Let $(X, Y)$ be jointly normal with zero means, unit variances, and correlation $\rho \in (-1, 1)$. For constants $k$ and $b$: 1. Compute $P(Y > kX + b)$ in closed form. 2. Compute $P(Y > kX + b \mid X > x_0)$ in closed form. 3. Give the special cases when $b = 0$ and when $x_0 = 0$, and explain th…

Gaussian Wedge Probability

Probability · Hard · Free problem

Let $(X, Y)$ be jointly normal with zero means, unit variances, and correlation $\rho \in (-1, 1)$.

For constants $k$ and $b$:

Compute $P(Y > kX + b)$ in closed form.

Compute $P(Y > kX + b \mid X > x_0)$ in closed form.

Give the special cases when $b = 0$ and when $x_0 = 0$, and explain the geometric symmetry that arises when $k = -1$.

Hints

The variable $Y - kX$ is a linear combination of jointly normal variables -- what is its distribution?
For the conditional probability, use $Y \mid X = x \sim N(\rho x,\, 1 - \rho^2)$ and integrate $P(Y > kx + b \mid X = x)$ against $\phi(x)$ over $x > x_0$. The result involves the bivariate normal CDF $\Phi_2$.
Apply the identity $\int_a^{\infty} \Phi(\alpha + \beta x)\phi(x)\,dx = \Phi_2(-a, \alpha/\sqrt{1+\beta^2};\, \beta/\sqrt{1+\beta^2})$ with $\alpha = -b/\sqrt{1-\rho^2}$ and $\beta = (\rho - k)/\sqrt{1-\rho^2}$.

Worked Solution

How to Think About It: The region $Y > kX + b$ is a half-plane in the $(X, Y)$ plane, and $(X, Y)$ lives on a bivariate normal. Part 1 boils down to recognizing that $Y - kX$ is itself normal, so the unconditional probability is just a univariate $\Phi$ evaluation. Part 2 is harder: you condition on $X > x_0$ and integrate out $X$ using the conditional distribution $Y \mid X$. The resulting integral has a clean closed form in terms of the bivariate normal CDF $\Phi_2$. Before diving in, note the sanity check: when $b = 0$ the boundary passes through the origin (the mean of the joint distribution), and by symmetry the unconditional probability should be exactly

/2$.

Quick Estimate: Take $\rho = 0.5$, $k = 1$, $b = 1$. Then $Y - kX \sim N(0, 1 + 1 - 2(0.5)(1)) = N(0, 1)$. So $P(Y > X + 1) = \Phi(-1) \approx 0.159$. Makes sense -- you need $Y$ to beat $X$ by more than 1, and when $\rho > 0$ they tend to move together, making large gaps rarer.

Approach: For Part 1, analyze $Y - kX$ as a linear combination of jointly normal variables. For Part 2, use the law of total probability with the conditional $Y \mid X = x$, then apply a standard identity relating the resulting integral to $\Phi_2$.

Formal Solution:

Part 1: Unconditional probability

Since $(X, Y)$ is bivariate normal, the linear combination $W = Y - kX$ is normal with:

$E[W] = 0, \quad \text{Var}(W) = 1 + k^2 - 2k\rho$

Let $\sigma_W = \sqrt{1 + k^2 - 2k\rho}$. Then:

$P(Y > kX + b) = P(W > b) = \Phi\!\left(\frac{-b}{\sigma_W}\right) = \Phi\!\left(\frac{-b}{\sqrt{1 + k^2 - 2k\rho}}\right)$

Part 2: Conditional probability given $X > x_0$

The conditional distribution of $Y$ given $X = x$ is:

$Y \mid X = x \;\sim\; N\!\left(\rho x,\; 1 - \rho^2\right)$

So the conditional exceedance probability at a fixed $x$ is:

$P(Y > kx + b \mid X = x) = \Phi\!\left(\frac{\rho x - kx - b}{\sqrt{1 - \rho^2}}\right) = \Phi\!\left(\frac{(\rho - k)x - b}{\sqrt{1 - \rho^2}}\right)$

Now integrate over the conditioning event $X > x_0$:

$P(Y > kX + b \mid X > x_0) = \frac{1}{\Phi(-x_0)} \int_{x_0}^{\infty} \Phi\!\left(\frac{(\rho - k)x - b}{\sqrt{1 - \rho^2}}\right) \phi(x)\, dx$

where $\phi$ is the standard normal density. We apply the identity:

$\int_a^{\infty} \Phi(\alpha + \beta x)\,\phi(x)\,dx = \Phi_2\!\left(-a,\; \frac{\alpha}{\sqrt{1 + \beta^2}};\; \frac{\beta}{\sqrt{1 + \beta^2}}\right)$

with $a = x_0$, $\alpha = \frac{-b}{\sqrt{1 - \rho^2}}$, and $\beta = \frac{\rho - k}{\sqrt{1 - \rho^2}}$. Computing the required quantities:

