# Re: [R] Simulate phi-coefficient (correlation between dichotomous vars)

From: Bliese, Paul D LTC USAMH <paul.bliese_at_us.army.mil>
Date: Tue 27 Sep 2005 - 16:57:44 EST

Newsgroup members,

I appreciate the help on this topic.

David Duffy provided a solution (below) that was quite helpful, and came close to what I needed. It did a great job creating two vectors of dichotomous variables with a known correlation (what I referred to as a phi-coefficient).

My situation is a bit more complicated and I'm not sure it is easily solved. The problem is that I must assume one of the vectors is constant and generate one or more vectors that covary with the constant vector.

In a continuous example I could use the following code that I got from the S-PLUS newsgroup year ago:

sample.cor<-function (x, rho)
{

y <- (rho * (x - mean(x)))/sqrt(var(x)) + sqrt(1 - rho^2) *

rnorm(length(x))
cat("Sample corr = ", cor(x, y), "\n")     return(y)
}

X<-rnorm(100) #a constant vector
Y1<-sample.cor(X,.30) # a new vector that correlates with X .30 Y2<-sample.cor(X,.45) # a second vector that correlates with X .45

I can, of course, have X be a vector of zeros and ones, and I can dichotomize Y1 and Y2, but the program always returns a phi-coefficient correlation lower than the continuous correlation. Mathematically, I guess this is expected because the phi-coefficient is partially a function of the percentage of positive responses. This, in turn, explains Pearson's (1900) interest in the whole area of tetrachoric correlations -- a tetrachoric correlation being the Pearson product moment correlation that would have been observed had two dichotomously scored variables been measured on a continuous scale (Pearson, 1900).

Appreciate any additional input or possible solutions.

Paul

> From: "Bliese, Paul D LTC USAMH" <paul.bliese@us.army.mil>
>
> Given a sample of zeros and ones, for example:
> > VECTOR1<-rep(c(1,0),c(15,10))
> How would I create a new sample (VECTOR2) also containing zeros and
> ones, in which the phi-coefficient between the two sample vectors was
> drawn from a population with a known phi-coefficient value?
>
> I know there are ways to do this with normally distributed numbers
(for
> example the mvrnorm function in MASS), but am stumped when dealing
with
> dichotomous variables.
>
> Paul

One way is to sample from the 2x2 table with the specified means and pearson
correlation (phi):

for a fourfold table, a b

```                      c d
```

with marginal proportions p1 and p2
cov <- phi * sqrt(p1*(1-p1)*p2*(1-p2))
```a <- p1*p2 + cov
b <- p1*(1-p2) - cov
c <- (1-p1)*p2 - cov
d <- (1-p1)*(1-p2) + cov
```

expand.grid(0:1,0:1)[sample(1:4, size=25, replace=TRUE, prob=c(a,b,c,d)),]

David.

```| David Duffy (MBBS PhD)                                         ,-_|\
| email: davidD@qimr.edu.au  ph: INT+61+7+3362-0217 fax: -0101  /     *
```
| Epidemiology Unit, Queensland Institute of Medical Research \_,-._/ | 300 Herston Rd, Brisbane, Queensland 4029, Australia GPG 4D0B994A v

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