SMHS SurvivalAnalysis - Revision history

Dinov: /* References */

2026-02-10T21:27:24Z

‎References

Dinov: /* Additional Examples */

2025-12-08T19:41:01Z

‎Additional Examples

Dinov: /* Additional Examples */

2025-12-08T19:28:58Z

‎Additional Examples

Dinov at 19:18, 8 December 2025

2025-12-08T19:18:36Z

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Dinov: /* Theory */

2022-11-18T18:50:33Z

‎Theory

Dinov: /* Theory */

2022-11-18T18:45:45Z

‎Theory

Dinov: /* Theory */

2022-11-18T18:42:24Z

‎Theory

Dinov: /* Hazard function models */

2014-11-19T20:45:22Z

‎Hazard function models

Dinov: /* Example */

2014-11-19T19:23:48Z

‎Example

Dinov: /* Example */

2014-11-19T19:22:45Z

‎Example

@@ Line 578: / Line 578: @@
 ===References===
 * [http://en.wikipedia.org/wiki/Survival_analysis  Survival Analysis Wikipedia]
 * [http://en.wikipedia.org/wiki/Hazard_ratio  Hazard Ratio Wikipedia]
 <hr>
-* SOCR Home page: http://www.socr.umich.edu
+* SOCR Home page: https://socr.umich.edu
-{{translate|pageName=http://wiki.socr.umich.edu/index.php?title=SMHS_SurvivalAnalysis}}
+{{translate|pageName=https://wiki.socr.umich.edu/index.php?title=SMHS_SurvivalAnalysis}}

@@ Line 547: / Line 547: @@
         library(cmprsk)
         library(survival)
-        data(mgus2)
-        # Competing risks regression
+       # Create a numeric covariate matrix
-        crr_model <- crr(mgus2$ptime, mgus2$pstat, cov1 = mgus2[, c("age", "sex")])
+        covariates <- model.matrix(~ age + sex, data = mgus2)[, -1]  # remove intercept column
         summary(crr_model)
         </pre>
@@ Line 557: / Line 563: @@
       <pre>
       library(survival)
       data(lung)
       # Fit Cox model
-      cox_model <- coxph(Surv(time, status) ~ age + sex + ph.ecog, data = lung)
+      cox_model <- coxph(Surv(time, status) ~ age + sex + ph.ecog, data = lung2)
       summary(cox_model)
       # Plot survival curves stratified by sex
-      surv_fit <- survfit(Surv(time, status) ~ sex, data = lung)
+      surv_fit <- survfit(Surv(time, status) ~ sex, data = lung2)
       plot(surv_fit, col = c("blue", "red"), xlab = "Time (days)", ylab = "Survival Probability")
       legend("topright", c("Male", "Female"), col = c("blue", "red"), lty = 1)

@@ Line 546: / Line 546: @@
         <pre>
         library(cmprsk)
-        data(mgus)
+       library(survival)
         # Competing risks regression
-        crr_model <- crr(mgus$ptime, mgus$pstat, cov1 = mgus[, c("age", "sex")])
+        crr_model <- crr(mgus2$ptime, mgus2$pstat, cov1 = mgus2[, c("age", "sex")])
         summary(crr_model)
         </pre>

