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TL;DR

Events A and B are independent if \( P(A \cap B) = P(A)P(B) \). For random variables, X and Y are independent if their joint distribution factors: \( f(x,y) = f_X(x)f_Y(y) \).

By Valenke Exam Prep Team·Last updated 2026-06-03

Statistical Independence

Events A and B are independent if \( P(A \cap B) = P(A)P(B) \). For random variables, X and Y are independent if their joint distribution factors: \( f(x,y) = f_X(x)f_Y(y) \).

Why it matters for interviews

Independence assumptions simplify calculations enormously but are often violated in financial data. Knowing when independence holds (and when it fails) is critical for correct probability reasoning in interviews.

Definition and Mathematical Foundation

Events A and B are independent if \( P(A \cap B) = P(A)P(B) \). For random variables, X and Y are independent if their joint distribution factors: \( f(x,y) = f_X(x)f_Y(y) \).

Application in Quantitative Finance

Independence assumptions simplify calculations enormously but are often violated in financial data. Knowing when independence holds (and when it fails) is critical for correct probability reasoning in interviews.

Related Terms

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Frequently Asked Questions

Does uncorrelation imply independence?
No. Uncorrelated means \( \text{Cov}(X,Y) = 0 \), which only rules out linear dependence. Nonlinear dependencies can exist. The exception: for jointly normal variables, uncorrelation does imply independence.
What is conditional independence?
X and Y are conditionally independent given Z if \( P(X,Y|Z) = P(X|Z)P(Y|Z) \). This is weaker than marginal independence and is fundamental to graphical models and Bayesian networks.
Why do quant interviewers test independence so heavily?
Many probability formulas (e.g., variance of sums, MGF of sums) simplify dramatically under independence. Interviewers test whether candidates know when to apply these simplifications and when dependence invalidates them.