常见的可靠性指标及其概率解释

失效分布和平均寿命

剩余寿命

具有年龄t的产品从t开始继续使用下去直到失效为止所经历的时间，记为 ξt ξ t <script type="math/tex" id="MathJax-Element-1">\xi_t</script>

Ft(x)=P(ξt⩽x)=P(ξ⩽t+x|ξ>t)=F(t+x)−F(t)1−F(x) F t ( x ) = P ( ξ t ⩽ x ) = P ( ξ ⩽ t + x | ξ > t ) = F ( t + x ) − F ( t ) 1 − F ( x )
<script type="math/tex; mode=display" id="MathJax-Element-2"> \begin{aligned} F_t(x) &= P(\xi_t \leqslant x) = P(\xi \leqslant t + x | \xi > t) \\ &= \frac{F(t + x) - F(t)}{1 - F(x)} \end{aligned} </script>

平均寿命

MTTF或MTBF

E(ξ)=∫∞0tf(t)dt E ( ξ ) = ∫ 0 ∞ t f ( t ) d t
<script type="math/tex; mode=display" id="MathJax-Element-3"> E(\xi) = \int_{0}^{\infty} t f(t) dt </script>

可靠度和可靠寿命

要求 R(t)=R R ( t ) = R <script type="math/tex" id="MathJax-Element-4">R(t) = R</script>，求相应的时间t，这个时间称为可靠寿命 tR t R <script type="math/tex" id="MathJax-Element-5">t_R</script>

  • R=0.5 R = 0.5 <script type="math/tex" id="MathJax-Element-6">R = 0.5</script>时 t0.5 t 0.5 <script type="math/tex" id="MathJax-Element-7">t_{0.5}</script>称为中位寿命
  • R=e−1 R = e − 1 <script type="math/tex" id="MathJax-Element-8">R = e^{-1}</script>时 te−1 t e − 1 <script type="math/tex" id="MathJax-Element-9">t_{e^{-1}}</script>称为特征寿命

数据的初步整理分析

直方图

分为频数（r）、频率（f）、累计频率（F）、失效率（λ）直方图

组数： k=1+3.3lg(n) k = 1 + 3.3 l g ( n ) <script type="math/tex" id="MathJax-Element-10">k = 1 + 3.3lg(n)</script>

经验分布函数

Fn(t)=⎧⎩⎨0in1,t<t1,ti⩽t<ti+1,t⩾tn F n ( t ) = { 0 , t < t 1 i n , t i ⩽ t < t i + 1 1 , t ⩾ t n
<script type="math/tex; mode=display" id="MathJax-Element-11"> F_n(t) = \begin{cases} 0 &, t < t_1 \\ \frac{i}{n} &, t_i \leqslant t < t_{i + 1} \\ 1 &, t \geqslant t_n \end{cases} </script>

完全样本

样本容量较大时： Fn(ti)=in F n ( t i ) = i n <script type="math/tex" id="MathJax-Element-12">F_n(t_i) = \frac{i}{n}</script>

样本容量较小（ n⩽20 n ⩽ 20 <script type="math/tex" id="MathJax-Element-13">n \leqslant 20</script>）时：

  • 海森公式： Fn(ti)=i−0.5n F n ( t i ) = i − 0.5 n <script type="math/tex" id="MathJax-Element-14">F_n(t_i) = \frac{i - 0.5}{n}</script>
  • 数学期望公式： Fn(ti)=in+1 F n ( t i ) = i n + 1 <script type="math/tex" id="MathJax-Element-15">F_n(t_i) = \frac{i}{n + 1}</script>
  • 近似中位秩公式： Fn(ti)=i−0.3n+0.4 F n ( t i ) = i − 0.3 n + 0.4 <script type="math/tex" id="MathJax-Element-16">F_n(t_i) = \frac{i - 0.3}{n + 0.4}</script>

当 i=n=1 i = n = 1 <script type="math/tex" id="MathJax-Element-17">i = n = 1</script>时它们都等于0.5

删失样本

用乘积限估计：

R^(t)=⎧⎩⎨⎪⎪1∏ji=1(n−in−i+1)δi0,t<t1,tj⩽t<tj+1,t⩾tn R ^ ( t ) = { 1 , t < t 1 ∏ i = 1 j ( n − i n − i + 1 ) δ i , t j ⩽ t < t j + 1 0 , t ⩾ t n
<script type="math/tex; mode=display" id="MathJax-Element-18"> \hat{R}(t) = \begin{cases} 1 &, t < t_1 \\ \prod_{i = 1}^{j} (\frac{n - i}{n - i + 1})^{\delta_i} &, t_j \leqslant t < t_{j + 1} \\ 0 &, t \geqslant t_n \end{cases} </script>

随机截尾寿命试验的可靠度函数计算

残存比率法

残存比率法是由概率乘法公式得来的，因此它适用于样本量较大的情况

定义产品在时间区间 (ti−1,ti) ( t i − 1 , t i ) <script type="math/tex" id="MathJax-Element-19">(t_{i-1}, t_i)</script>内的残存概率 S(ti) S ( t i ) <script type="math/tex" id="MathJax-Element-20">S(t_i)</script>，它是一个条件概率。表示在 ti−1 t i − 1 <script type="math/tex" id="MathJax-Element-21">t_{i-1}</script>时刻能完好工作的产品继续工作至 ti t i <script type="math/tex" id="MathJax-Element-22">t_i</script>时刻尚能完好工作的概率

S(ti)=P(ξ>ti|ξ>ti−1)=R(ti)R(ti−1) S ( t i ) = P ( ξ > t i | ξ > t i − 1 ) = R ( t i ) R ( t i − 1 )
<script type="math/tex; mode=display" id="MathJax-Element-23"> \begin{aligned} S(t_i) &= P(\xi > t_i | \xi > t_{i - 1}) \\ &= \frac{R(t_i)}{R(t_{i - 1})} \end{aligned} </script>

产品在某时刻ti的可靠度

R(ti)=∏j=1iS(tj) R ( t i ) = ∏ j = 1 i S ( t j )
<script type="math/tex; mode=display" id="MathJax-Element-24"> \begin{aligned} R(t_i) &= \prod_{j = 1}^{i} S(t_j) \end{aligned} </script>

S(ti) S ( t i ) <script type="math/tex" id="MathJax-Element-25">S(t_i)</script>可以由样本估计

平均秩次法

平均秩次法可用于样本量较小的情况，它采用了中位秩公式

算出平均秩次i，带入中位秩公式： Fn(tk)=i−0.3n+0.4 F n ( t k ) = i − 0.3 n + 0.4 <script type="math/tex" id="MathJax-Element-26">F_n(t_k) = \frac{i - 0.3}{n + 0.4}</script>

例如：如果F2秩次为2，它前面和后面的排列总共有24种；如果F2秩次为3，它前面和后面的排列总共有6种，F2的平均秩次由2、3的种类加权平均得 Fn(t2)=6∗3+24∗26+24=2.2 F n ( t 2 ) = 6 ∗ 3 + 24 ∗ 2 6 + 24 = 2.2 <script type="math/tex" id="MathJax-Element-27">F_n(t_2) = \frac{6 * 3 + 24 * 2}{6 + 24} = 2.2</script>

样本量多时用增量公式： ΔAk=Ak−Ak−1=n+1−Ak−1n−i+2 Δ A k = A k − A k − 1 = n + 1 − A k − 1 n − i + 2 <script type="math/tex" id="MathJax-Element-28">\Delta A_k = A_k - A_{k - 1} = \frac{n + 1 - A_{k - 1}}{n - i + 2}</script>

顺序统计量的应用

第k个顺序统计量的分布

设总体 ξ ξ <script type="math/tex" id="MathJax-Element-29">\xi</script>的分布函数为 F(t) F ( t ) <script type="math/tex" id="MathJax-Element-30">F(t)</script>，密度函数为 f(t) f ( t ) <script type="math/tex" id="MathJax-Element-31">f(t)</script>，则容量为n的子样，其第k个顺序统计量 T(k) T ( k ) <script type="math/tex" id="MathJax-Element-32">T(k)</script>的分布密度函数为：

fT(k)(t)=nCk−1n−1[F(t)]k−1[1−F(t)]n−kf(t) f T ( k ) ( t ) = n C n − 1 k − 1 [ F ( t ) ] k − 1 [ 1 − F ( t ) ] n − k f ( t )
<script type="math/tex; mode=display" id="MathJax-Element-33"> f_{T(k)}(t) = n C_{n - 1}^{k - 1} [F(t)]^{k - 1} [1 - F(t)]^{n - k} f(t) </script>

