精算模型1 一元生存分析2 参数生存模型
精算模型1 一元生存分析2 參數(shù)生存模型
- 均勻分布 (de Moivre 1724)
- 指數(shù)分布
- Gompertz分布 (1825)
- Makeham分布 (1860)
- Weibull分布 (Frechet,1927; Weibull 1951)
- Gamma函數(shù)
- Weibull分布的基本生存函數(shù)
這一講介紹幾個(gè)常用的剩余壽命TTT的分布。
均勻分布 (de Moivre 1724)
假設(shè)www表示極限年齡,則T~U(0,w)T \sim U(0,w)T~U(0,w),
fT(t)=1wI0≤t≤wf_T(t) = \frac{1}{w}I_{0 \le t \le w}fT?(t)=w1?I0≤t≤w?
特點(diǎn):第一個(gè)壽命的連續(xù)概率模型;剩余壽命均勻分布,隨著年齡增長(zhǎng)危險(xiǎn)率上升,達(dá)到極限年齡時(shí)必死無(wú)疑;
適用性:長(zhǎng)時(shí)間區(qū)間不適用。
性質(zhì):
S(t)=w?twS(t)=\frac{w-t}{w}S(t)=ww?t?
指數(shù)分布
假設(shè)T~f(t)T \sim f(t)T~f(t),
fT(t)=1θe?1θt,t>0f_T(t)=\frac{1}{\theta}e^{-\frac{1}{\theta}t},t>0fT?(t)=θ1?e?θ1?t,t>0
特點(diǎn):常值死亡力(危險(xiǎn)率函數(shù)為常數(shù));一段時(shí)間內(nèi)的死亡概率與當(dāng)前年齡無(wú)關(guān);是Gamma分布與Weibull分布的特例;
適用性:一般用在一年、一年以?xún)?nèi)的年齡區(qū)間。
性質(zhì)
S(t)=e?1θtS(t)=e^{-\frac{1}{\theta}t}S(t)=e?θ1?t
Gompertz分布 (1825)
Gompertz分布通過(guò)直接定義危險(xiǎn)率函數(shù)得到:
h(t)=Bct,t≥0,c>1,B>0h(t) = Bc^t,t \ge 0, c >1 ,B>0h(t)=Bct,t≥0,c>1,B>0
它的適用性不強(qiáng),因?yàn)橄嚓P(guān)的生存分析基本函數(shù)的形式非常復(fù)雜,生存函數(shù)稍微簡(jiǎn)單一點(diǎn)
S(t)=exp?(Bln?c(1?ct))S(t)=\exp \left( \frac{B}{\ln c}(1-c^t) \right)S(t)=exp(lncB?(1?ct))
Makeham分布 (1860)
Makeham分布是對(duì)Gompertz分布的修正,Gompertz分布用冪函數(shù)對(duì)與年齡相關(guān)的危險(xiǎn)率進(jìn)行建模,但沒(méi)有考慮到所有年齡段共有的一些死亡風(fēng)險(xiǎn),于是Makeham分布的危險(xiǎn)率函數(shù)修正為
h(t)=A+Bct,t≥0,c>1,B>0,A>?Bh(t) = A+Bc^t,t \ge 0, c >1 ,B>0, A>-Bh(t)=A+Bct,t≥0,c>1,B>0,A>?B
這個(gè)形式比Gompertz分布的形式還要復(fù)雜一點(diǎn),因此相關(guān)的生存分析基本函數(shù)的形式也非常復(fù)雜,生存函數(shù)為
S(t)=exp?(Bln?c(1?ct)?At)S(t)=\exp \left( \frac{B}{\ln c}(1-c^t) -At\right)S(t)=exp(lncB?(1?ct)?At)
Weibull分布 (Frechet,1927; Weibull 1951)
Weibull分布參數(shù)為θ,γ\theta,\gammaθ,γ,概率密度函數(shù)為
f(x)=γθxγ?1e?1θxγ,x>0,γ,θ>0f(x)=\frac{\gamma}{\theta}x^{\gamma-1}e^{-\frac{1}{\theta}x^{\gamma}},x>0,\gamma,\theta>0f(x)=θγ?xγ?1e?θ1?xγ,x>0,γ,θ>0
Gamma函數(shù)
Gamma函數(shù)是階乘在實(shí)數(shù)域的延拓,它由如下的積分形式定義:
Γ(x)=∫0∞tx?1e?tdt,x>0\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}dt,x>0Γ(x)=∫0∞?tx?1e?tdt,x>0
Gamma函數(shù)及其相關(guān)計(jì)算技巧在概率統(tǒng)計(jì)中非常重要,
性質(zhì)
證明
這里用概率論的思路給出性質(zhì)3的簡(jiǎn)單證明,也可以查閱任何一本數(shù)學(xué)分析的教材,學(xué)習(xí)用分析的思路證明性質(zhì)3的方法。
We try to proof it using probability theory. Suppose X1X_1X1?, X2X_2X2?, … , XnX_nXn? independently follow Poisson distribution with mean of each is 1. Define Sn=∑i=1nXiS_n = \sum_{i=1}^n X_iSn?=∑i=1n?Xi?, then E(Sn)=Var(Sn)=nE(S_n)=Var(S_n)=nE(Sn?)=Var(Sn?)=n. So
P(Sn=n)=e?nnnn!P(S_n = n) = \frac{e^{-n}n^n}{n!} P(Sn?=n)=n!e?nnn?
