How to Find the Probability That X₂ Is Larger Than X₁

To find the probability that one random quantity exceeds another, look at their difference: $X_2 > X_1$ exactly when $X_2 - X_1 > 0$. For normal variables, this turns the comparison into a normal-tail probability. The answer depends not only on the two means and variances, but also on whether the variables are dependent.

If the variables are independent and normally distributed, the probability is

$$P(X_2 > X_1) = \Phi\left(\frac{\mu_2-\mu_1}{\sqrt{\sigma_1^2+\sigma_2^2}}\right),$$

where $\mu_i$ and $\sigma_i^2$ are the mean and variance of $X_i$, and $\Phi$ is the standard normal cumulative distribution function. The formula changes slightly when the variables are correlated.

Why the difference gives the answer

Define $D=X_2-X_1$. Then the comparison is simply

$$P(X_2>X_1)=P(D>0).$$

When $X_1$ and $X_2$ are independent normal variables, their difference is normal as well. Its mean is $\mu_2-\mu_1$, and its variance is the sum of their variances:

$$D \sim \mathcal{N}(\mu_2-\mu_1,\;\sigma_1^2+\sigma_2^2).$$

Standardizing $D$ gives a standard normal variable. The cutoff $D=0$ becomes the standardized value $-(\mu_2-\mu_1)/\sqrt{\sigma_1^2+\sigma_2^2}$. Using the symmetry of the standard normal distribution yields the formula above.

Here, $\mathcal{N}(\mu,\sigma^2)$ uses the convention that the second parameter is the variance, not the standard deviation.

Worked example: independent normal variables

Suppose

$$X_1 \sim \mathcal{N}(10,4), \qquad X_2 \sim \mathcal{N}(12,9),$$

and assume they are independent. Their means are $10$ and $12$, and their variances are $4$ and $9$. Thus $D=X_2-X_1$ has mean $2$ and variance $13$:

$$D \sim \mathcal{N}(2,13).$$

The standard deviation of $D$ is $\sqrt{13}$. Therefore,

$$\begin{aligned} P(X_2>X_1) &=P(D>0)\\ &=\Phi\left(\frac{2}{\sqrt{13}}\right)\\ &\approx \Phi(0.555)\\ &\approx 0.710. \end{aligned}$$

So $X_2$ is larger than $X_1$ about 71% of the time under these assumptions. Notice that this is not the probability that $X_2$ is positive; it is the probability that its realized value exceeds the realized value of $X_1$.

If the variables are correlated

If $X_1$ and $X_2$ are jointly normal but not necessarily independent, their difference is still normal. Its variance includes the covariance:

$$\operatorname{Var}(X_2-X_1)=\sigma_1^2+\sigma_2^2-2\operatorname{Cov}(X_1,X_2).$$

Consequently,

$$P(X_2>X_1)=\Phi\left(\frac{\mu_2-\mu_1}{\sqrt{\sigma_1^2+\sigma_2^2-2\operatorname{Cov}(X_1,X_2)}}\right),$$

provided the difference has positive variance. If the correlation is $\rho$, then $\operatorname{Cov}(X_1,X_2)=\rho\sigma_1\sigma_2$, so the denominator is

$$\sqrt{\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2}.$$

For example, if both variables have variance $1$ and correlation $0.5$, the difference has variance $1+1-2(0.5)=1$. Positive correlation reduces the variability of the difference in this case. Ignoring that dependence would give a different probability.

What if the variables are not normal?

Knowing only the means and variances is generally not enough to determine $P(X_2>X_1)$ for nonnormal variables. You need information about their distributions and dependence. If they are independent and have continuous densities, one general expression is

$$P(X_2>X_1)=\int_{-\infty}^{\infty} f_2(x)F_1(x)\,dx,$$

where $f_2$ is the density of $X_2$ and $F_1$ is the cumulative distribution function of $X_1$. This adds up the probability that $X_2$ takes a value near $x$ and $X_1$ is below that value.

A useful special case does not require normality: if $X_1$ and $X_2$ are independent, identically distributed, and continuous, symmetry gives $P(X_2>X_1)=1/2$. For independent, identically distributed variables that can tie, the correct statement is $P(X_2>X_1)=P(X_1>X_2)=(1-P(X_1=X_2))/2$.

Common mistakes

• Adding standard deviations instead of variances. For independent variables, add $\sigma_1^2$ and $\sigma_2^2$, then take the square root.
• Assuming independence without justification. If the variables are correlated, include their covariance when they are jointly normal.
• Using only the difference of means. The mean difference sets the center, but the spread also affects the probability.
• Confusing the event with a one-variable probability. $P(X_2>X_1)$ compares two random outcomes; it is not generally the same as $P(X_2>0)$.
• Applying the normal formula to arbitrary distributions. The difference is normal if the variables are jointly normal; independent normal marginals are one common sufficient case. Normal marginal distributions alone do not guarantee a jointly normal pair.