@@ Line 9: / Line 9: @@
 ===Theory===
-'''Risk''' is the expected value of an undesirable outcome. To estimate the risk, we combine (weight-average) the probabilities of various possible events (outcomes) and some assessment of the corresponding harm into a single value, the risk. For a simple binary possibility of ''accident'' vs. ''no-accident'' situation, the Risk = (probability of the accident occurring) $\times $ (expected loss in case of the accident). For instance, probability of accident for certain sport activity may be equal to \(0.001\), which would be associated with a loss of \(\$ 10,000\) healthcare costs, then total risk of playing the sport is a loss of \(0.001\times \$10,000=\$10\).
+'''Risk''' is the expected value of an undesirable outcome. To estimate the risk, we combine (weight-average) the probabilities of various possible events (outcomes) and some assessment of the corresponding harm into a single value, the risk. For a simple binary possibility of ''accident'' vs. ''no-accident'' situation, the Risk = (probability of the accident occurring) \(\times \) (expected loss in case of the accident). For instance, probability of accident for certain sport activity may be equal to \(0.001\), which would be associated with a loss of \(\$ 10,000\) healthcare costs, then total risk of playing the sport is a loss of \(0.001\times \$10,000=\$10\).
 In general, when there are multiple types of negative outcomes (e.g., types of potential injuries), the total risk is the weighted-average of all different accidents (outcomes):
 $$Risk = \sum_{\text{For all accidents}} {(\text{probability of the accident occurring}) \times (\text{expected loss in case of the accident})}.$$
-For example, in certain clinical decision making, treatment X may lead to an adverse event A with probability of 0.01, which is associated with a (monetary) loss of $\$1,000$, and a different probability of 0.000001 of an adverse event of type B, associated with a loss of $\$200,000$, then the total risk (loss expectancy) would be:
+For example, in certain clinical decision making, treatment X may lead to an adverse event A with probability of \(0.01\), which is associated with a (monetary) loss of \(\$1,000\), and a different probability of \(0.000001\) of an adverse event of type B, associated with a loss of \(\$200,000\), then the total risk (loss expectancy) would be:
 $$Risk= (0.01\times \$1,000) + (0.000001\times \$200,000) = \$10.2.$$
 ====Survival function====
 The '''survival function''', is defined as
-$S(t) = P(T > t), $ where $t$ is time, $T$ is a random variable denoting the time of event (e.g., death), and $P$ is the probability function. Hence, the survival function is the probability that the time of event is bigger than some specified time $t$. It is common to assume $S(0) = 1$, as the possibility of immediate death is unlikely, most of the time. If $u \ge t$, then knowing that $T > u$ yields $T > t$, ie.g., the survival function is decreasing. Thus, $S(t) \ge S(u)$ for all $u \ge t$ -- for all subjects, survival to a later age is only possible if subjects have survived all previous younger ages. The survival function usually tends to zero as $T$ (age) increases.
+\(S(t) = P(T > t), \) where \(t\) is time, \(T\) is a random variable denoting the time of event (e.g., death), and \(P\) is the probability function. Hence, the survival function is the probability that the time of event is bigger than some specified time \(t\). It is common to assume \(S(0) = 1\), as the possibility of immediate death is unlikely, most of the time. If \(u \ge t\), then knowing that \(T > u\) yields \(T > t\), e.g., the survival function is decreasing. Thus, \(S(t) \ge S(u)\) for all \(u \ge t\) -- for all subjects, survival to a later age is only possible if subjects have survived all previous younger ages. The survival function usually tends to zero as \(T\) (age) increases.
 ====Hazard Ratios====
@@ Line 35: / Line 35: @@
 The Mantel-Haenszel method reports hazard ratios that are further from 1.0  (so the reported hazard ratio is too large when the hazard ratio is greater than 1.0, and too small when the hazard ratio is less than 1.0):
 # Compute the [http://www.biostat.au.dk/teaching/survival/F2004/survival-2004-1.pdf total variance, V];
-# Compute $K = \frac{O1 - E1}{V}$, where $O1$ - is the total ''observed'' number of events in group1, $E1$ is the total ''expected'' number of events in group1. You'd get the same value of $K$ if you used the other group;
+# Compute \(K = \frac{O1 - E1}{V}\), where \(O1\) - is the total ''observed'' number of events in group1, \(E1\) is the total ''expected'' number of events in group1. You'd get the same value of \(K\) if you used the other group;
-# The hazard ratio equals $e^K$;
+# The hazard ratio equals \(e^K\);
-# The 95% confidence interval of the hazard ratio is: ($e^{K - \frac{1.96}{\sqrt{V}}}$, $e^{K + \frac{1.96}{\sqrt{V}}}$).
+# The 95% confidence interval of the hazard ratio is: (\(e^{K - \frac{1.96}{\sqrt{V}}}$, $e^{K + \frac{1.96}{\sqrt{V}}}\)).
 =====The LogRank method=====
 The LogRank method is also referred to as O/E method and reports values that are closer to 1.0 than the true Hazard Ratio, especially when the hazard ratio is large or the sample size is large. When there are ties, both methods are less accurate. The logrank methods tend to report hazard ratios that are even closer to 1.0 (so the reported hazard ratio is too small when the hazard ratio is greater than 1.0, and too large when the hazard ratio is less than 1.0):
-# As part of the Kaplan-Meier calculations, compute the number of observed events (deaths, usually) in each group ($O_a$ and $O_b$), and the number of expected events assuming a null hypothesis of no difference in survival ($E_a$ and $E_b$);
+# As part of the Kaplan-Meier calculations, compute the number of observed events (deaths, usually) in each group ($O_a$ and $O_b$), and the number of expected events assuming a null hypothesis of no difference in survival (\(E_a\) and \(E_b\));
-# The hazard ratio then is: $HR= \frac{\frac{O_a}{E_a}}{\frac{O_b}{E_b}}$;
+# The hazard ratio then is: \(HR= \frac{\frac{O_a}{E_a}}{\frac{O_b}{E_b}}\);
-# The standard error of the natural logarithm of the hazard ratio is: $\ln{HR}=\sqrt{\frac{1}{E_a} + \frac{1}{E_b}}$;
+# The standard error of the natural logarithm of the hazard ratio is: \(\ln{HR}=\sqrt{\frac{1}{E_a} + \frac{1}{E_b}}\);
-# The lower and upper limits of the 95% confidence interval of the hazard ratio are: $e^{\big\{ \ln(O_a)-\ln(E_a) \pm 1.96 \sqrt{\frac{1}{E_a} + \frac{1}{E_b}}\big\}}$. Recall that $HR=\frac{O_a E_b}{O_b E_a}$, and under the null hypothesis $E_a=E_b$, so $\ln(HR)= \ln(O_a)-\ln(E_a).$
+# The lower and upper limits of the 95% confidence interval of the hazard ratio are: \(e^{\big\{ \ln(O_a)-\ln(E_a) \pm 1.96 \sqrt{\frac{1}{E_a} + \frac{1}{E_b}}\big\}}\). Recall that \(HR=\frac{O_a E_b}{O_b E_a}\), and under the null hypothesis \(E_a=E_b\), so \(\ln(HR)= \ln(O_a)-\ln(E_a).\)
 ====Survival analysis goals====