According CLT,
Sn?nn→dN(0,1)\frac{S_n-n}{\sqrt{n}} \to_d N(0,1) n?Sn??n?→d?N(0,1)
which means ?n∈N\forall n \in \mathbb{N}?n∈N, ??>0\forall \epsilon >0??>0, ?δ>0\exists \delta>0?δ>0 such that ?x∈B(n,δ)\forall x \in B(n,\delta)?x∈B(n,δ),
∣P(Sn=n)?[F(0)?F(?1x)]∣<?2|P(S_n=n)-[F(0)-F(-\frac{1}{\sqrt{x}})]|<\frac{\epsilon}{2} ∣P(Sn?=n)?[F(0)?F(?x?1?)]∣<2??
Here F(x)F(x)F(x) is the CDF of standard normal distribution.Since P(Sn=n)=P(n?1<Sn≤n)=P(?1/n<Sn≤0)P(S_n=n) = P(n-1<S_n\le n) = P(-1/\sqrt{n}<S_n \le 0)P(Sn?=n)=P(n?1<Sn?≤n)=P(?1/n?<Sn?≤0), this is approximately F(0)?F(?1/n)F(0)-F(-1/\sqrt{n})F(0)?F(?1/n?) according to convergence in distribution. Notice F(x)F(x)F(x) is continuous, so ?x∈B(n,δ)\forall x \in B(n,\delta)?x∈B(n,δ),
∣[F(0)?F(?1x)]?[F(0)?F(?1n)]∣<?2|[F(0)-F(-\frac{1}{\sqrt{x}})]-[F(0)-F(-\frac{1}{\sqrt{n}})]|<\frac{\epsilon}{2} ∣[F(0)?F(?x?1?)]?[F(0)?F(?n?1?)]∣<2??
So
∣P(Sn=n)?[F(0)?F(?1n)]∣≤∣P(Sn=n)?[F(0)?F(?1x)]∣?∣F(x)?[F(0)?F(?1n)]∣<?|P(S_n=n)-[F(0)-F(-\frac{1}{\sqrt{n}})] | \\ \le |P(S_n=n)-[F(0)-F(-\frac{1}{\sqrt{x}})]| -|F(x)-[F(0)-F(-\frac{1}{\sqrt{n}})]| <\epsilon ∣P(Sn?=n)?[F(0)?F(?n?1?)]∣≤∣P(Sn?=n)?[F(0)?F(?x?1?)]∣?∣F(x)?[F(0)?F(?n?1?)]∣<?
Notice F(x)F(x)F(x) is also bounded, so ?M>0\exists M>0?M>0 such that ?x,F(x)≤M\forall x,\ F(x) \le M?x,?F(x)≤M
∣P(Sn=n)F(0)?F(?1n)?1∣<?M|\frac{P(S_n=n)}{F(0)-F(-\frac{1}{\sqrt{n}})}-1| <\frac{\epsilon}{M} ∣F(0)?F(?n?1?)P(Sn?=n)??1∣<M??
F(0)?F(?1n)=∫?1n012πe?x22dx=∫?1n012π(1?x22+o(x3))dx=12πn?112πn3+o(1n2)F(0)-F(-\frac{1}{\sqrt{n}}) = \int_{-\frac{1}{\sqrt{n}}}^0 \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} dx \\ = \int_{-\frac{1}{\sqrt{n}}}^0 \frac{1}{\sqrt{2 \pi}}( 1-\frac{x^2}{2}+o(x^3)) dx \\ = \frac{1}{\sqrt{2 \pi n}} - \frac{1}{\sqrt{12 \pi n^3}} + o(\frac{1}{n^2}) F(0)?F(?n?1?)=∫?n?1?0?2π?1?e?2x2?dx=∫?n?1?0?2π?1?(1?2x2?+o(x3))dx=2πn?1??12πn3?1?+o(n21?)
According to the two equations and inequality, when n is large enough, we can ignore ?112πn3+o(1n2)- \frac{1}{\sqrt{12 \pi n^3}} + o(\frac{1}{n^2})?12πn3?1?+o(n21?), so
e?nnnn!→12π\(zhòng)frac{e^{-n}n^n}{n!} \to \frac{1}{\sqrt{2\pi}}n!e?nnn?→2π?1?
證畢
Weibull分布的基本生存函數(shù)
性質(zhì)
S(x)=e?1θxγ,x≥0S(x)=e^{-\frac{1}{\theta}x^{\gamma}},x \ge 0S(x)=e?θ1?xγ,x≥0
總結(jié)
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