Probability of failure

A structure will fail if subjected to a load greater then its own resistance:

failure := load > resistance


We can assume that the load and the resistance are independent.

By means of probability density functions (pdf) and cumulative density functions (cdf) of the load and of the resistance, is it correct to say that the probability of failure can be calculated by: $$p_{failure} = \int_{-\infty}^\infty PDF_{load} * CDF_{resistance}$$ ?

I'm trying to teach myself statistics, but sometimes without a good reference it's difficult to know what to look for. What this "tool" would be called by the way? I'm sure there's a proper name for this.

• The 2 methods proposed below by Dilip Sarwate and SeanEaster tested and compared. Implemented using python scipy. Results match perfectly. In terms of computing resources, there is a big difference: the convolution method takes about 70 times longer to run. – Raf Nov 27 '15 at 11:20
• I should have mentioned that this will be slow in you solve the convolution numerically—edited to reflect. – Sean Easter Nov 27 '15 at 15:49

Let $X$ denote the resistance and $Y$ the load. Then, \begin{align} P\{Y > X\} &= \int_{y=-\infty}^\infty \int_{x=-\infty}^y f_{X,Y}(x,y) \,\mathrm dx \,\mathrm dy\\ &= \int_{y=-\infty}^\infty \int_{x=-\infty}^y f_{X}(x)f_{Y}(y) \,\mathrm dx \,\mathrm dy & \scriptstyle{\text{because}~X~\text{and} ~Y~\text{are independent}}\\ &= \int_{y=-\infty}^\infty f_{Y}(y)\left[ \int_{x=-\infty}^y f_{X}(x) \,\mathrm dx\right] \,\mathrm dy\\ &= \int_{y=-\infty}^\infty f_{Y}(y)F_{X}(y) \,\mathrm dy\\ \text{that is}, \qquad p_{failure} &= \int_{-\infty}^\infty PDF_{load} \times CDF_{resistance} \end{align} which is the formula that you are asking about, without needing to worry about convolutions, cross-correlations, complex numbers, and the like as in Sean Easter's answer.

As a practical matter, $X$ and $Y$ are likely to take on nonnegative values only, in which case the above integral need only be on the positive real line.

• Nice. That's exactly how I derived that equation. I wasn't sure though whether it was correct or if there was a more rigorous way of doing it. Convolutions and cross-correlations still elude my understanding. About the domain, I'm actually setting a threshold and only integrating over the non-negligible densities. – Raf Nov 26 '15 at 18:15
• Quite nice! Adding that to my bag of tricks +1 – Sean Easter Nov 26 '15 at 19:14

Rephrased, the probability of failure is equivalent to the probability that resistance - load is less than zero. What you're looking for is the distribution of the difference of random variables.

Since these are independent, you can use convolution to solve for their difference. But it's applied to the densities, not a cumulative density. Also, the convolution is itself an infinite integral. Let $X$ represent load, $Y$ resistance. You'd want to convolve $p_{X}(-t)$ and $p_{Y}(t)$, called the cross-correlation in signal processing:

$$p_{Y-X}(\tau) = p_x(-\tau) \ast p_Y(\tau)= \int_{-\infty}^{\infty}p_{X}(t)p_{Y}(\tau + t)dt$$

Strictly, cross-correlation is equivalent to the convolution of $p_X^*(-\tau)$ and $p_Y(\tau)$, where the asterisk is the complex conjugate. Since densities are real-valued, $p_X^*(-\tau) = p_X(-\tau)$ and there's no need to worry.

The probability of failure is the probability that the difference is less than zero, which you can find by integrating the density of the differences up to zero: $\int_{-\infty}^0p_{Y-X}(\tau)d\tau$. (I.e., the CDF of the difference.) You can do all of this numerically, but the more you can do analytically, the more efficient it will be.

• As you've noted, this convolution will give the density function. From that I fail to see how to get the $p_{failure}$, which is a real value since both load and resistance are fixed distribution. – Raf Nov 26 '15 at 18:11
• Ah, I realize I left out the last step: integrate the density up to 0. – Sean Easter Nov 26 '15 at 18:16
• If you find the time to update your answer to include that final step, you'll get a +1 from me. – Silverfish Nov 26 '15 at 22:28
• @Silverfish So edited, now that my desk is no longer a makeshift Thanksgiving and my desktop again has a place to sit :) – Sean Easter Nov 27 '15 at 15:47
• Perhaps if you note that $\tau$ occurs only in the argument of $p_Y$ so that the integration w.r.t. $\tau$ that you propose ends up replacing $p_Y$ by the CDF of $Y$ evaluated at $t$, you will end up at the same point that my answer did. – Dilip Sarwate Nov 27 '15 at 15:54