顺序统计量的联合分布、经验分布函数的分布

略

各种截尾样本的联合分布

fT1,T2,...,Tr(t1,t2,...,tr)= f T 1 , T 2 , . . . , T r ( t 1 , t 2 , . . . , t r ) = <script type="math/tex" id="MathJax-Element-34"> f_{T_1, T_2, ..., T_r}(t_1, t_2, ..., t_r) = </script>

  • 完全寿命试验： n!∏ni=1f(ti) n ! ∏ i = 1 n f ( t i ) <script type="math/tex" id="MathJax-Element-35"> n! \prod_{i = 1}^{n} f(t_i) </script>
  • 定数截尾试验： n!(n−r)![∏ri=1f(ti)][1−F(tr)]n−r n ! ( n − r ) ! [ ∏ i = 1 r f ( t i ) ] [ 1 − F ( t r ) ] n − r <script type="math/tex" id="MathJax-Element-36"> \frac{n!}{(n - r)!} [\prod_{i = 1}^{r} f(t_i)] [1 - F(t_r)]^{n - r} </script>（失效的时间为 ti t i <script type="math/tex" id="MathJax-Element-37">t_i</script>，顺序已定；未失效的为 tr t r <script type="math/tex" id="MathJax-Element-38">t_r</script>，(n - r)个排列）
  • 定时截尾试验： n!(n−r)![∏ri=1f(ti)][1−F(t0)]n−r n ! ( n − r ) ! [ ∏ i = 1 r f ( t i ) ] [ 1 − F ( t 0 ) ] n − r <script type="math/tex" id="MathJax-Element-39"> \frac{n!}{(n - r)!} [\prod_{i = 1}^{r} f(t_i)] [1 - F(t_0)]^{n - r} </script>（把 tr t r <script type="math/tex" id="MathJax-Element-40">t_r</script>换为 t0 t 0 <script type="math/tex" id="MathJax-Element-41">t_0</script>）
  • 随机截尾试验： C∏ri=1f(ti)∏ki=1[1−F(τi)] C ∏ i = 1 r f ( t i ) ∏ i = 1 k [ 1 − F ( τ i ) ] <script type="math/tex" id="MathJax-Element-42"> C \prod_{i = 1}^{r} f(t_i) \prod_{i = 1}^{k} [1 - F(\tau_i)] </script>（C不能用一个统一的表达式表示，此处暂以C代之，因为在实际应用中，其大小无影响）

常见失效分布

二项分布

n次伯努利实验成功次数

  • P(X=x)=Cxnpx(1−p)n−x P ( X = x ) = C n x p x ( 1 − p ) n − x <script type="math/tex" id="MathJax-Element-43"> P(X = x) = C_n^x p^x (1 - p)^{n - x} </script>
  • E(X)=np E ( X ) = n p <script type="math/tex" id="MathJax-Element-44"> E(X) = np </script>
  • D(X)=np(1−p) D ( X ) = n p ( 1 − p ) <script type="math/tex" id="MathJax-Element-45"> D(X) = np(1 - p) </script>

当n很大，p很小时近似于泊松分布， λ=np λ = n p <script type="math/tex" id="MathJax-Element-46">\lambda = np</script>

二项分布的和服从贝塔分布

负二项分布

预定成功次数s，求试验次数

  • P(X=x)=Cs−1x−1ps(1−p)x−s P ( X = x ) = C x − 1 s − 1 p s ( 1 − p ) x − s <script type="math/tex" id="MathJax-Element-47"> P(X = x) = C_{x - 1}^{s - 1} p^s (1 - p)^{x - s} </script>（最后一次为成功，前面的抽 s−1 s − 1 <script type="math/tex" id="MathJax-Element-48">s - 1</script>次成功）
  • E(X)=sp E ( X ) = s p <script type="math/tex" id="MathJax-Element-49"> E(X) = \frac{s}{p} </script>
  • D(X)=s(1−p)p2 D ( X ) = s ( 1 − p ) p 2 <script type="math/tex" id="MathJax-Element-50"> D(X) = \frac{s(1 - p)}{p^2} </script>

超几何分布

N个产品有D个次品，抽n个，求次品数

  • P(X=x)=CxDCn−xN−DCnN P ( X = x ) = C D x C N − D n − x C N n <script type="math/tex" id="MathJax-Element-51"> P(X = x) = \frac{C_D^x C_{N-D}^{n-x}}{C_N^n} </script>
  • E(X)=nDN E ( X ) = n D N <script type="math/tex" id="MathJax-Element-52"> E(X) = n \frac{D}{N} </script>
  • D(X)=N−nN−1nDNN−DN D ( X ) = N − n N − 1 n D N N − D N <script type="math/tex" id="MathJax-Element-53"> D(X) = \frac{N - n}{N - 1} n \frac{D}{N} \frac{N - D}{N} </script>

当N很大，n很小时近似于二项分布， p=DN p = D N <script type="math/tex" id="MathJax-Element-54">p = \frac{D}{N}</script>

泊松分布

一段时间内事件发生的次数

  • P(X=x)=λxx!e−λ P ( X = x ) = λ x x ! e − λ <script type="math/tex" id="MathJax-Element-55"> P(X = x) = \frac{\lambda^x}{x!} e^{-\lambda} </script>
  • E(X)=λ E ( X ) = λ <script type="math/tex" id="MathJax-Element-56"> E(X) = \lambda </script>
  • D(X)=λ D ( X ) = λ <script type="math/tex" id="MathJax-Element-57"> D(X) = \lambda </script>

贝塔分布

成功s次，失败f次，成功概率的分布

  • f(x)=1B(s,f)xs−1(1−x)f−1=Ix(s,f),0<x<1,s>0,f>0 f ( x ) = 1 B ( s , f ) x s − 1 ( 1 − x ) f − 1 = I x ( s , f ) , 0 < x < 1 , s > 0 , f > 0 <script type="math/tex" id="MathJax-Element-58"> f(x) = \frac{1}{B(s, f)} x^{s - 1} (1 - x)^{f - 1} = I_x(s, f), 0 < x < 1, s > 0, f > 0 </script>
  • E(x)=ss+f E ( x ) = s s + f <script type="math/tex" id="MathJax-Element-59"> E(x) = \frac{s}{s + f} </script>
  • D(x)=sf(s+f)2(s+f+1) D ( x ) = s f ( s + f ) 2 ( s + f + 1 ) <script type="math/tex" id="MathJax-Element-60"> D(x) = \frac{sf}{(s + f)^2 (s + f + 1)} </script>

理解

其中B是贝塔函数

B(p,q)=∫10tp−1(1−t)q−1dt B ( p , q ) = ∫ 0 1 t p − 1 ( 1 − t ) q − 1 d t
<script type="math/tex; mode=display" id="MathJax-Element-61"> B(p, q) = \int_0^1 t^{p - 1} (1 - t)^{q - 1} dt </script>
B(p,q)=Γ(p)Γ(q)Γ(p+q)=B(q,p) B ( p , q ) = Γ ( p ) Γ ( q ) Γ ( p + q ) = B ( q , p )
<script type="math/tex; mode=display" id="MathJax-Element-62"> B(p, q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p + q)} = B(q, p) </script>

不完全贝塔函数（积分上限换为x）

Bx(p,q)=∫x0tp−1(1−t)q−1dt B x ( p , q ) = ∫ 0 x t p − 1 ( 1 − t ) q − 1 d t
<script type="math/tex; mode=display" id="MathJax-Element-63"> B_x(p, q) = \int_0^x t^{p - 1} (1 - t)^{q - 1} dt </script>

正则贝塔函数

Ix(p,q)=Bx(p,q)B(p,q) I x ( p , q ) = B x ( p , q ) B ( p , q )
<script type="math/tex; mode=display" id="MathJax-Element-64"> I_x(p, q) = \frac{B_x(p, q)}{B(p, q)} </script>