$1 + \beta^2 = \frac{1 + k^2 - 2k\rho}{1 - \rho^2}$

$\frac{\alpha}{\sqrt{1+\beta^2}} = \frac{-b}{\sqrt{1 + k^2 - 2k\rho}}, \qquad \frac{\beta}{\sqrt{1+\beta^2}} = \frac{\rho - k}{\sqrt{1 + k^2 - 2k\rho}}$

Define $\sigma_W = \sqrt{1 + k^2 - 2k\rho}$ and $\tilde{\rho} = \frac{\rho - k}{\sigma_W}$. Then:

$\boxed{P(Y > kX + b \mid X > x_0) = \frac{\Phi_2\!\left(-x_0,\; \frac{-b}{\sigma_W};\; \tilde{\rho}\right)}{\Phi(-x_0)}}$

where $\Phi_2(a, b; r)$ denotes the standard bivariate normal CDF $P(Z_1 \leq a,\, Z_2 \leq b)$ with correlation $r$.

Part 3: Special cases and geometric symmetry

*Special case $b = 0$:* The unconditional probability becomes $\Phi(0) = 1/2$. This is immediate: $W = Y - kX$ has mean zero, and $P(W > 0) = 1/2$ for any zero-mean normal. Geometrically, the boundary $Y = kX$ passes through the mean of the distribution, so by the symmetry of the Gaussian density about any line through its center, each half-plane has equal probability.

For the conditional probability with $b = 0$:

$P(Y > kX \mid X > x_0) = \frac{\Phi_2(-x_0, 0;\, \tilde{\rho})}{\Phi(-x_0)}$

which is no longer

/2$ in general because conditioning on $X > x_0$ breaks the symmetry.

*Special case $x_0 = 0$:* Since $\Phi(0) = 1/2$, the conditional probability simplifies to:

$P(Y > kX + b \mid X > 0) = 2\,\Phi_2\!\left(0,\; \frac{-b}{\sigma_W};\; \tilde{\rho}\right)$

*Geometric symmetry when $k = -1$:* When $k = -1$ the boundary is the line $Y = -X + b$, equivalently $X + Y = b$. The variable $S = X + Y \sim N(0, 2 + 2\rho)$, so the unconditional probability is $\Phi(-b/\sqrt{2 + 2\rho})$. The key geometric point is that the line $X + Y = b$ is perpendicular to the direction $y = x$. For a standard bivariate normal, the density contours are ellipses whose axes of symmetry lie along $y = x$ and $y = -x$ (these are the eigenvectors of the covariance matrix with eigenvalues

+ \rho$ and

- \rho$). The boundary $X + Y = b$ is aligned with one axis of symmetry. When $b = 0$ this line passes through the center of the ellipses, giving $P = 1/2$ by exact reflection symmetry. For $b \neq 0$, the probability depends only on $b/\sqrt{2 + 2\rho}$, which is the signed distance from the mean measured in units of $\text{std}(X + Y)$.

Answer:

$P(Y > kX + b) = \Phi\!\left(\frac{-b}{\sqrt{1 + k^2 - 2k\rho}}\right)$

$P(Y > kX + b \mid X > x_0) = \frac{\Phi_2\!\left(-x_0,\; \frac{-b}{\sqrt{1+k^2-2k\rho}};\; \frac{\rho-k}{\sqrt{1+k^2-2k\rho}}\right)}{\Phi(-x_0)}$

When $b = 0$: the unconditional probability is exactly

/2$. When $x_0 = 0$: multiply the bivariate CDF by 2. When $k = -1$: the boundary aligns with a symmetry axis of the Gaussian ellipse, reducing everything to a one-dimensional problem in $X + Y$.

Intuition

This problem is really about the geometry of slicing a Gaussian blob with a hyperplane. Unconditionally, any half-plane probability for a bivariate normal reduces to a univariate $\Phi$ evaluation because the signed distance from the mean to the boundary, measured in the right units, is all that matters. The conditional version is richer: restricting $X > x_0$ carves out a wedge-shaped region (hence "Gaussian wedge"), and the probability of landing in the intersection of two half-planes is precisely what $\Phi_2$ was invented to compute. The auxiliary correlation $\tilde{\rho} = (\rho - k)/\sqrt{1 + k^2 - 2k\rho}$ captures how the original correlation $\rho$ and the slope $k$ of the boundary interact.

In practice this shows up constantly: any time you ask "what is the probability that a correlated pair of returns both exceed thresholds?" or "given that one factor is extreme, what is the probability the other breaches a barrier?" you are computing a Gaussian wedge probability. The $k = -1$ symmetry case is particularly useful -- it tells you that for sum-type constraints ($X + Y > b$), the problem collapses to one dimension along the principal axis, which is a fast mental shortcut in interviews and on the desk.

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