Gaussian Half-Space Probability
Let $X, Y \sim N(0, 1)$ be jointly normal with correlation $\rho \in (-1, 1)$. For constants $k \in \mathbb{R}$ and $b \in \mathbb{R}$:
- Compute $P(Y > kX + b)$ in closed form.
- Compute $P(Y > kX + b \mid X > x_0)$ in closed form by conditioning on $X$ and using the conditional normal distribution.
Hints
- For part 1, define $W = Y - kX$ and compute its mean and variance. What distribution does $W$ follow?
- For part 2, recall that $Y \mid X = x \sim N(\rho x, 1 - \rho^2)$. Compute $P(Y > kx + b \mid X = x)$ as a $\Phi$ expression, then integrate over $x > x_0$.
- The integral $\int_{x_0}^{\infty} \Phi(\alpha + \beta x) \phi(x) \, dx$ evaluates to a bivariate normal CDF $\Phi_2$. Look up the standard identity.
Worked Solution
How to Think About It: The event $\{Y > kX + b\}$ is a half-space in the $(X, Y)$ plane. Since $(X, Y)$ is bivariate normal, any linear function of $(X, Y)$ is univariate normal. So $P(Y > kX + b) = P(Y - kX > b)$, and $Y - kX$ is normal -- we just need its mean and variance. For part 2, conditioning on $X > x_0$ means we first condition on a specific $X = x$, use the conditional distribution $Y \mid X = x$, and then integrate over $x > x_0$.
Quick Estimate: Take $\rho = 0$, $k = 1$, $b = 0$. Then $Y - X \sim N(0, 2)$, so $P(Y > X) = P(Y - X > 0) = 1/2$ by symmetry. That checks out. If $\rho = 0.5$, $k = 1$, $b = 0$: $\text{Var}(Y - X) = 1 + 1 - 2(0.5)(1) = 1$, so $Y - X \sim N(0, 1)$ and $P(Y > X) = 1/2$ again. If $b = 1$: $P(Y - X > 1) = 1 - \Phi(1) \approx 0.159$.
Formal Solution:
Part 1: $P(Y > kX + b)$
Define $W = Y - kX$. Since $X$ and $Y$ are jointly normal, $W$ is normal with:
$$E[W] = E[Y] - kE[X] = 0$$
$$\text{Var}(W) = \text{Var}(Y) + k^2\text{Var}(X) - 2k\text{Cov}(X, Y) = 1 + k^2 - 2k\rho$$
So $W \sim N(0, 1 + k^2 - 2k\rho)$. Let $\sigma_W = \sqrt{1 + k^2 - 2k\rho}$.
$$P(Y > kX + b) = P(W > b) = 1 - \Phi\!\left(\frac{b}{\sigma_W}\right) = \Phi\!\left(\frac{-b}{\sqrt{1 + k^2 - 2k\rho}}\right)$$
Part 2: $P(Y > kX + b \mid X > x_0)$
The conditional distribution of $Y$ given $X = x$ is:
$$Y \mid X = x \sim N\!\left(\rho x, \, 1 - \rho^2\right)$$
So for a fixed $x$:
$$P(Y > kx + b \mid X = x) = P\!\left(\frac{Y - \rho x}{\sqrt{1 - \rho^2}} > \frac{kx + b - \rho x}{\sqrt{1 - \rho^2}}\right) = \Phi\!\left(\frac{\rho x - kx - b}{\sqrt{1 - \rho^2}}\right)$$
$$= \Phi\!\left(\frac{(\rho - k)x - b}{\sqrt{1 - \rho^2}}\right)$$
Now integrate over $X > x_0$ using the law of total probability:
$$P(Y > kX + b \mid X > x_0) = \frac{P(Y > kX + b, \, X > x_0)}{P(X > x_0)}$$
The denominator is $P(X > x_0) = \Phi(-x_0)$.
The numerator is:
$$P(Y > kX + b, \, X > x_0) = \int_{x_0}^{\infty} \Phi\!\left(\frac{(\rho - k)x - b}{\sqrt{1 - \rho^2}}\right) \phi(x) \, dx$$
where $\phi(x)$ is the standard normal PDF.
This integral can be evaluated in closed form using the identity for $\int \Phi(\alpha + \beta x)\phi(x)\,dx$. Specifically, for $\alpha, \beta \in \mathbb{R}$:
$$\int_{x_0}^{\infty} \Phi(\alpha + \beta x) \phi(x) \, dx = \Phi_2\!\left(-x_0, \, \alpha + \beta \cdot 0, \, \frac{\beta}{\sqrt{1 + \beta^2}}\right) \text{ ... (via bivariate normal CDF)}$$
More precisely, using the standard result: for jointly normal $(X, Z)$ where $Z = \alpha + \beta X + \sqrt{1}\cdot\epsilon$ with $\epsilon$ independent standard normal, the joint event $\{X > x_0, Z > 0\}$ is a bivariate normal probability.
In our case, set $\alpha = -b/\sqrt{1-\rho^2}$ and $\beta = (\rho - k)/\sqrt{1-\rho^2}$. Then:
$$P(Y > kX + b, \, X > x_0) = \Phi_2\!\left(-x_0, \, \frac{-b}{\sqrt{1 + k^2 - 2k\rho}}, \, \frac{\rho - k}{\sqrt{1 + k^2 - 2k\rho}}\right)$$
where $\Phi_2(a, b, r)$ is the standard bivariate normal CDF $P(X_1 > -a, X_2 > -b)$ with correlation $r$.
So the final answer is:
$$P(Y > kX + b \mid X > x_0) = \frac{\Phi_2\!\left(-x_0, \, \frac{-b}{\sqrt{1+k^2-2k\rho}}, \, \frac{\rho-k}{\sqrt{1+k^2-2k\rho}}\right)}{\Phi(-x_0)}$$
Answer:
- $P(Y > kX + b) = \Phi\!\left(\dfrac{-b}{\sqrt{1 + k^2 - 2k\rho}}\right)$
- $P(Y > kX + b \mid X > x_0) = \dfrac{\Phi_2\!\left(-x_0, \, \frac{-b}{\sqrt{1+k^2-2k\rho}}, \, \frac{\rho-k}{\sqrt{1+k^2-2k\rho}}\right)}{\Phi(-x_0)}$
where $\Phi_2(a, b, r)$ denotes the bivariate normal CDF with correlation $r$.
Intuition
Part 1 is a clean application of the fact that linear combinations of jointly normal variables are normal. Whenever you see an event like $\{Y > kX + b\}$ with Gaussian variables, your first move should be to form the difference $W = Y - kX$ and compute its distribution. The answer is always a single $\Phi$ evaluation.
Part 2 is trickier because conditioning on $X > x_0$ (an inequality, not an equality) means you cannot just plug in a single value of $X$. You have to integrate the conditional probability over the truncated distribution of $X$. This integral happens to reduce to a bivariate normal CDF, which is a standard result but not one most people have memorized. In practice, this type of computation arises in credit risk (probability that a correlated obligor defaults given that another has breached a threshold), in options pricing (barrier options with correlated assets), and in signal processing (detection probabilities with correlated noise).