和二项分布的关系（失败次数在f次以内的概率）

∑i=0fCinpn−i(1−p)i=IR(n−f,f+1) ∑ i = 0 f C n i p n − i ( 1 − p ) i = I R ( n − f , f + 1 )
<script type="math/tex; mode=display" id="MathJax-Element-65"> \sum_{i = 0}^{f} C_n^i p^{n - i} (1 - p)^i = I_R(n - f, f + 1) </script>

指数分布

如果产品所受到的应力冲击服从强度为 λ λ <script type="math/tex" id="MathJax-Element-66">\lambda</script>的泊松过程，且产品受到一次冲击就故障，则产品故障次数服从泊松分布，产品寿命服从指数分布（一次事件发生的时间）

  • f(x)=λe−λx f ( x ) = λ e − λ x <script type="math/tex" id="MathJax-Element-67"> f(x) = \lambda e^{-\lambda x} </script>
  • F(x)=1−e−λx F ( x ) = 1 − e − λ x <script type="math/tex" id="MathJax-Element-68"> F(x) = 1 - e^{-\lambda x} </script>
  • E(x)=1λ=θ E ( x ) = 1 λ = θ <script type="math/tex" id="MathJax-Element-69"> E(x) = \frac{1}{\lambda} = \theta </script>
  • D(x)=1λ2=θ2 D ( x ) = 1 λ 2 = θ 2 <script type="math/tex" id="MathJax-Element-70"> D(x) = \frac{1}{\lambda^2} = \theta^2 </script>

伽马分布

产品受到k次冲击才故障，产品寿命服从伽马分布（k次事件发生的时间）

  • f(x)=λkΓ(k)xk−1e−λx f ( x ) = λ k Γ ( k ) x k − 1 e − λ x <script type="math/tex" id="MathJax-Element-71"> f(x) = \frac{\lambda^k}{\Gamma(k)} x^{k - 1} e^{-\lambda x} </script>
  • F(x)=1−e−λx∑k−1i=0(λx)ii! F ( x ) = 1 − e − λ x ∑ i = 0 k − 1 ( λ x ) i i ! <script type="math/tex" id="MathJax-Element-72"> F(x) = 1 - e^{-\lambda x} \sum_{i = 0}^{k - 1} \frac{(\lambda x)^i}{i!} </script>
  • E(x)=kλ E ( x ) = k λ <script type="math/tex" id="MathJax-Element-73"> E(x) = \frac{k}{\lambda} </script>
  • D(x)=kλ2 D ( x ) = k λ 2 <script type="math/tex" id="MathJax-Element-74"> D(x) = \frac{k}{\lambda^2} </script>

理解

Γ(k)1f(x)=∫∞0yk−1e−ydy=∫∞0(λx)k−1e−λxd(λx)=∫∞0λkxk−1e−λxdx=∫∞0λkΓ(k)xk−1e−λxdx=λkΓ(k)xk−1e−λx Γ ( k ) = ∫ 0 ∞ y k − 1 e − y d y = ∫ 0 ∞ ( λ x ) k − 1 e − λ x d ( λ x ) = ∫ 0 ∞ λ k x k − 1 e − λ x d x 1 = ∫ 0 ∞ λ k Γ ( k ) x k − 1 e − λ x d x f ( x ) = λ k Γ ( k ) x k − 1 e − λ x
<script type="math/tex; mode=display" id="MathJax-Element-75"> \begin{aligned} \Gamma(k) &= \int_{0}^{\infty} y^{k - 1} e^{-y} dy \\ &= \int_{0}^{\infty} (\lambda x)^{k - 1} e^{-\lambda x} d(\lambda x) \\ &= \int_{0}^{\infty} \lambda^k x^{k - 1} e^{-\lambda x} dx \\ 1 &= \int_{0}^{\infty} \frac{\lambda^k}{\Gamma(k)} x^{k - 1} e^{-\lambda x} dx \\ f(x) &= \frac{\lambda^k}{\Gamma(k)} x^{k - 1} e^{-\lambda x} \end{aligned} </script>

正态分布

很多独立同分布的随机变量相加服从正态分布

  • f(x)=1σ2π√e−(x−μ)22σ2=1σϕ(x−μσ) f ( x ) = 1 σ 2 π e − ( x − μ ) 2 2 σ 2 = 1 σ ϕ ( x − μ σ ) <script type="math/tex" id="MathJax-Element-76"> f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}} = \frac{1}{\sigma} \phi(\frac{x - \mu}{\sigma}) </script>
  • F(x)=Φ(x−μσ) F ( x ) = Φ ( x − μ σ ) <script type="math/tex" id="MathJax-Element-77"> F(x) = \Phi(\frac{x - \mu}{\sigma}) </script>
  • λ(x)=1σϕ(x−μσ)1−Φ(x−μσ) λ ( x ) = 1 σ ϕ ( x − μ σ ) 1 − Φ ( x − μ σ ) <script type="math/tex" id="MathJax-Element-78"> \lambda(x) = \frac{\frac{1}{\sigma} \phi(\frac{x - \mu}{\sigma})}{1 - \Phi(\frac{x - \mu}{\sigma})} </script>
  • E(x)=μ E ( x ) = μ <script type="math/tex" id="MathJax-Element-79"> E(x) = \mu </script>
  • D(x)=σ2 D ( x ) = σ 2 <script type="math/tex" id="MathJax-Element-80"> D(x) = \sigma^2 </script>

描述寿命时，x从0开始，当 μ>3σ μ > 3 σ <script type="math/tex" id="MathJax-Element-81">\mu > 3\sigma</script>时差别很小，若不满足，则除以 x>0 x > 0 <script type="math/tex" id="MathJax-Element-82">x > 0</script>部分的面积来满足积分为1（截尾正态分布）

对数正态分布

很多独立同分布的随机变量相乘服从正态分布（ ln(x) l n ( x ) <script type="math/tex" id="MathJax-Element-83">ln(x)</script>服从正态分布）

μ μ <script type="math/tex" id="MathJax-Element-84">\mu</script>、 σ σ <script type="math/tex" id="MathJax-Element-85">\sigma</script>是 ln(x) l n ( x ) <script type="math/tex" id="MathJax-Element-86">ln(x)</script>的均值、标准差

  • f(x)=1xσ2π√e−[ln(x)−μ]22σ2=1xσϕ(ln(x)−μσ) f ( x ) = 1 x σ 2 π e − [ l n ( x ) − μ ] 2 2 σ 2 = 1 x σ ϕ ( l n ( x ) − μ σ ) <script type="math/tex" id="MathJax-Element-87"> f(x) = \frac{1}{x \sigma \sqrt{2\pi}} e^{-\frac{[ln(x) - \mu]^2}{2\sigma^2}} = \frac{1}{x\sigma} \phi(\frac{ln(x) - \mu}{\sigma}) </script>
  • F(x)=Φ(ln(x)−μσ) F ( x ) = Φ ( l n ( x ) − μ σ ) <script type="math/tex" id="MathJax-Element-88"> F(x) = \Phi(\frac{ln(x) - \mu}{\sigma}) </script>
  • λ(x)=1xσϕ(ln(x)−μσ)1−Φ(ln(x)−μσ) λ ( x ) = 1 x σ ϕ ( l n ( x ) − μ σ ) 1 − Φ ( l n ( x ) − μ σ ) <script type="math/tex" id="MathJax-Element-89"> \lambda(x) = \frac{\frac{1}{x\sigma} \phi(\frac{ln(x) - \mu}{\sigma})}{1 - \Phi(\frac{ln(x) - \mu}{\sigma})} </script>
  • E(x)=eμ+σ22 E ( x ) = e μ + σ 2 2 <script type="math/tex" id="MathJax-Element-90"> E(x) = e^{\mu + \frac{\sigma^2}{2}} </script>
  • D(x)=E2(x)(eσ2−1) D ( x ) = E 2 ( x ) ( e σ 2 − 1 ) <script type="math/tex" id="MathJax-Element-91"> D(x) = E^2(x) (e^{\sigma^2} - 1) </script>

