 at 15:35
  • $\begingroup$ @JMS Excellent answer! $\endgroup$
    – user333
    Commented Jul 12, 2011 at 20:53

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")

         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
  • 3
    $\begingroup$ Nice answer, though from an aesthetic standpoint (not a practical one) a probit regression model might be even nicer. The probit model is equivalent to starting with a bivariate Gaussian R.V. and thresholding one of them (to zero or 1). Really it just involves substituting the Gaussian cumulative normal ("probit") function for the logit used in logistic regression. Practically this should give the same performance (and computationally it's slower since normcdf is expensive to evaluate (1+e^x)^-1), but it's nice to think about a Gaussian with one of the variables censored ("rounded"). $\endgroup$
    – jpillow
    Commented Jul 11, 2011 at 6:44
  • $\begingroup$ @jpillow: Thank you for your comment. I will think on it asap! $\endgroup$
    – ocram
    Commented Jul 11, 2011 at 7:04
  • 1
    $\begingroup$ What's nice about the probit / Gaussian copula model is that the parameters take the form of a covariance matrix between the two quantities (one of which is then binarized into 0 and 1). So it's nice from the standpoint of interpretability (but not so nice from the standpoint of computational convenience). $\endgroup$
    – jpillow
    Commented Jul 11, 2011 at 8:25

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.


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