X and Y are jointly normal with mean 0, standard deviations 1 and 2, and correlation 0.6. Given X = 1.5, what is the probability that Y exceeds 3?

X and Y are jointly normal with mean 0, standard deviations 1 and 2, and correlation 0.6. Given X = 1.5, what is the probability that Y exceeds 3?

Approach: The conditional law of Y given X is normal with a mean that scales by the correlation times the ratio of standard deviations, and a variance shrunk by the factor 1 minus the squared correlation.

0.2266. The conditional distribution is Y given X = x, normal with mean rho (sigma_Y/sigma_X) x and variance sigma_Y^2 (1 - rho^2). Here the correlation of 0.6 gives a conditional mean of 0.6 * 2 * 1.5 = 1.8, and the variance reduction factor 1 - 0.36 = 0.64 gives a conditional variance of 2.56 and a conditional standard deviation of 1.6. The standard normal tail then gives P(Y > 3 | X = 1.5) = P(Z > (3 - 1.8)/1.6) = P(Z > 0.75) = 0.2266. Unconditionally the answer would be P(Z > 1.5) = 0.0668. Observing X therefore more than triples the probability.

Follow-up: What is E[Y | X > 1.5] for the same pair?

Key concepts: conditional distribution, correlation, variance reduction factor, standard normal tail.