威布尔分布

极值分布

n个环寿命为 Ti T i <script type="math/tex" id="MathJax-Element-92">T_i</script>

  • 最小极值分布： FT1=1−[1−F(t)]n F T 1 = 1 − [ 1 − F ( t ) ] n <script type="math/tex" id="MathJax-Element-93">F_{T_1} = 1 - [1 - F(t)]^n</script>
  • 最大极值分布： FTn=[F(t)]n F T n = [ F ( t ) ] n <script type="math/tex" id="MathJax-Element-94">F_{T_n} = [F(t)]^n</script>

当 n→∞ n → ∞ <script type="math/tex" id="MathJax-Element-95">n \to \infty</script>时它们趋向于渐进分布，根据 F(t) F ( t ) <script type="math/tex" id="MathJax-Element-96">F(t)</script>不同，有三种较典型的渐近分布

极小值分布
  • I型： Fs(t)=1−e−et−γη,η>0 F s ( t ) = 1 − e − e t − γ η , η > 0 <script type="math/tex" id="MathJax-Element-97"> F_s(t) = 1 - e^{-e^{\frac{t - \gamma}{\eta}}}, \eta > 0 </script>（极值分布）
  • II型： Fs(t)=1−e−[−t−γη]−m,−∞<t⩽γ,η>0,m>0 F s ( t ) = 1 − e − [ − t − γ η ] − m , − ∞ < t ⩽ γ , η > 0 , m > 0 <script type="math/tex" id="MathJax-Element-98"> F_s(t) = 1 - e^{-[-\frac{t - \gamma}{\eta}]^{-m}}, -\infty < t \leqslant \gamma, \eta > 0, m > 0 </script>
  • III型： Fs(t)=1−e−[t−γη]m,γ⩽t<∞,η>0,m>0 F s ( t ) = 1 − e − [ t − γ η ] m , γ ⩽ t < ∞ , η > 0 , m > 0 <script type="math/tex" id="MathJax-Element-99"> F_s(t) = 1 - e^{-[\frac{t - \gamma}{\eta}]^{m}}, \gamma \leqslant t < \infty, \eta > 0, m > 0 </script>（三参数威布尔分布）
极大值分布
  • I型： FL(t)=e−e−t−γη,η>0 F L ( t ) = e − e − t − γ η , η > 0 <script type="math/tex" id="MathJax-Element-100"> F_L(t) = e^{-e^{-\frac{t - \gamma}{\eta}}}, \eta > 0 </script>
  • II型： FL(t)=e−[t−γη]−m,γ⩽t<∞,η>0,m>0 F L ( t ) = e − [ t − γ η ] − m , γ ⩽ t < ∞ , η > 0 , m > 0 <script type="math/tex" id="MathJax-Element-101"> F_L(t) = e^{-[\frac{t - \gamma}{\eta}]^{-m}}, \gamma \leqslant t < \infty, \eta > 0, m > 0 </script>（Frechet分布）
  • III型： FL(t)=e−[−t−γη]m,−∞<t⩽γ,η>0,m>0 F L ( t ) = e − [ − t − γ η ] m , − ∞ < t ⩽ γ , η > 0 , m > 0 <script type="math/tex" id="MathJax-Element-102"> F_L(t) = e^{-[-\frac{t - \gamma}{\eta}]^m}, -\infty < t \leqslant \gamma, \eta > 0, m > 0 </script>

威布尔分布

一般 t0=ηm t 0 = η m <script type="math/tex" id="MathJax-Element-103">t_0 = \eta^m</script>

  • F(t)=1−e−(t−γ)mt0 F ( t ) = 1 − e − ( t − γ ) m t 0 <script type="math/tex" id="MathJax-Element-104"> F(t) = 1 - e^{-\frac{(t - \gamma)^m}{t_0}} </script>
  • f(t)=mt0(t−γ)m−1e−(t−γ)mt0 f ( t ) = m t 0 ( t − γ ) m − 1 e − ( t − γ ) m t 0 <script type="math/tex" id="MathJax-Element-105"> f(t) = \frac{m}{t_0} (t - \gamma)^{m - 1} e^{-\frac{(t - \gamma)^m}{t_0}} </script>
  • λ(t)=mt0(t−γ)m−1 λ ( t ) = m t 0 ( t − γ ) m − 1 <script type="math/tex" id="MathJax-Element-106"> \lambda(t) = \frac{m}{t_0} (t - \gamma)^{m - 1} </script>
  • E(t)=γ+ηΓ(1+1m) E ( t ) = γ + η Γ ( 1 + 1 m ) <script type="math/tex" id="MathJax-Element-107"> E(t) = \gamma + \eta \Gamma(1 + \frac{1}{m}) </script>
  • D(t)=η2[Γ(1+2m)−Γ2(1+1m)] D ( t ) = η 2 [ Γ ( 1 + 2 m ) − Γ 2 ( 1 + 1 m ) ] <script type="math/tex" id="MathJax-Element-108"> D(t) = \eta^2 [\Gamma(1 + \frac{2}{m}) - \Gamma^2(1 + \frac{1}{m})] </script>

m是形状参数， m>1 m > 1 <script type="math/tex" id="MathJax-Element-109">m > 1</script>时失效率递增， m=1 m = 1 <script type="math/tex" id="MathJax-Element-110">m = 1</script>时失效率恒定， m<1 m < 1 <script type="math/tex" id="MathJax-Element-111">m < 1</script>时失效率递减

t0 t 0 <script type="math/tex" id="MathJax-Element-112">t_0</script>或 η η <script type="math/tex" id="MathJax-Element-113">\eta</script>是尺度参数，往往与工作条件负载的大小有关，负载大的，相应的尺度参数要小

γ γ <script type="math/tex" id="MathJax-Element-114">\gamma</script>是位置参数，平移函数

特征寿命： te−1=γ+η t e − 1 = γ + η <script type="math/tex" id="MathJax-Element-115">t_{e^{-1}} = \gamma + \eta</script>

BS分布、逆高斯分布

略

可靠性指标的点估计

用来估计总体参数的实值统计量称为点估计量。点估计是样本的函数，因而是随机变量。当样本值给定后，得到的是参数的单值

点估计的优良性

  • 均方差： MSEθ=E(θ^−θ)2=D(θ^)+[E(θ^)−θ]2 M S E θ = E ( θ ^ − θ ) 2 = D ( θ ^ ) + [ E ( θ ^ ) − θ ] 2 <script type="math/tex" id="MathJax-Element-116"> MSE_{\theta} = E(\hat{\theta} - \theta)^2 = D(\hat{\theta}) + [E(\hat{\theta}) - \theta]^2 </script>
  • 无偏性： E(θ^)=θ E ( θ ^ ) = θ <script type="math/tex" id="MathJax-Element-117"> E(\hat{\theta}) = \theta </script>，消除系统误差
  • 一致性： limn→∞P(|θ^−θ|>ε)=0 lim n → ∞ P ( | θ ^ − θ | > ε ) = 0 <script type="math/tex" id="MathJax-Element-118"> \lim_{n \to \infty} P(|\hat{\theta} - \theta| > \varepsilon) = 0 </script>，即 θ^→Pθ θ ^ → P θ <script type="math/tex" id="MathJax-Element-119"> \hat{\theta} \overset{P}{\rightarrow} \theta </script>，当样本信息足够多时，估计的不确定性可减小到任意小
  • 方差： D(θ^) D ( θ ^ ) <script type="math/tex" id="MathJax-Element-120"> D(\hat{\theta}) </script>

矩估计

如果某参数 θ θ <script type="math/tex" id="MathJax-Element-121">\theta</script>可以表示为总体前r阶矩的函数，则可以用前r阶矩估计 θ θ <script type="math/tex" id="MathJax-Element-122">\theta</script>

原则：（1）涉及到的矩的阶数尽可能小，常用的矩估计一般只涉及一二阶矩；（2）所用估计最好是（最小）充分统计量的函数

极大似然估计（MLE）

当样本x给定后，可考虑对不同的 θ θ <script type="math/tex" id="MathJax-Element-123">\theta</script>，概率密度如何变化，它代表试验出现该组样本的相对可能性大小。选取一个使样本观测值结果出现的概率达到最大的值作为 θ θ <script type="math/tex" id="MathJax-Element-124">\theta</script>的估计值