@@ Line 9: / Line 9: @@
 ===Theory===
-'''Risk''' is the expected value of an undesirable outcome. To estimate the risk, we combine (weight-average) the probabilities of various possible events (outcomes) and some assessment of the corresponding harm into a single value, the risk. For a simple binary possibility of ''accident'' vs. ''no-accident'' situation, the Risk = (probability of the accident occurring) $\times $ (expected loss in case of the accident). For instance, probability of accident for certain sport activity may be equal to 0.001, which would be associated with a loss of $\$ 10,000$ healthcare costs, then total risk of playing the sport is a loss of $0.001\times \$10,000=\$10$.
+'''Risk''' is the expected value of an undesirable outcome. To estimate the risk, we combine (weight-average) the probabilities of various possible events (outcomes) and some assessment of the corresponding harm into a single value, the risk. For a simple binary possibility of ''accident'' vs. ''no-accident'' situation, the Risk = (probability of the accident occurring) $\times $ (expected loss in case of the accident). For instance, probability of accident for certain sport activity may be equal to \(0.001\), which would be associated with a loss of \(\$ 10,000\) healthcare costs, then total risk of playing the sport is a loss of \(0.001\times \$10,000=\$10\).
 In general, when there are multiple types of negative outcomes (e.g., types of potential injuries), the total risk is the weighted-average of all different accidents (outcomes):

@@ Line 217: / Line 217: @@
 The second subject was lost to follow up, and cases 4 and 5 were alive after 12 months. The remaining 2 cases (1 and 3) died at times 4 and 7 months, respectively. To estimate the likelihood of a patient to survive 1 year (12 months) we break the 12 months into intervals anchored at points of dying (in this case 4 and 7 months):
-$$ S(12) = P(\text{surviving intervals [0,4] and [4,7]})=$$
+$$ S(12) = P(\text{surviving intervals [0,4] and (4,7] and and (7,12]})=$$
-$$P(T>4)\times P(T>7|T>4)=$$
+$$P(T>4)\times P(T>7|T>4)\times P(T>12|T>7)=$$
-$$P(\text{surviving interval [0,4]})\times P(\text{surviving interval [4,7]}|\text{survived interval [0,4]})$$
+$P(\text{surviving interval [0,4]})\times$ $P(\text{surviving interval (4,7]}|\text{survived interval [0,4]})$ $\times P(\text{surviving interval (7,12]}|\text{survived interval [0,7]})$
-:: Note the single censored case at time 6 months! $P(\text{surviving intervals [0,4] and [4,7]})=4/5 \times 2/3= 0.53$, as 4 of the five cases make it beyond time 4, and only 2 cases make it beyond the 7th month. Thus, $ S(12)=0.53$ which is between the probability of surviving 12 months = 40% (we end with 2 survivors at the end) and 60% (3/5) (only 2 are known to have died during the year).
 <center>[[Image:SMHS_SurvivalAnalysis_Fig1.c.png|500px]]</center>

@@ Line 235: / Line 235: @@
 |}
 </center>
-::: to estimate $S(10)$: $S(10)=P(T>10)=P(T>10,T>8,T>5,T>2)=$ $P(T>10│T>8)P(T>8│T>5)P(T>5│T>2)P(T>2)=$ $7/7 * 7/8 * 9/9 * 9/10=0.7875$. So, $\hat{S}_{KM}(10)=0.7875$.
+* To estimate $S(10)$: $S(10)=P(T>10)=P(T>10,T>8,T>5,T>2)=$ $P(T>10│T>8)P(T>8│T>5)P(T>5│T>2)P(T>2)=$ $7/7 * 7/8 * 9/9 * 9/10=0.7875$. So, $\hat{S}_{KM}(10)=0.7875$.
-::: to estimate $S(20)$: $S(20)=P(T>20)=P(T>20,T>15,T>10,T>8,T>5,T>2)=$ $P(T>20│T>15)P(T>15│T>10)P(T>10│T>8)P(T>8│T>5)P(T>5│T>2)P(T>2)=$ $6/6 * 5/7 * 7/7 * 7/8 * 9/9 * 9/10=0.5625$. So, $\hat{S}_{KM}(20)=0.5625$.
+* To estimate $S(20)$: $S(20)=P(T>20)=P(T>20,T>15,T>10,T>8,T>5,T>2)=$ $P(T>20│T>15)P(T>15│T>10)P(T>10│T>8)P(T>8│T>5)P(T>5│T>2)P(T>2)=$ $6/6 * 5/7 * 7/7 * 7/8 * 9/9 * 9/10=0.5625$. So, $\hat{S}_{KM}(20)=0.5625$.
 <center>[[Image:SMHS_SurvivalAnalysis_Fig1.d.png|500px]]</center>