Probability One Normal Exceeds Another

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Problem Statement

If X \sim N(10, 4) and Y \sim N(8, 9) are independent, what is P(X > Y)?

Solution

This problem requires finding the distribution of the difference of two independent normal random variables, then computing a probability using standardization.

Step 1: Reformulate the Problem

We want P(X > Y), which is equivalent to:

P(X > Y) = P(X - Y > 0)

Step 2: Find the Distribution of X - Y

A key property: linear combinations of independent normal random variables are also normally distributed.

For independent normals:

  • X \sim N(\mu_X, \sigma_X^2)

  • Y \sim N(\mu_Y, \sigma_Y^2)

We have:

X - Y \sim N(\mu_X - \mu_Y, \sigma_X^2 + \sigma_Y^2)

Important: Variances add, not subtract! This is because:

\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(-Y) = \text{Var}(X) + (-1)^2 \text{Var}(Y) = \sigma_X^2 + \sigma_Y^2

In our case:

  • X \sim N(10, 4): \mu_X = 10, \sigma_X^2 = 4

  • Y \sim N(8, 9): \mu_Y = 8, \sigma_Y^2 = 9

Therefore:

X - Y \sim N(10 - 8, 4 + 9) = N(2, 13)

Step 3: Standardize to Compute the Probability

We want P(X - Y > 0) where X - Y \sim N(2, 13).

To use standard normal tables, we standardize. For W \sim N(\mu, \sigma^2):

Z = \frac{W - \mu}{\sigma} \sim N(0, 1)

Here, W = X - Y, \mu = 2, and \sigma = \sqrt{13}:

P(X - Y > 0) = P\left(\frac{X - Y - 2}{\sqrt{13}} > \frac{0 - 2}{\sqrt{13}}\right) = P\left(Z > -\frac{2}{\sqrt{13}}\right)

where Z \sim N(0, 1) is standard normal.

Step 4: Use Symmetry of Standard Normal

Since the standard normal is symmetric:

P\left(Z > -\frac{2}{\sqrt{13}}\right) = P\left(Z < \frac{2}{\sqrt{13}}\right) = \Phi\left(\frac{2}{\sqrt{13}}\right)

where \Phi is the standard normal cumulative distribution function.

Step 5: Evaluate Numerically

\frac{2}{\sqrt{13}} = \frac{2}{3.606} \approx 0.555

Using standard normal tables or computational tools:

\Phi(0.555) \approx 0.71

Final Answer

P(X > Y) = \Phi\left(\frac{2}{\sqrt{13}}\right) \approx 0.71

Alternative Expression

We can also write this as:

P(X > Y) = \Phi\left(\frac{\mu_X - \mu_Y}{\sqrt{\sigma_X^2 + \sigma_Y^2}}\right) = \Phi\left(\frac{10 - 8}{\sqrt{4 + 9}}\right) = \Phi\left(\frac{2}{\sqrt{13}}\right)

Key Insights

  1. Difference of Independent Normals: For independent X \sim N(\mu_X, \sigma_X^2) and Y \sim N(\mu_Y, \sigma_Y^2):
X - Y \sim N(\mu_X - \mu_Y, \sigma_X^2 + \sigma_Y^2)

Note that variances add, not subtract!

  1. Variance Addition: This is a common mistake. Even though we’re subtracting Y, the variance of X - Y is \sigma_X^2 + \sigma_Y^2, not \sigma_X^2 - \sigma_Y^2. This is because \text{Var}(aX + bY) = a^2\text{Var}(X) + b^2\text{Var}(Y) for independent variables.

  2. General Formula: For P(X > Y) where X \sim N(\mu_X, \sigma_X^2) and Y \sim N(\mu_Y, \sigma_Y^2) are independent:

P(X > Y) = \Phi\left(\frac{\mu_X - \mu_Y}{\sqrt{\sigma_X^2 + \sigma_Y^2}}\right)
  1. Intuition: The probability depends on:

  • The difference in means (\mu_X - \mu_Y): larger difference favors X

  • The combined variance (\sigma_X^2 + \sigma_Y^2): larger variance makes the outcome more uncertain

  1. Symmetric Case: If \mu_X = \mu_Y, then P(X > Y) = \frac{1}{2} regardless of variances (by symmetry).

Applications

  • A/B Testing: Probability that treatment A outperforms treatment B.

  • Quality Control: Probability that one production line produces higher quality than another.

  • Financial Analysis: Probability that one asset’s return exceeds another’s.

  • Sports Analytics: Probability that one team/player outperforms another.