将x带入联合分布函数得到似然函数 L(x;θ) L ( x ; θ ) <script type="math/tex" id="MathJax-Element-125">L(x; \theta)</script>，为了方便可取对数，求 θ θ <script type="math/tex" id="MathJax-Element-126">\theta</script>使L最大（求导，令导数为0解方程）

在一些情况下不能得到完全寿命数据，其似然函数为截尾样本的联合分布： n!(n−f)![∏ri=1f(ti)][1−F(tr)]n−r n ! ( n − f ) ! [ ∏ i = 1 r f ( t i ) ] [ 1 − F ( t r ) ] n − r <script type="math/tex" id="MathJax-Element-127"> \frac{n!}{(n - f)!} [\prod_{i = 1}^{r} f(t_i)] [1 - F(t_r)]^{n - r} </script>

指数分布的可靠性指标估计

形式都是 θ^=Tr,λ^=1θ^ θ ^ = T r , λ ^ = 1 θ ^ <script type="math/tex" id="MathJax-Element-128"> \hat{\theta} = \frac{T}{r}, \hat{\lambda} = \frac{1}{\hat{\theta}} </script>

  • 完全样本： T=∑ni=1ti T = ∑ i = 1 n t i <script type="math/tex" id="MathJax-Element-129"> T = \sum_{i = 1}^n t_i </script>
  • 无替换定数截尾： T=∑ri=1ti+(n−r)tr T = ∑ i = 1 r t i + ( n − r ) t r <script type="math/tex" id="MathJax-Element-130"> T = \sum_{i = 1}^r t_i + (n - r) t_r </script>
  • 无替换定时截尾： T=∑ri=1ti+(n−r)t0 T = ∑ i = 1 r t i + ( n − r ) t 0 <script type="math/tex" id="MathJax-Element-131"> T = \sum_{i = 1}^r t_i + (n - r) t_0 </script>
  • 有替换定数截尾： T=ntr T = n t r <script type="math/tex" id="MathJax-Element-132"> T = n t_r </script>
  • 有替换定时截尾： T=nt0 T = n t 0 <script type="math/tex" id="MathJax-Element-133"> T = n t_0 </script>
  • 随机截尾： T=∑ri=1ti+∑ki=1τi T = ∑ i = 1 r t i + ∑ i = 1 k τ i <script type="math/tex" id="MathJax-Element-134"> T = \sum_{i = 1}^r t_i + \sum_{i = 1}^k \tau_i </script>

最小二乘估计

F(t)经过某种变换G后，使得G(F(t))与 φ(t) φ ( t ) <script type="math/tex" id="MathJax-Element-135">\varphi(t)</script>呈线性关系，其中 φ φ <script type="math/tex" id="MathJax-Element-136">\varphi</script>为某种变换，所有未知参数包含在系数中，那么此时可以采用最小二乘估计方法得到参数点估计

例如：威布尔分布 F(t)=1−e−tmt0 F ( t ) = 1 − e − t m t 0 <script type="math/tex" id="MathJax-Element-137"> F(t) = 1 - e^{-\frac{t^m}{t_0}} </script>，经变换得 ln(ln(11−F(t)))=−ln(t0)+mln(t)=b0+b1x l n ( l n ( 1 1 − F ( t ) ) ) = − l n ( t 0 ) + m l n ( t ) = b 0 + b 1 x <script type="math/tex" id="MathJax-Element-138"> ln(ln(\frac{1}{1 - F(t)})) = -ln(t_0) + m ln(t) = b_0 + b_1 x </script>，由最小二乘法得 b0,b1 b 0 , b 1 <script type="math/tex" id="MathJax-Element-139">b_0, b_1</script>，变换得 t0,m t 0 , m <script type="math/tex" id="MathJax-Element-140">t_0, m</script>

线性估计

分类：

  • 最优线性无偏估计（BLUE）
  • 最优线性不变估计（BLIE）
  • 简单线性无偏估计（GLUE）
  • 简单线性不变估计（GLIE）

位置-尺度分布族

设X为位置-尺度分布族随机变量，其分布函数可以表示为

P(X<x)=F(x−μσ) P ( X < x ) = F ( x − μ σ )
<script type="math/tex; mode=display" id="MathJax-Element-141"> P(X < x) = F(\frac{x - \mu}{\sigma}) </script>

式中 μ μ <script type="math/tex" id="MathJax-Element-142">\mu</script>为位置参数， σ σ <script type="math/tex" id="MathJax-Element-143">\sigma</script>为尺度参数，分布函数可以完全由位置参数和尺度参数确定

正态分布、极值分布、指数分布是位置-尺度分布族，威布尔分布、对数正态分布是对数位置-尺度分布族

Y0=Y−μσ Y 0 = Y − μ σ <script type="math/tex" id="MathJax-Element-144"> Y^0 = \frac{Y - \mu}{\sigma} </script>为标准位置-刻度族随机变量

线性估计就是用通用方法估计得 μ,σ μ , σ <script type="math/tex" id="MathJax-Element-145">\mu, \sigma</script>，再变换得到未知参数

可靠性指标的置信限估计

P(θ^L<θ<θ^U)=1−α P ( θ ^ L < θ < θ ^ U ) = 1 − α
<script type="math/tex; mode=display" id="MathJax-Element-146"> P(\hat{\theta}_L < \theta < \hat{\theta}_U) = 1 - \alpha </script>

其中 θ^L θ ^ L <script type="math/tex" id="MathJax-Element-147">\hat{\theta}_L</script>是置信下限， θ^U θ ^ U <script type="math/tex" id="MathJax-Element-148">\hat{\theta}_U</script>是置信上限， α α <script type="math/tex" id="MathJax-Element-149">\alpha</script>是显著度， γ=1−α γ = 1 − α <script type="math/tex" id="MathJax-Element-150">\gamma = 1 - \alpha</script>是置信水平

P(θ^L<θ)=1−α P ( θ ^ L < θ ) = 1 − α
<script type="math/tex; mode=display" id="MathJax-Element-151"> P(\hat{\theta}_L < \theta) = 1 - \alpha </script>

其中 θ^L θ ^ L <script type="math/tex" id="MathJax-Element-152">\hat{\theta}_L</script>是单侧置信下限，同理有单侧置信上限

参数 θ θ <script type="math/tex" id="MathJax-Element-153">\theta</script>是固定的数，因此区间估计的含义为“置信区间覆盖参数真值”而不是“参数真值在该区间内取值”的概率

置信限根据样本不同而不同，因此置信区间描述的是“置信区间覆盖参数真值”可信程度，而不是“由一次样本得到的置信区间覆盖参数真值”的概率

枢轴量法

设 G=G(X,η) G = G ( X , η ) <script type="math/tex" id="MathJax-Element-154">G = G(X, \eta)</script>是样本X和未知参数 η η <script type="math/tex" id="MathJax-Element-155">\eta</script>的一个函数，如果G的分布与 η η <script type="math/tex" id="MathJax-Element-156">\eta</script>无关，则这样的函数称为枢轴量。枢轴量的分布不依赖未知参数，保证最终求得的置信区间覆盖参数真值的概率不依赖于未知参数

统计量是不包含未知参数的样本的函数，其分布可以包含未知量；枢轴量是包含未知参数的用来估计参数的样本的函数，其分布已知，不包含未知参数

正态分布参数的区间估计

σ σ <script type="math/tex" id="MathJax-Element-157">\sigma</script>已知， μ μ <script type="math/tex" id="MathJax-Element-158">\mu</script>的置信区间

