Correlated Variables Example

Problem

A component is checked against both yield and ultimate failure. Yield strength Sy and ultimate strength Su are correlated with rho = 0.85. If they are incorrectly assumed independent, how does the failure probability change?

Given

• Sy ~ N(350, 25^2) MPa
• Su ~ N(480, 35^2) MPa
• Correlation: rho = 0.85
• Applied stress: 300 MPa (deterministic)

Method

Compute the probability that BOTH Sy < 300 AND Su < 300 (parallel system — both must fail for the component to fail by both criteria). Compare the independent and correlated cases.

Solution (illustrative)

P(Sy < 300) = Phi((300-350)/25) = Phi(-2.0) = 0.0228
P(Su < 300) = Phi((300-480)/35) = Phi(-5.14) ≈ 1.4e-7

If independent:
P(both) = 0.0228 * 1.4e-7 ≈ 3.2e-9

If correlated (rho = 0.85):
The joint probability is higher than the product
because both strengths tend to be low together.
The exact value requires bivariate normal integration.

Interpretation

For a parallel system (both must fail), positive correlation increases the joint failure probability above the independent estimate. For a series system (either fails), positive correlation decreases the system failure probability below the independent estimate. Ignoring correlation can give misleading results in either direction.