TITLE: From correlation coefficient to conditional probability
QUESTION [9 upvotes]: In the best-selling book Thinking Fast and Slow (p. 205), Daniel Kahneman (a Nobel Prize winner in Economics) makes the following claim: 'Suppose you consider many pairs of firms. The two firms in each pair are generally similar, but the CEO of one of them is better than the other. How often will you find that the firm with the stronger CEO is the more successful of the two? . . . A correlation of .30 [between stronger CEO and more successful firm] implies that you would find the stronger CEO leading the stronger firm in about 60% of the pairs.'
How is the 'about 60%' computed under plausible assumptions?

REPLY [4 votes]: For a randomly selected firm let $X$ be the CEO quality and $Y$ be the firm success (however these are measured). The author's assertion follows under the assumption that $(X,Y)$ has a bivariate normal distribution with correlation $\rho=0.3$.
If $(X_1,Y_1)$ and $(X_2,Y_2)$ are measured for independently selected firms, then the difference $(X_1-X_2,Y_1-Y_2)$ is also bivariate normal, with mean zero and the same correlation $\rho$. Kahneman's claim is that $P(Y_1>Y_2\mid X_1>X_2)\approx 0.6$. This follows from a fact(*) about bivariate normal variables:
If $(A,B)$ are bivariate normal with means $\mu_A$ and $\mu_B$ respectively, and correlation $\rho$, then
$$P(B>\mu_B\mid A>\mu_A)=\frac12 + \frac{\arcsin\rho}\pi.$$
If $\rho=0.3$ the RHS works out to $0.59698668$.
(*) The fact can be deduced from this result.