X¯=1n∑i=1nXiZ=X¯−μσ0n√∼N(μ,σ20n)∼N(0,1) X ¯ = 1 n ∑ i = 1 n X i ∼ N ( μ , σ 0 2 n ) Z = X ¯ − μ σ 0 n ∼ N ( 0 , 1 )
<script type="math/tex; mode=display" id="MathJax-Element-159"> \begin{aligned} \bar{X} = \frac{1}{n} \sum_{i = 1}^n X_i &\sim N(\mu, \frac{\sigma_0^2}{n}) \\ Z = \frac{\bar{X} - \mu}{\frac{\sigma_0}{\sqrt{n}}} &\sim N(0, 1) \end{aligned} </script>
P{Zα2<P{X¯−σ0n−−√Z1−α2<Z<Z1−α2}=1−αμ<X¯−σ0n−−√Zα2}=1−α P { Z α 2 < Z < Z 1 − α 2 } = 1 − α P { X ¯ − σ 0 n Z 1 − α 2 < μ < X ¯ − σ 0 n Z α 2 } = 1 − α
<script type="math/tex; mode=display" id="MathJax-Element-160"> \begin{aligned} P\{Z_{\frac{\alpha}{2}} < &Z < Z_{1 - \frac{\alpha}{2}}\} = 1 - \alpha \\ P\{\bar{X} - \frac{\sigma_0}{\sqrt{n}} Z_{1 - \frac{\alpha}{2}} < &\mu < \bar{X} - \frac{\sigma_0}{\sqrt{n}} Z_{\frac{\alpha}{2}}\} = 1 - \alpha \end{aligned} </script>

μ μ <script type="math/tex" id="MathJax-Element-161">\mu</script>已知， σ σ <script type="math/tex" id="MathJax-Element-162">\sigma</script>的置信区间

Xi−μ0σχ2=(n−1)S2σ2=1σ2∑i=1n(Xi−μ0)2∼N(0,1)∼χ2(n) X i − μ 0 σ ∼ N ( 0 , 1 ) χ 2 = ( n − 1 ) S 2 σ 2 = 1 σ 2 ∑ i = 1 n ( X i − μ 0 ) 2 ∼ χ 2 ( n )
<script type="math/tex; mode=display" id="MathJax-Element-163"> \begin{aligned} \frac{X_i - \mu_0}{\sigma} &\sim N(0, 1) \\ \chi^2 = \frac{(n - 1) S^2}{\sigma^2} = \frac{1}{\sigma^2} \sum_{i = 1}^n (X_i - \mu_0)^2 &\sim \chi^2(n) \end{aligned} </script>
P{χ2α2(n)<P{(n−1)S2χ21−α2(n)<χ2<χ21−α2(n)}=1−ασ2<(n−1)S2χ2α2(n)}=1−α P { χ α 2 2 ( n ) < χ 2 < χ 1 − α 2 2 ( n ) } = 1 − α P { ( n − 1 ) S 2 χ 1 − α 2 2 ( n ) < σ 2 < ( n − 1 ) S 2 χ α 2 2 ( n ) } = 1 − α
<script type="math/tex; mode=display" id="MathJax-Element-164"> \begin{aligned} P\{\chi_{\frac{\alpha}{2}}^2(n) < &\chi^2 < \chi_{1 - \frac{\alpha}{2}}^2(n)\} = 1 - \alpha \\ P\{\frac{(n - 1) S^2}{\chi_{1 - \frac{\alpha}{2}}^2(n)} < &\sigma^2 < \frac{(n - 1) S^2}{\chi_{\frac{\alpha}{2}}^2(n)}\} = 1 - \alpha \\ \end{aligned} </script>

μ μ <script type="math/tex" id="MathJax-Element-165">\mu</script>、 σ σ <script type="math/tex" id="MathJax-Element-166">\sigma</script>未知时的置信区间

T=X¯−μSn√χ2=(n−1)S2σ2=1σ2∑i=1n(Xi−X¯2)∼t(n−1)∼χ2(n−1) T = X ¯ − μ S n ∼ t ( n − 1 ) χ 2 = ( n − 1 ) S 2 σ 2 = 1 σ 2 ∑ i = 1 n ( X i − X ¯ 2 ) ∼ χ 2 ( n − 1 )
<script type="math/tex; mode=display" id="MathJax-Element-167"> \begin{aligned} T = \frac{\bar{X} - \mu}{\frac{S}{\sqrt{n}}} &\sim t(n - 1) \\ \chi^2 = \frac{(n - 1) S^2}{\sigma^2} = \frac{1}{\sigma^2} \sum_{i = 1}^n (X_i - \bar{X}^2) &\sim \chi^2(n - 1) \end{aligned} </script>
P{X¯−Sn−−√t1−α2(n−1)<P{(n−1)S2χ21−α2(n−1)<μ<X¯−Sn−−√tα2(n−1)}=1−ασ2<(n−1)S2χ2α2(n−1)}=1−α P { X ¯ − S n t 1 − α 2 ( n − 1 ) < μ < X ¯ − S n t α 2 ( n − 1 ) } = 1 − α P { ( n − 1 ) S 2 χ 1 − α 2 2 ( n − 1 ) < σ 2 < ( n − 1 ) S 2 χ α 2 2 ( n − 1 ) } = 1 − α
<script type="math/tex; mode=display" id="MathJax-Element-168"> \begin{aligned} P\{\bar{X} - \frac{S}{\sqrt{n}} t_{1 - \frac{\alpha}{2}}(n - 1) < &\mu < \bar{X} - \frac{S}{\sqrt{n}} t_{\frac{\alpha}{2}}(n - 1)\} = 1 - \alpha \\ P\{\frac{(n - 1) S^2}{\chi_{1 - \frac{\alpha}{2}}^2(n - 1)} < &\sigma^2 < \frac{(n - 1) S^2}{\chi_{\frac{\alpha}{2}}^2(n - 1)}\} = 1 - \alpha \end{aligned} </script>

指数分布参数的区间估计

定数截尾

H=2Tθ∼χ2(2r) H = 2 T θ ∼ χ 2 ( 2 r )
<script type="math/tex; mode=display" id="MathJax-Element-169"> H = \frac{2T}{\theta} \sim \chi^2(2r) </script>
P{2Tχ21−α2(2r)<θ<2Tχ2α2(2r)}=1−α P { 2 T χ 1 − α 2 2 ( 2 r ) < θ < 2 T χ α 2 2 ( 2 r ) } = 1 − α
<script type="math/tex; mode=display" id="MathJax-Element-170"> P\{\frac{2T}{\chi_{1 - \frac{\alpha}{2}}^2(2r)} < \theta < \frac{2T}{\chi_{\frac{\alpha}{2}}^2(2r)}\} = 1 - \alpha </script>

定时截尾

P{2Tχ21−α2(2r+2)<θ<2Tχ2α2(2r)}=1−α P { 2 T χ 1 − α 2 2 ( 2 r + 2 ) < θ < 2 T χ α 2 2 ( 2 r ) } = 1 − α
<script type="math/tex; mode=display" id="MathJax-Element-171"> P\{\frac{2T}{\chi_{1 - \frac{\alpha}{2}}^2(2r + 2)} < \theta < \frac{2T}{\chi_{\frac{\alpha}{2}}^2(2r)}\} = 1 - \alpha </script>

二项分布

p的置信下限 pL p L <script type="math/tex" id="MathJax-Element-172">p_L</script>由下式确定

∑x=0n−fCxnpn−xL(1−pL)x=α ∑ x = 0 n − f C n x p L n − x ( 1 − p L ) x = α
<script type="math/tex; mode=display" id="MathJax-Element-173"> \sum_{x = 0}^{n - f} C_n^x p_L^{n - x} (1 - p_L)^x = \alpha </script>

p的置信上限 pU p U <script type="math/tex" id="MathJax-Element-174">p_U</script>由下式确定

∑x=0fCxn(1−pU)n−xpxU=α ∑ x = 0 f C n x ( 1 − p U ) n − x p U x = α
<script type="math/tex; mode=display" id="MathJax-Element-175"> \sum_{x = 0}^{f} C_n^x (1 - p_U)^{n - x} p_U^x = \alpha </script>

增函数确定下限，减函数确定上限

可靠度

p为失效概率， R=1−p R = 1 − p <script type="math/tex" id="MathJax-Element-176">R = 1 - p</script>为可靠度，置信下限 RL R L <script type="math/tex" id="MathJax-Element-177">R_L</script>

∑x=0fCxnRn−xL(1−RL)x=α ∑ x = 0 f C n x R L n − x ( 1 − R L ) x = α
<script type="math/tex; mode=display" id="MathJax-Element-178"> \sum_{x = 0}^{f} C_n^x R_L^{n - x} (1 - R_L)^x = \alpha </script>

