ML Interview Q Series: Bivariate Normal Probabilities: Deriving P(Y>X|X>0) and P(Y/X≤1) using Arctangent.
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Question
Let the random vector (X, Y) have a standard bivariate normal distribution with correlation coefficient ρ where −1 < ρ < 1. Find the probabilities P(Y > X | X > 0) and P(Y/X ≤ 1).
Short Compact solution
Using polar coordinates and symmetry properties of the standard bivariate normal, one obtains:
Comprehensive Explanation
The pair (X, Y) follows a standard bivariate normal distribution. This means X and Y are jointly normal with mean 0, variance 1 each, and correlation ρ. Such a distribution has the joint probability density function in two variables x and y with the usual bivariate normal form. However, the standard form and correlation structure introduce symmetry properties that allow us to evaluate conditional probabilities more elegantly than directly tackling the double integral in rectangular coordinates.
Key Insight: Relation to a N(0,1) Random Variable Z
A useful trick is to represent Y (conditional on X) as ρ X + sqrt(1−ρ²) Z, where Z is independent of X and is distributed standard normal. This representation preserves the correlation structure between X and Y and simplifies events like Y > X or Y ≥ X.
Probability P(Y > X | X > 0)
We first note that conditioning on X > 0 is equivalent to restricting x to the positive real axis. Then:
P(Y > X | X > 0) = 2 P(Y > X, X > 0)
since P(X > 0) = 1/2 in the standard normal distribution for X alone.
Under the representation Y = ρ X + sqrt(1−ρ²) Z, the event Y > X is the same as ρ X + sqrt(1−ρ²) Z > X, which rearranges to Z > (1−ρ)/ sqrt(1−ρ²} * X, for each fixed X > 0. Integrating over all X > 0 while considering the distribution of Z and X together yields an integral that can be tackled nicely via polar coordinates or symmetry arguments in the two-dimensional plane. After carrying out the integral, the resulting closed-form expression becomes:
P(Y > X | X > 0) = 1/2 − (1/π) arctg((1−ρ) / sqrt(1−ρ²}).
Probability P(Y/X ≤ 1)
The second probability P(Y/X ≤ 1) involves the ratio Y/X. We can split it into two events based on the sign of X:
The event Y ≤ X given X > 0
The event Y ≥ X given X < 0
By symmetry and again using Z = (Y − ρX)/ sqrt(1−ρ²}, one can transform these conditions on Y, X, and Z. Summing the relevant integrals (one over X > 0 and the other over X < 0) and simplifying leads to:
P(Y/X ≤ 1) = 1/2 + (1/π) arctg((1−ρ) / sqrt(1−ρ²}).
Intuitive Geometric Understanding
If we view the plane of (x, y), the line y = x divides the plane into two regions. Because (X, Y) are jointly normal with correlation ρ, the probability of lying in certain sectors depends on the interplay of symmetry and the shape of the bivariate normal contours. The sign constraints X > 0 and X < 0 along with the correlation factor ρ can shift probabilities away from a naive 1/2 guess and thus yield these arctg-based corrections.
Potential Numerical Computation
In a practical setting—for example, if ρ is a known correlation—we can easily compute these probabilities using standard libraries in Python. One could approximate the required integrals directly or plug in the closed-form expressions.
import numpy as np
def p_y_greater_x_given_xpos(rho):
return 0.5 - (1/np.pi)*np.arctan((1 - rho)/np.sqrt(1 - rho**2))
def p_y_over_x_le_one(rho):
return 0.5 + (1/np.pi)*np.arctan((1 - rho)/np.sqrt(1 - rho**2))
# Example usage:
rho_example = 0.3
print("P(Y > X | X > 0) =", p_y_greater_x_given_xpos(rho_example))
print("P(Y/X <= 1) =", p_y_over_x_le_one(rho_example))
These formulae match our closed-form derivations, and they are straightforward to compute for any ρ in (−1, 1).
Follow-up Question: How does the correlation ρ affect these probabilities?
When ρ is close to 1, the variables X and Y are strongly positively correlated. This means Y ≈ X in a random sense, but typically they move in the same direction with little spread. In that scenario:
The event Y > X | X > 0 is less likely because Y is generally near X. The correction term from arctg(...) becomes small or may even shift sign, reducing or increasing the probability subtly.
For P(Y/X ≤ 1), if X and Y are strongly positively correlated, then Y and X will be close to each other (with Y not too frequently exceeding X significantly). That changes the probability distribution of Y/X.
When ρ is close to −1, X and Y move in opposite directions. If X > 0, Y is more likely to be negative, which makes Y < X more probable. For the ratio Y/X ≤ 1, a negative correlation also influences the geometry in a different manner, typically increasing the proportion of times Y ≤ X.
Follow-up Question: Could we arrive at the same results via symmetry without polar coordinates?
Yes, there is a symmetry argument that can simplify the integrals. Instead of performing a direct polar integration, one can exploit:
The fact that for a standard bivariate normal with correlation ρ, if (X, Y) has that distribution, then (−X, −Y) also has the same distribution.
By splitting the plane into regions and leveraging the symmetrical shape of the joint density around the origin, certain integrals reduce to simpler expressions.
This approach often boils down to carefully applying transformations Z = (Y − ρX)/ sqrt(1−ρ²} and X itself, then integrating over half-planes or using known results about bivariate normals crossing certain lines.
Follow-up Question: Are there any edge cases as ρ approaches ±1?
As ρ → +1, the distribution of Y given X converges to Y ≈ X (with small variance around that line). In that extreme, the event Y > X | X > 0 has probability tending to 1/2, since Y and X are almost the same. One can check the limit of the expression 1/2 − (1/π) arctg((1−ρ)/ sqrt(1−ρ²}) as ρ → 1 and verify consistency. A similar check applies as ρ → −1. These limits align with the intuition that if Y ≈ −X, the geometry of Y/X ≤ 1 changes drastically.
Follow-up Question: How would you verify these results with a Monte Carlo simulation?
One can simulate large samples of (X, Y) from a bivariate normal with correlation ρ using a standard library (e.g., NumPy’s random.multivariate_normal). Then compute the empirical frequency of (Y > X, X > 0) and (Y/X ≤ 1), compare them with the formula-based values, and observe convergence as sample size grows. This numerical check is a common approach in validating closed-form probability results in practical ML or data science pipelines.
Below are additional follow-up questions
How do these probabilities change if X and Y are not marginally standard normal but still jointly normal with correlation ρ?
When X and Y are both normal with arbitrary means (say μ_X and μ_Y) and standard deviations (say σ_X and σ_Y) but the same correlation ρ, we can transform X and Y to the standard normal setting by defining X' = (X − μ_X)/σ_X and Y' = (Y − μ_Y)/σ_Y. The pair (X', Y') then follows a standard bivariate normal with correlation ρ. Consequently, the events Y > X or Y/X ≤ 1 in terms of X and Y translate into equivalent events in terms of X' and Y'. Specifically:
Y > X becomes (Y'σ_Y + μ_Y) > (X'σ_X + μ_X), which is an affine transformation but preserves the correlation structure.
Y/X ≤ 1 becomes ((Y'σ_Y + μ_Y) / (X'σ_X + μ_X)) ≤ 1, which similarly can be reformulated in terms of X' and Y'.
Since the correlation remains ρ after standardization, the same formula applies, but you have to shift and scale your data to standard normals first if you want to apply these closed-form expressions directly. A subtlety arises when X is not guaranteed to be positive in its original scale—if you condition on X > 0 in the original unstandardized variables, you must carefully translate that condition into X' > (−μ_X)/σ_X in the standardized domain.
A real-world pitfall is forgetting to standardize X and Y consistently before applying the known results for standard bivariate normals. Failing to account for different variances or nonzero means can lead to incorrect probability estimates.
How would missing data in X or Y affect estimation of these probabilities in practice?
In real-world settings, you might not observe all pairs (X, Y). Missing data in X or Y could lead to biased estimates of the correlation ρ if the missingness is not completely random. For instance, if large positive X values are systematically missing, then your sample correlation might be underestimated or overestimated, depending on how Y is recorded. Consequently:
The estimated correlation ρ_hat might deviate from the true ρ, impacting probabilities computed from the arctg(...) formulas.
If there is a systematic pattern to the missingness—e.g., always missing Y when X is above a certain threshold—your conditional events (like Y > X | X > 0) become unrepresentative of the true population.
Dealing with missing data typically involves techniques like multiple imputation or maximum likelihood estimation under a joint normal model, each requiring careful validation. If these steps are overlooked or done improperly, the final probabilities can be far off from the true values.
What if we only observe X when X > 0 (a truncated sample) but still want to estimate P(Y > X | X > 0)?
In a truncated data setting, suppose we only record pairs (X, Y) for which X > 0, and discard or never see data for X ≤ 0. The distribution of (X, Y) in this truncated sample is no longer the same as the original bivariate normal because the range of X is restricted to positive values. Hence, the sample correlation computed from this truncated sample is not the same as the original correlation in the full dataset. This truncation can create an artificial dependence structure between X and Y that does not represent the entire population.
To properly compute P(Y > X | X > 0) under truncation, you need to fit the truncated bivariate normal model, which has a known but more complex likelihood function. Once you estimate the correlation of the truncated distribution, you can then incorporate that into the conditional probability. One major pitfall is to naïvely plug the truncated-sample correlation into the same arctg(...) formula derived for the full, untruncated scenario. Doing so neglects the selection bias caused by observing only X > 0. Specialized maximum likelihood methods or Bayesian approaches with truncated distributions are necessary to get correct estimates.
How do outliers in the data affect the calculation of these probabilities?
Outliers—particularly in one variable—can heavily distort the sample correlation ρ when data come from real-world processes that might deviate from exact normality. For example, if Y has rare but extreme values, the empirical correlation can shift dramatically away from the theoretical correlation we would expect under a “clean” bivariate normal. This shift in correlation translates directly into inaccuracies when one uses the arctg(...) expressions.
In a robust statistics framework, one might use robust correlation estimates (e.g., Spearman’s rank correlation or Kendall’s tau) instead of the Pearson correlation if outliers are common. However, the standard formula for P(Y > X | X > 0) and P(Y/X ≤ 1) is strictly derived for the Pearson correlation under a true bivariate normal distribution. Hence, if the data do not meet these conditions due to outliers or heavier tails (like a Student-t distribution), the derived formula becomes approximate or invalid. A practical solution involves modeling the data with a more suitable distribution or applying robust methods to dampen outlier impact.
How does the probability P(Y > X | X > 0) behave if we consider instead the event Y ≥ X, and does this strict vs. non-strict inequality matter?
Mathematically, the difference between Y > X and Y ≥ X is negligible for continuous distributions like the bivariate normal because the probability that Y = X exactly is zero. Hence:
P(Y > X | X > 0) = P(Y ≥ X | X > 0)
for the continuous bivariate normal distribution. But in discrete or mixed distributions (for instance, if X and Y took integer values or had a point mass at some location), there would be a difference. For continuous bivariate normal, no real difference arises. This nuance is important in practical applications where variables might be quantized (e.g., measuring X and Y in discrete bins). In purely continuous settings, treating strict or non-strict comparisons will give you the same numeric probability.
Are these formula-based probabilities directly applicable to elliptical distributions beyond the normal?
While many elliptical distributions (like Student-t) share contours that are concentric “ellipses” and have correlation-type parameters, the specific formula involving arctg(...) is special to the normal family. In elliptical distributions, the geometry might look similar, but the joint density and tail behaviors differ. For instance, if (X, Y) followed a bivariate Student-t with correlation ρ, the region Y > X does not integrate to the same closed form. The difference arises from heavier tails and a different characteristic function.
Therefore, if you attempt to use these formulas for elliptical distributions like Student-t or Pearson Type II/Type VII, your result will likely be an approximation that ignores heavier (or lighter) tail effects. If you must handle such distributions, you either need specialized formulae (if they exist) or rely on numerical integration or simulation. Overlooking this distinction is a common pitfall that leads to erroneous probability assessments when the actual data exhibit heavy tails.
In a high-dimensional setting, could these results for two variables be extended to multiple comparisons among multiple correlated variables?
In many real ML tasks—like portfolio risk in finance or multi-dimensional feature spaces in ML—one might have more than two variables. You may want to compare one variable to several others simultaneously. For instance, you might ask: “What is the probability that Y > X1, Y > X2, and so on?” In a multivariate normal context, each subset of variables is itself jointly normal, and one could try to expand the polar coordinate approach or use known results for the distribution of linear combinations of correlated normals. However, closed-form expressions akin to the simple arctg(...) formulas typically do not exist in higher dimensions.
Instead, we often rely on:
Numerical methods: multi-dimensional integration techniques or Monte Carlo simulations.
Analytical approximations: for instance, using special function expansions or copula-based approximations for high dimensions.
A major pitfall is attempting to naïvely “stack” two-dimensional results for multiple comparisons without respecting the joint dependence structure among all variables. Bivariate methods might incorrectly assume independence among partial correlations. Consequently, in high-dimensional scenarios, the simple 2D approach does not straightforwardly extend, and specialized high-dimensional correlation or covariance structure analyses are essential.
How would you incorporate prior knowledge about ρ in a Bayesian framework for estimating P(Y > X | X > 0)?
In a Bayesian context, instead of treating ρ as a fixed unknown parameter and estimating it via maximum likelihood, you place a prior distribution on ρ (typically constrained to −1 < ρ < 1). You also specify priors for the mean and variance of X and Y if needed. The posterior distribution then captures our uncertainty about ρ (and other parameters) after seeing data.
To compute probabilities such as P(Y > X | X > 0), you would:
Draw posterior samples of (ρ, μ_X, μ_Y, σ_X, σ_Y) if dealing with non-standard normals.
For each draw, transform to the standard normal scenario if needed, then calculate the theoretical probability using the arctg(...) expression.
Average those computed probabilities across all posterior samples to get the posterior predictive distribution for P(Y > X | X > 0).
The main benefit is that this approach naturally accounts for parameter uncertainty. A pitfall is computational complexity, as sampling a posterior in high dimensions can be slow. Also, if the prior is not well chosen, it can unduly influence the posterior correlation estimate. But for many practical ML or data science tasks, a Bayesian approach yields robust and interpretable inferences about these conditional probabilities.