Estimación de S(t) no parámetrica
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2026-02-17
Estimación de la función de supervivencia \(S(t)\) con el estimador de Kaplan Meier
Ejemplo: La información de 12 personas acerca de la duración en su primer empleo se presenta a continuación (3 personas declararon permanecer en su primer empleo)
Tiempo en el 1er. empleo
| i | Tiempo |
|---|---|
| 1 | 0.4 |
| 2 | 0.9 |
| 3 | 1.1 |
| 4 | 1.9 |
| 5 | 2.0 |
| 6 | 3.3 |
| 7 | 5.3 |
| 8 | 5.8 |
| 9 | 14.0 |
Tiempo en el primer empleo (aún permanecen trabajando en el primer empleo)
| i | Tiempo |
|---|---|
| 1 | 1.5 |
| 2 | 3.0 |
| 3 | 6.2 |
Cálculo Manual
| \(i\) | \(t_i\) | \(n_i\) | \(d_i\) | (\(1-\frac{d_i}{n_i}\)) | \(\widehat{S}(t)=\prod\limits_{t_{(i)}\leq t}(1-\frac{d_i}{n_i})\) | \[ \widetilde{S}(t)=\prod_{t_{(i)}\leq t}\exp{(-\frac{d_i}{n_i})} \] |
|---|---|---|---|---|---|---|
| 1 | 0.4 | 12 | 1 | \(1-\frac{1}{12}=\frac{11}{12}\) | \(\frac{11}{12}=\) 0.917 | \(\exp{[-(\frac{1}{12})]}=\) 0.920 |
| 2 | 0.9 | 11 | 1 | \(1-\frac{1}{11}=\frac{10}{11}\) | \(\frac{11}{12}\frac{10}{11}=\frac{10}{12}=\frac{5}{6}=\) 0.833 | \(\exp{-[(\frac{1}{12}+\frac{1}{11})]}=\) 0.840 |
| 3 | 1.1 | 10 | 1 | \(1-\frac{1}{10}=\frac{9}{10}\) | \(\frac{9}{10}\frac{10}{12}=\frac{9}{12}=\frac{3}{4}=\) 0.750 | \(\exp{-[(\frac{1}{12}+\frac{1}{11}+\frac{1}{10})]}=\) 0.760 |
| 4 | 1.5+ | 9 | 0 | \(\bf{1-\frac{0}{9}=1}\) | \(\bf{\frac{3}{4}*1}\)=0.750 | \(\bf{\exp{-[(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+0)]}=}\) 0.760 |
| 5 | 1.9 | 8 | 1 | \(1-\frac{1}{8}=\frac{7}{8}\) | \(\frac{3}{4}\frac{7}{8}=\frac{21}{32}=\) 0.656 | \(\exp{[-(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{8})]}=\) 0.671 |
| 6 | 2.0 | 7 | 1 | \(1-\frac{1}{7}=\frac{6}{7}\) | \(\frac{21}{32}\frac{6}{7}=\frac{9}{16}=\) 0.562 | \(\exp{[-(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{8}+\frac{1}{7})]}=\) 0.582 |
| 7 | 3.0+ | 6 | 0 | \(\bf{1-\frac{0}{6}=1}\) | \(\bf{\frac{9}{16}*1=}\) 0.562 | \(\bf{\exp{[-(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{8}+\frac{1}{7}+0)]}=}\) 0.582 |
| 8 | 3.3 | 5 | 1 | \(1-\frac{1}{5}=\frac{4}{5}\) | \(\frac{9}{16}\frac{4}{5}=\frac{9}{20}=\) 0.450 | \(\exp{[-(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{8}+\frac{1}{7}+\frac{1}{5})]}=\) 0.476 |
| 9 | 5.3 | 4 | 1 | \(1-\frac{1}{4}=\frac{3}{4}\) | \(\frac{9}{20}\frac{3}{4}=\frac{27}{80}=\) 0.338 | \(\exp{[-(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{8}+\frac{1}{7}+\frac{1}{5}+\frac{1}{4})]}=\) 0.371 |
| 10 | 5.8 | 3 | 1 | \(1-\frac{1}{3}=\frac{2}{3}\) | \(\frac{27}{80}\frac{2}{3}=\frac{9}{40}=\) 0.225 | \(\exp{[-(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{8}+\frac{1}{7}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3})]}=\) 0.266 |
| 11 | 6.2+ | 2 | 1 | \(\bf{1-\frac{0}{2}=1}\) | \(\bf{\frac{9}{40}*1=}\) 0.225 | \(\bf{\exp{[-(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{8}+\frac{1}{7}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+0)]}=}\) 0.098 |
| 12 | 14.0 | 1 | 1 | \(1-\frac{1}{1}=0\) | \(\frac{9}{40}*0=\) 0.000 | \(\exp{[-(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{8}+\frac{1}{7}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{1})]}=\) 0.098 |
Usando la paquetería survival de R
library(survival)
library(survminer)
time<-c(1.5,3.0,6.2,0.4,0.9,1.1,1.9,2.0,3.3,5.3,5.8,14)