特殊情况：当 f=0 f = 0 <script type="math/tex" id="MathJax-Element-179">f = 0</script>时

RL=α1n R L = α 1 n
<script type="math/tex; mode=display" id="MathJax-Element-180"> R_L = \alpha^{\frac{1}{n}} </script>

基于渐近正态性的置信限估计、基于线性估计的置信限估计、Bootstrap方法

略

应力强度干涉与容限系数

应力强度干涉模型

应力s和强度 δ δ <script type="math/tex" id="MathJax-Element-181">\delta</script>都是随机变量，应力分布与强度分布发生干涉时结构将可能失效

可靠度

R=P(δ>s)=P(δ−s>0)=∫∞−∞[∫∞sfδ(δ)dδ]fs(s)ds R = P ( δ > s ) = P ( δ − s > 0 ) = ∫ − ∞ ∞ [ ∫ s ∞ f δ ( δ ) d δ ] f s ( s ) d s
<script type="math/tex; mode=display" id="MathJax-Element-182"> \begin{aligned} R &= P(\delta > s) = P(\delta - s > 0) \\ &= \int_{-\infty}^{\infty} [\int_s^{\infty} f_{\delta}(\delta) d\delta] f_s(s) ds \end{aligned} </script>

应力强度干涉可靠度点估计

正态分布

y=δ−s∼N(μδ−μs,σ2δ+σ2s) y = δ − s ∼ N ( μ δ − μ s , σ δ 2 + σ s 2 )
<script type="math/tex; mode=display" id="MathJax-Element-183"> y = \delta - s \sim N(\mu_{\delta} - \mu_s, \sigma_{\delta}^2 + \sigma_s^2) </script>
R=P(y>0)=1−Φ(0−μyσy)=Φ(μyσy) R = P ( y > 0 ) = 1 − Φ ( 0 − μ y σ y ) = Φ ( μ y σ y )
<script type="math/tex; mode=display" id="MathJax-Element-184"> R = P(y > 0) = 1 - \Phi(\frac{0 - \mu_y}{\sigma_y}) = \Phi(\frac{\mu_y}{\sigma_y}) </script>

对数正态分布

z=−ln(δ^)−ln(s^)σ2lnδ+σ2lns−−−−−−−−√ z = − l n ( δ ^ ) − l n ( s ^ ) σ l n δ 2 + σ l n s 2
<script type="math/tex; mode=display" id="MathJax-Element-185"> z = -\frac{ln(\hat{\delta}) - ln(\hat{s})}{\sqrt{\sigma_{ln \delta}^2 + \sigma_{ln s}^2}} </script>
R=∫∞zΦ(z)dz R = ∫ z ∞ Φ ( z ) d z
<script type="math/tex; mode=display" id="MathJax-Element-186"> R = \int_z^{\infty} \Phi(z) dz </script>

指数分布

R=λsλs+λδ=θδθδ+θs R = λ s λ s + λ δ = θ δ θ δ + θ s
<script type="math/tex; mode=display" id="MathJax-Element-187"> R = \frac{\lambda_s}{\lambda_s + \lambda_{\delta}} = \frac{\theta_{\delta}}{\theta_{\delta} + \theta_s} </script>

应力强度干涉可靠度区间估计

略

（单侧）容限系数法

单侧容限系数的原理是用k取代 X¯+uRσ^ X ¯ + u R σ ^ <script type="math/tex" id="MathJax-Element-188">\bar{X} + u_R \hat{\sigma}</script>中的 uR u R <script type="math/tex" id="MathJax-Element-189">u_R</script>，使由 X¯+kσ^ X ¯ + k σ ^ <script type="math/tex" id="MathJax-Element-190">\bar{X} + k \hat{\sigma}</script>估计出的百分位值以一定的置信度 γ γ <script type="math/tex" id="MathJax-Element-191">\gamma</script>小于真值 μ+uRσ^ μ + u R σ ^ <script type="math/tex" id="MathJax-Element-192">\mu + u_R \hat{\sigma}</script>

统计抽样法

模拟抽样，略

系统可靠性综合评估

系统综合：利用系统及组成设备的试验信息对系统可靠性指标进行综合评估（点估计和置信下限）

系统可靠性模型

  • 串联系统： R(t)=∏mi=1Ri(t) R ( t ) = ∏ i = 1 m R i ( t ) <script type="math/tex" id="MathJax-Element-193"> R(t) = \prod_{i = 1}^{m} R_i(t) </script>
  • 并联系统： R(t)=1−∏mi=1(1−Ri(t)) R ( t ) = 1 − ∏ i = 1 m ( 1 − R i ( t ) ) <script type="math/tex" id="MathJax-Element-194"> R(t) = 1 - \prod_{i = 1}^{m} (1 - R_i(t)) </script>
  • 表决系统： R(t)=∑mi=kCim[R1(t)]i[1−R1(t)]m−i R ( t ) = ∑ i = k m C m i [ R 1 ( t ) ] i [ 1 − R 1 ( t ) ] m − i <script type="math/tex" id="MathJax-Element-195"> R(t) = \sum_{i = k}^m C_m^i [R_1(t)]^i [1 - R_1(t)]^{m - i} </script>

  • 旁联系统（冷储备，转换完全可靠）：

    R(t)MTTF=1−∫t0f1(t)∗f2(t)∗...∗fm(t)dt=∑i=1mE(Ti) R ( t ) = 1 − ∫ 0 t f 1 ( t ) ∗ f 2 ( t ) ∗ . . . ∗ f m ( t ) d t M T T F = ∑ i = 1 m E ( T i )
    <script type="math/tex; mode=display" id="MathJax-Element-196"> \begin{aligned} R(t) &= 1 - \int_0^t f_1(t) * f_2(t) * ... * f_m(t) dt \\ MTTF &= \sum_{i = 1}^m E(T_i) \end{aligned} </script>

    如果每个部件都服从同一 λ λ <script type="math/tex" id="MathJax-Element-197">\lambda</script>的指数分布（系统失效时间服从伽马分布）

    R(t)MTTF=e−λt∑k=0m−1(λt)kk!=mλ R ( t ) = e − λ t ∑ k = 0 m − 1 ( λ t ) k k ! M T T F = m λ
    <script type="math/tex; mode=display" id="MathJax-Element-198"> \begin{aligned} R(t) &= e^{-\lambda t} \sum_{k = 0}^{m - 1} \frac{(\lambda t)^k}{k!} \\ MTTF &= \frac{m}{\lambda} \end{aligned} </script>

    理解卷积

    其实就是X、Y联合分布密度在 X+Y=t X + Y = t <script type="math/tex" id="MathJax-Element-199">X + Y = t</script>的第二类曲线积分

    fX+Y(t)FX+Y(t)=∫X+Y=tfX+Y(x,y)dx=∫t0fX(x)fY(t−x)dx=fX(x)∗fY(y)=∬X+Y<tfX+Y(x,y)dxdy=∫t0[∫t−x0fY(y)dy]fX(x)dx=∫t0fX(x)FY(t−x)dx=fX(x)∗FY(y) f X + Y ( t ) = ∫ X + Y = t f X + Y ( x , y ) d x = ∫ 0 t f X ( x ) f Y ( t − x ) d x = f X ( x ) ∗ f Y ( y ) F X + Y ( t ) = ∬ X + Y < t f X + Y ( x , y ) d x d y = ∫ 0 t [ ∫ 0 t − x f Y ( y ) d y ] f X ( x ) d x = ∫ 0 t f X ( x ) F Y ( t − x ) d x = f X ( x ) ∗ F Y ( y )
    <script type="math/tex; mode=display" id="MathJax-Element-200"> \begin{aligned} f_{X + Y}(t) &= \int_{X + Y = t} f_{X + Y}(x, y) dx \\ &= \int_0^t f_X(x) f_Y(t - x) dx \\ &= f_X(x) * f_Y(y) \\ F_{X + Y}(t) &= \iint_{X + Y < t} f_{X + Y}(x, y) dxdy \\ &= \int_0^t [\int_0^{t - x} f_Y(y) dy] f_X(x) dx \\ &= \int_0^t f_X(x) F_Y(t - x) dx \\ &= f_X(x) * F_Y(y) \end{aligned} </script>

