Template:Stress-Strength Confidence Intervals

Confidence Intervals on the Probability
Both the stress and strength distributions can be either estimated from actual data or specified by engineers based on engineering knowledge or existing references. Based on the source of the distribution, there are two types of variation associated with the calculated probability.

Variation in Model Parameters

If both the stress and strength distributions are estimated from data sets, then there are uncertainties associated with the estimated distribution parameters. These uncertainties will cause some degree of variation of the probability calculated from the stress-strength analysis. Therefore, we can use these uncertainties to estimate the confidence intervals on the calculated probability. To get the confidence intervals, we first calculate the variance of the reliability, $$R$$, using: $$Var\left[ R \right]=\int_{0}^{\infty }{Var\left[ {{f}_{1}}(x) \right]}{{\left[ {{R}_{2}}(x) \right]}^{2}}dx+{{\int_{0}^{\infty }{\left[ {{f}_{1}}(x) \right]}}^{2}}Var\left[ {{R}_{2}}(x) \right]dx$$

Variance of $${{f}_{1}}(x)$$ and $${{R}_{2}}(x)$$ can be estimated from the Fisher Information Matrix. For details, please see Confidence Bounds.

Once the variance of the expected reliability is obtained, the two-sided confidence intervals can be calculated using:

$$[\frac{R}{R+(1-R)w},\frac{R}{R+(1-R)/w}]$$

where:
 * CL is the confidence level
 * $$\alpha$$ = 1-CL
 * $$w=\exp \{{{z}_{1-\alpha /2}}\sqrt{Var(R)}/[R(1-R)]\}$$
 * $${{Z}_{1-\alpha /2}}$$ is the $$1-\alpha/2$$ percentile of a standard normal distribution.

If the upper bound (U) and lower bound (L) are not infinite and 0, the above calculated variance of $$ R $$ is adjusted by $${{\left[ {1}/{\left( {{F}_{1}}(U)-{{F}_{1}}(L) \right)}\; \right]}^{2}}$$.

Variation in Probability Values

Assume the distributions for stress and strength are known. From the stress-strength equation:

$$R=P[Stress\le Strength]=\int_{0}^{\infty }{{{f}_{Stress}}(x)\cdot {{R}_{Strength}}(x)}dx$$

we know the calculated reliability is the expected value of the probability that a strength value is larger than a stress value. Since both strength and stress are random variables from their distributions, the reliability is also a random variable. This can be explained using the following example. Let's first assume stress is a fixed value of 567. The reliability then is: $$R(567)=\Pr (Strength>567)={{R}_{2}}(567)$$ This is the reliability when the stress value is 567 and when the strength distribution is given. If stress is not a fixed value, instead it follows a distribution, then it can take values other than 567. For instance, it can be a value of 700. Therefore, we get another reliability value of $${R}(700)$$. Since stress is a random variable, for any stress value $${x}_{i}$$, there is a reliability value of $$R({{x}_{i}})$$ calculated from the strength distribution. We will end up with many $$R({{x}_{i}})$$s or $$R_{2}({{x}_{i}})$$s. From these $$R({{x}_{i}})$$s, we can get the mean and variance of the reliability. In fact, its mean is the result from:

$$R=P[Stress\le Strength]=\int_{0}^{\infty }{{{f}_{Stress}}(x)\cdot {{R}_{Strength}}(x)}dx$$

and its variance is:

$$\begin{align} & Var\left[ R \right]=Var\left[ {{R}_{2}}({{X}_{1}}) \right]=E\left[ {{R}_{2}}{{\left( {{X}_{1}} \right)}^{2}} \right]-{{\left( E\left[ {{R}_{2}}\left( {{X}_{1}} \right) \right] \right)}^{2}} \\ & =\int_{0}^{\infty }{{{f}_{1}}(x){{\left[ {{R}_{2}}(x) \right]}^{2}}dx}-{{\left( E\left[ {{R}_{2}}\left( {{X}_{1}} \right) \right] \right)}^{2}} \\ & =\int_{0}^{\infty }{{{f}_{1}}(x){{\left[ {{R}_{2}}(x) \right]}^{2}}dx}-{{\left( R \right)}^{2}} \\ \end{align}$$

where R is the expected value of the reliability.

Once the variance of the expected reliability is obtained, the two-sided confidence intervals can be calculated using:

$$[\frac{R}{R+(1-R)w},\frac{R}{R+(1-R)/w}]$$

where:
 * CL is the confidence level
 * $$\alpha$$ = 1-CL
 * $$w=\exp \{{{z}_{1-\alpha /2}}\sqrt{Var(R)}/[R(1-R)]\}$$
 * $${{Z}_{1-\alpha /2}}$$ is the $$1-\alpha/2$$ percentile of a standard normal distribution

If the upper bound (U) and lower bound (L) are not infinite and 0, the above calculated variance of R is adjusted by $${{\left[ {1}/{\left( {{F}_{1}}(U)-{{F}_{1}}(L) \right)}\; \right]}^{2}}$$.

Example 1: