Probability Product of Normals is Positive

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Problem Statement

If X, Y \sim N(0, 1) are independent, what is P(XY > 0)?

Solution

This problem requires understanding when the product of two numbers is positive, which occurs when both numbers have the same sign.

Step 1: Identify When the Product is Positive

The product XY > 0 when:

  • Both X > 0 and Y > 0 (both positive), OR

  • Both X < 0 and Y < 0 (both negative)

These are mutually exclusive events, so:

P(XY > 0) = P(X > 0, Y > 0) + P(X < 0, Y < 0)

Step 2: Use Independence

Since X and Y are independent, the joint probability equals the product of marginal probabilities:

P(X > 0, Y > 0) = P(X > 0) \cdot P(Y > 0)
P(X < 0, Y < 0) = P(X < 0) \cdot P(Y < 0)

Step 3: Compute Individual Probabilities

For a standard normal distribution N(0, 1), by symmetry:

  • P(X > 0) = \frac{1}{2} (half the probability mass is above the mean)

  • P(X < 0) = \frac{1}{2} (half the probability mass is below the mean)

  • P(X = 0) = 0 (continuous distribution)

Similarly for Y:

  • P(Y > 0) = \frac{1}{2}

  • P(Y < 0) = \frac{1}{2}

Step 4: Calculate the Final Probability

P(XY > 0) = P(X > 0, Y > 0) + P(X < 0, Y < 0)
= P(X > 0) \cdot P(Y > 0) + P(X < 0) \cdot P(Y < 0)
= \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Final Answer

P(XY > 0) = \frac{1}{2} or 0.5

Alternative Approach: Geometric Interpretation

We can visualize this in the (X, Y) plane:

  • The condition XY > 0 corresponds to the first and third quadrants.

  • Each quadrant has probability \frac{1}{4} (by independence and symmetry).

  • Therefore, P(XY > 0) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.

Key Insights

  1. Sign Analysis: The product of two numbers is positive when they have the same sign, negative when they have opposite signs.

  2. Symmetry of Standard Normal: For X \sim N(0, 1), the distribution is symmetric about 0, so P(X > 0) = P(X < 0) = \frac{1}{2}.

  3. Independence Simplifies: Independence allows us to multiply probabilities: P(X \in A, Y \in B) = P(X \in A) \cdot P(Y \in B).

  4. Generalization: For independent X, Y \sim N(\mu, \sigma^2) (not necessarily standard normal), if \mu \neq 0, the calculation becomes more complex, but the principle remains the same.

  5. Correlated Case: If X and Y are correlated, the calculation changes because P(X > 0, Y > 0) \neq P(X > 0) \cdot P(Y > 0).

Applications

  • Sign Agreement: Probability that two independent measurements have the same sign.

  • Correlation Testing: Understanding when two variables move in the same direction.

  • Decision Making: Probability that two independent factors both favor a particular outcome.