Generate a Gaussian and a binary random variables with predefined correlation I'm trying to generate a dataset with a pre-defined correlation between a normally distributed variable and a binary variable.
The method I had originally thought of was the following:


*

*Generate $X \sim Norm(0,1)$

*Generate $Y \sim Norm(0,1)$

*Generate $Q = \rho X + \sqrt{1-\rho^2}Y$, this will be the log-odds of success

*Generate $P = 1-\frac{1}{exp(Q) + 1}$, this is the probability of success

*Generate $U = Unif(0,1)$

*Generate $T = I(U < P) $


This method ensures that $Corr(X,P) = \rho$, however $Corr(X,T) \ne\rho$.
An alternative algorithm, replacing step 3 with:


*Generate $Q = \rho X$


provides similar results for $-1<\rho<1$, but in this second algorithm, we can take $\rho$ to be any value in $\mathbb{R}$, and still we have control of the correlation between $X$ and $T$.
I ran this algorithm through R using different values of $\rho$ and plotted $\rho$ against $Corr(X,T)$ (using the default methods in cor.test):

After altering $\rho$, it appears that the correlation between the continuous and binary variable is bounded, approximately $-0.8 \le Corr(X,T) \le 0.8$. While trying to figure out a relationship between $\rho$ and $Corr(X,T)$, I thought it looked like an arctangent and so the red line is the plot of $1.6*tan^{-1}(\rho)/\pi$, which doesn't quite match. when using the first algorithm (and limiting $-1<\rho<1$), the relationship between $\rho$ and $Corr(X,T)$ appears to be linear, with $Corr(X,T) = 0.43\rho$
My first question is whether there is any literature or sources on explicitly finding the relationship between $\rho$ and $Corr(X,T)$? That way I can predefine this correlation, rather than $\rho$. And my second is whether this is the best way to simulate this kind of data? Note that in the work that I'm doing, there is a causal relationship between X and T (X -> T)
 A: To generate such a pair $(B,Y)$ with $B$ Bernoulli (with parameter $p$) and $Y$ normal, why not begin with a suitable binormal variable $(X,Y)$ and define $B$ to be the indicator that $X$ exceeds its $1-p$ quantile?  By centering $(X,Y)$ at the origin and standardizing its marginals, the only question concerns what correlation $r$ should hold between $X$ and $Y$ so that the correlation between $B$ and $Y$ will be a given value $\rho$.
To this end, express $Y = r X + \sqrt{1-r^2}Z$ for independent standard Normal variables $X$ and $Z$.  Set $x_0$ to be the $1-p$ quantile of $X$, so that $\Phi(x_0)=1-p$.  (As is conventional, $\Phi$ is the standard Normal distribution and $\phi$ will be its density.)
Since the variance of $B$ is $p(1-p)$ and the variance of $Y$ is $1$, and $Y$ has zero mean, the correlation between $B$ and $Y$ is
$$\eqalign{
\rho&=\operatorname{Cor}(B,X) = \frac{E[BY] - E[B]E[Y]}{\sqrt{p(1-p)}\sqrt{1}}\\
&= \frac{E[B(rX+\sqrt{1-r^2}Z)]-0} {\sqrt{p(1-p)}} \\
&= \frac{rE[X\mid X \ge x_0]\Pr(X \ge x_0)}{\sqrt{p(1-p)}}.
}$$
The conditional expectation is readily computed by integration, giving
$$E[X\mid X \ge x_0]\Pr(X \ge x_0) = \frac{1}{\sqrt{2\pi}}\int_{x_0}^\infty x e^{-x^2/2}dx = \frac{e^{-x_0^2/2}}{\sqrt{2\pi}} = \phi(x_0),$$
whence
$$\rho = \frac{r \phi(x_0)}{\sqrt{p(1-p)}}.$$
Solve this for $r$: by setting
$$r = \frac{\rho \sqrt{p(1-p)}}{\phi(x_0)},$$
$B$ and $Y$ will have correlation $\rho$.
Note that since it's necessary that $1-r^2\ge 0$, any values of $\rho$ that cause $|r|$ to exceed $1$ will not be achievable in this fashion.  The figure plots feasible values of $r$ as a function of the desired correlation $\rho$ and Bernoulli parameter $p$: the contours range in increments of $1/10$ from $-1$ at the upper left through $+1$ at the upper right.