cc<-c(0,0,0,1,1,1,1,1,1,1,1,1)
Primer_empleo_plain<-
surv_summary(survfit(Surv(time,cc)~1, conf.type="plain",conf.int=0.95))
Primer_empleo_log<-
surv_summary(survfit(Surv(time,cc)~1, conf.type="log-log",conf.int=0.95))
knitr::kable(Primer_empleo_plain, caption = "Intervalos simples", digits = 3)| time | n.risk | n.event | n.censor | surv | std.err | upper | lower |
|---|---|---|---|---|---|---|---|
| 0.4 | 12 | 1 | 0 | 0.917 | 0.087 | 1.000 | 0.760 |
| 0.9 | 11 | 1 | 0 | 0.833 | 0.129 | 1.000 | 0.622 |
| 1.1 | 10 | 1 | 0 | 0.750 | 0.167 | 0.995 | 0.505 |
| 1.5 | 9 | 0 | 1 | 0.750 | 0.167 | 0.995 | 0.505 |
| 1.9 | 8 | 1 | 0 | 0.656 | 0.214 | 0.931 | 0.381 |
| 2.0 | 7 | 1 | 0 | 0.562 | 0.264 | 0.853 | 0.272 |
| 3.0 | 6 | 0 | 1 | 0.562 | 0.264 | 0.853 | 0.272 |
| 3.3 | 5 | 1 | 0 | 0.450 | 0.346 | 0.755 | 0.145 |
| 5.3 | 4 | 1 | 0 | 0.338 | 0.450 | 0.635 | 0.040 |
| 5.8 | 3 | 1 | 0 | 0.225 | 0.608 | 0.493 | 0.000 |
| 6.2 | 2 | 0 | 1 | 0.225 | 0.608 | 0.493 | 0.000 |
| 14.0 | 1 | 1 | 0 | 0.000 | Inf | NaN | NaN |
knitr::kable(Primer_empleo_log, caption = "Intervalos transformados", digits = 3)| time | n.risk | n.event | n.censor | surv | std.err | upper | lower |
|---|---|---|---|---|---|---|---|
| 0.4 | 12 | 1 | 0 | 0.917 | 0.087 | 0.988 | 0.539 |
| 0.9 | 11 | 1 | 0 | 0.833 | 0.129 | 0.956 | 0.482 |
| 1.1 | 10 | 1 | 0 | 0.750 | 0.167 | 0.912 | 0.408 |
| 1.5 | 9 | 0 | 1 | 0.750 | 0.167 | 0.912 | 0.408 |
| 1.9 | 8 | 1 | 0 | 0.656 | 0.214 | 0.856 | 0.320 |
| 2.0 | 7 | 1 | 0 | 0.562 | 0.264 | 0.791 | 0.244 |
| 3.0 | 6 | 0 | 1 | 0.562 | 0.264 | 0.791 | 0.244 |
| 3.3 | 5 | 1 | 0 | 0.450 | 0.346 | 0.710 | 0.155 |
| 5.3 | 4 | 1 | 0 | 0.338 | 0.450 | 0.618 | 0.086 |
| 5.8 | 3 | 1 | 0 | 0.225 | 0.608 | 0.511 | 0.036 |
| 6.2 | 2 | 0 | 1 | 0.225 | 0.608 | 0.511 | 0.036 |
| 14.0 | 1 | 1 | 0 | 0.000 | Inf | NA | NA |
ggsurvplot(Primer_empleo_plain, conf.int = FALSE)+ggtitle("S(t) Kaplan Meier")
ggsurvplot(Primer_empleo_plain, conf.int = TRUE)+ggtitle("S(t) Kaplan Meier, IC simples")
ggsurvplot(Primer_empleo_log, conf.int = TRUE)+ggtitle("S(t) Kaplan Meier, IC log-transformados")
El estimador de Nelson-Aalen
Primer_empleo_log_NA<-
surv_summary(survfit(Surv(time,cc)~1, , type= "fleming-harrington", conf.type="log-log",conf.int=0.95))
knitr::kable(Primer_empleo_log_NA, caption = "Intervalos transformados", digits = 3)| time | n.risk | n.event | n.censor | surv | std.err | upper | lower |
|---|---|---|---|---|---|---|---|
| 0.4 | 12 | 1 | 0 | 0.920 | 0.083 | 0.988 | 0.553 |
| 0.9 | 11 | 1 | 0 | 0.840 | 0.123 | 0.957 | 0.498 |
| 1.1 | 10 | 1 | 0 | 0.760 | 0.159 | 0.916 | 0.426 |
| 1.5 | 9 | 0 | 1 | 0.760 | 0.159 | 0.916 | 0.426 |
| 1.9 | 8 | 1 | 0 | 0.671 | 0.202 | 0.862 | 0.341 |
| 2.0 | 7 | 1 | 0 | 0.582 | 0.247 | 0.801 | 0.265 |
| 3.0 | 6 | 0 | 1 | 0.582 | 0.247 | 0.801 | 0.265 |
| 3.3 | 5 | 1 | 0 | 0.476 | 0.318 | 0.726 | 0.179 |
| 5.3 | 4 | 1 | 0 | 0.371 | 0.405 | 0.640 | 0.110 |
| 5.8 | 3 | 1 | 0 | 0.266 | 0.524 | 0.543 | 0.056 |
| 6.2 | 2 | 0 | 1 | 0.266 | 0.524 | 0.543 | 0.056 |
| 14.0 | 1 | 1 | 0 | 0.098 | 1.129 | 0.407 | 0.002 |
ggsurvplot(Primer_empleo_log_NA, conf.int = FALSE)+ggtitle("S(t) Nelson-Aalen")
ggsurvplot(Primer_empleo_log_NA, conf.int = TRUE)+ggtitle("S(t) Nelson-Aalen, IC simples")
ggsurvplot(Primer_empleo_log_NA, conf.int = TRUE)+ggtitle("S(t) Nelson-Aalen, IC log-transformados")