LM法

适合于部件（子系统）组成的成败型串联系统。

若部件j在 nj n j <script type="math/tex" id="MathJax-Element-201">n_j</script>次试验中有 rj r j <script type="math/tex" id="MathJax-Element-202">r_j</script>次失效， sj s j <script type="math/tex" id="MathJax-Element-203">s_j</script>次成功，则系统可靠度的极大似然估计为

R^s=∏j=1msjnj R ^ s = ∏ j = 1 m s j n j
<script type="math/tex; mode=display" id="MathJax-Element-204"> \hat{R}_s = \prod_{j = 1}^m \frac{s_j}{n_j} </script>

令 n∗=min(n1,n2,...,nm) n ∗ = m i n ( n 1 , n 2 , . . . , n m ) <script type="math/tex" id="MathJax-Element-205"> n^* = min(n_1, n_2, ..., n_m) </script>

s∗=n∗R^s s ∗ = n ∗ R ^ s
<script type="math/tex; mode=display" id="MathJax-Element-206"> s^* = n^* \hat{R}_s </script>

将 s∗ s ∗ <script type="math/tex" id="MathJax-Element-207">s^*</script>和 n∗ n ∗ <script type="math/tex" id="MathJax-Element-208">n^*</script>看作是整个系统进行 n∗ n ∗ <script type="math/tex" id="MathJax-Element-209">n^*</script>次成败试验，有 s∗ s ∗ <script type="math/tex" id="MathJax-Element-210">s^*</script>次成功。然后再以通常的二项分布参数的区间估计方法求得系统可靠性 Rs R s <script type="math/tex" id="MathJax-Element-211">R_s</script>的置信下限

如果 s∗ s ∗ <script type="math/tex" id="MathJax-Element-212">s^*</script>不是整数，可按 (n∗,[s∗]) ( n ∗ , [ s ∗ ] ) <script type="math/tex" id="MathJax-Element-213">(n^*, [s^*])</script>和 (n∗,[s∗]+1) ( n ∗ , [ s ∗ ] + 1 ) <script type="math/tex" id="MathJax-Element-214">(n^*, [s^*] + 1)</script>分别计算相应的置信下限，然后再插值，得到系统在 (n∗,s∗) ( n ∗ , s ∗ ) <script type="math/tex" id="MathJax-Element-215">(n^*, s^*)</script>条件下的近似置信下限

ILM法

克服LM法所得到的置信下限一般偏保守的缺点

1m1−R^0R^0=∑i=1m1ni1−R^iR^i 1 m 1 − R ^ 0 R ^ 0 = ∑ i = 1 m 1 n i 1 − R ^ i R ^ i
<script type="math/tex; mode=display" id="MathJax-Element-216"> \frac{1}{m} \frac{1 - \hat{R}_0}{\hat{R}_0} = \sum_{i = 1}^m \frac{1}{n_i} \frac{1 - \hat{R}_i}{\hat{R}_i} </script>

解出 m0 m 0 <script type="math/tex" id="MathJax-Element-217">m_0</script>，令 s∗=m0R^s s ∗ = m 0 R ^ s <script type="math/tex" id="MathJax-Element-218"> s^* = m_0 \hat{R}_s </script>

如果 s∗ s ∗ <script type="math/tex" id="MathJax-Element-219">s^*</script>不是整数，取最接近的整数替代，将 s∗ s ∗ <script type="math/tex" id="MathJax-Element-220">s^*</script>和 m0 m 0 <script type="math/tex" id="MathJax-Element-221">m_0</script>看作是整个系统进行 m0 m 0 <script type="math/tex" id="MathJax-Element-222">m_0</script>次成败试验，有 s∗ s ∗ <script type="math/tex" id="MathJax-Element-223">s^*</script>次成功

MML法

对来自成败型单元的子样数据，取极大似然估计下被估子样的方差等于二项分布的方差

R^iD(R^i)R^sD^(R^s)=sini=R^i(1−R^i)ni=R(R^1,R^2,...,R^m)=∑i=1m[∂R^s∂R^i]2D(R^i) R ^ i = s i n i D ( R ^ i ) = R ^ i ( 1 − R ^ i ) n i R ^ s = R ( R ^ 1 , R ^ 2 , . . . , R ^ m ) D ^ ( R ^ s ) = ∑ i = 1 m [ ∂ R ^ s ∂ R ^ i ] 2 D ( R ^ i )
<script type="math/tex; mode=display" id="MathJax-Element-224"> \begin{aligned} \hat{R}_i &= \frac{s_i}{n_i} \\ D(\hat{R}_i) &= \frac{\hat{R}_i (1 - \hat{R}_i)}{n_i} \\ \hat{R}_s &= R(\hat{R}_1, \hat{R}_2, ..., \hat{R}_m) \\ \hat{D}(\hat{R}_s) &= \sum_{i = 1}^m [\frac{\partial \hat{R}_s}{\partial \hat{R}_i}]^2 D(\hat{R}_i) \end{aligned} </script>

令

n^s^=R^s(1−R^s)D^(R^s)=n^R^s n ^ = R ^ s ( 1 − R ^ s ) D ^ ( R ^ s ) s ^ = n ^ R ^ s
<script type="math/tex; mode=display" id="MathJax-Element-225"> \begin{aligned} \hat{n} &= \frac{\hat{R}_s (1 - \hat{R}_s)}{\hat{D}(\hat{R}_s)} \\ \hat{s} &= \hat{n} \hat{R}_s \end{aligned} </script>

如果 n^,s^ n ^ , s ^ <script type="math/tex" id="MathJax-Element-226">\hat{n}, \hat{s}</script>不是整数，插值。看作是整个系统进行 n^ n ^ <script type="math/tex" id="MathJax-Element-227">\hat{n}</script>次成败试验，有 s^ s ^ <script type="math/tex" id="MathJax-Element-228">\hat{s}</script>次成功

对于串联系统解得

n^s^=∏mi=1nisi−1∑mi=11si−∑mi=11ni=n∏i=1msini n ^ = ∏ i = 1 m n i s i − 1 ∑ i = 1 m 1 s i − ∑ i = 1 m 1 n i s ^ = n ∏ i = 1 m s i n i
<script type="math/tex; mode=display" id="MathJax-Element-229"> \begin{aligned} \hat{n} &= \frac{\prod_{i = 1}^m \frac{n_i}{s_i} - 1}{\sum_{i = 1}^m \frac{1}{s_i} - \sum_{i = 1}^m \frac{1}{n_i}} \\ \hat{s} &= n \prod_{i = 1}^m \frac{s_i}{n_i} \end{aligned} </script>

IMML法

针对MML法在单元失败数为零时会出现冒进的情况

串联系统

n^s^=∏mi=11R^i−1∑mi=11niR^i−∑mi=11ni=n∏i=1msini n ^ = ∏ i = 1 m 1 R ^ i − 1 ∑ i = 1 m 1 n i R ^ i − ∑ i = 1 m 1 n i s ^ = n ∏ i = 1 m s i n i
<script type="math/tex; mode=display" id="MathJax-Element-230"> \begin{aligned} \hat{n} &= \frac{\prod_{i = 1}^m \frac{1}{\hat{R}_i} - 1}{\sum_{i = 1}^m \frac{1}{n_i \hat{R}_i} - \sum_{i = 1}^m \frac{1}{n_i}} \\ \hat{s} &= n \prod_{i = 1}^m \frac{s_i}{n_i} \end{aligned} </script>

R^i={siniRiL(0.5,n,0),sini<1,sini=1 R ^ i = { s i n i , s i n i < 1 R i L ( 0.5 , n , 0 ) , s i n i = 1
<script type="math/tex; mode=display" id="MathJax-Element-231"> \hat{R}_i = \begin{cases} \frac{s_i}{n_i} &, \frac{s_i}{n_i} < 1 \\ R_{iL}(0.5, n, 0) &, \frac{s_i}{n_i} = 1 \end{cases} </script>

实质对于零失效单元，用置信为0.5时的可靠度单侧置信下限值作为其点估计。这样意味着有一半的可能大于真值，也有一半的可能小于真值，其冒进风险极大降低

# 数学
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