---
title: "Recurrent events"
author: Klaus Holst & Thomas Scheike
date: "`r Sys.Date()`"
output:
  knitr:::html_vignette:
    fig_caption: yes
vignette: >
  %\VignetteIndexEntry{Recurrent events}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

```{r, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)
library(mets)
```

Overview 
========

For recurrent events data it is often of interest to compute basic descriptive
quantities as a first go at getting some basic understanding of the phenonmenon
studied. We here demonstrate how one can compute 

 * the marginal mean 
 * the variance 
 * the probability of exceeding k events

In addition several tools can be used for simulating recurrent events and
bivariate recurrent events data, in the case with a possible terminating event. 

For bivariate recurrent events we also compute summary measures that describe their
dependence such as 

 * the covariance 
 * directional dependence 
 * the bivariate probability of exceeding $(k_1,k_2)$ events
 


Simulation of recurrents events
==============================


We start by simulating some recurrent events data with two type of events with
cumulative hazards 

 * $\Lambda_1(t)$ (rate among survivors)
 * $\Lambda_2(t)$ (rate among survivors)
 * $\Lambda_D(t)$

 where we consider types 1 and 2 and with a rate of the terminal event given by 
 $\Lambda_D(t)$. We let the events be independent, but could also specify a random effects
 structure to generate dependence. 

 When simulating data we can impose various random-effects structures to generate dependence  

* Dependence=0: The intensities can be independent.

* Dependence=1:  We can one gamma distributed random effects $Z$.  Then the intensities are 
    + $Z \lambda_1(t)$
    + $Z \lambda_2(t)$
    + $Z \lambda_D(t)$

* Dependence=2: We can draw normally distributed random effects $Z_1,Z_2,Z_d$ were the variance (var.z) 
  and correlation can be specified (cor.mat).  Then the intensities are 
    + $\exp(Z_1) \lambda_1(t)$
    + $\exp(Z_2) \lambda_2(t)$
    + $\exp(Z_3) \lambda_D(t)$

* Dependence=3: We can draw gamma distributed random effects $Z_1,Z_2,Z_d$ were the sum-structure can be speicifed via a matrix
   cor.mat. We compute $\tilde Z_j = \sum_k  Z_k^{cor.mat(j,k)}$ for $j=1,2,3$.  
   Then the intensities are
    + $\tilde Z_1 \lambda_1(t)$
    + $\tilde Z_2 \lambda_2(t)$
    + $\tilde Z_3 \lambda_D(t)$


 We return to how to run the different set-ups later and start by simulating independent processes. 

Utility functions 
=================

We here mention two utility functions 

 * tie.breaker for breaking ties among jump-times which is expected in the functions below.
 * count.history that counts the number of jumps previous for each subject that is $N_1(t-)$ and $N_2(t-)$. 
 

Marginal Mean 
=============

 We start by estimating the marginal mean $E(N_1(t \wedge D))$ where $D$ is the timing of the terminal event. 

 This is based on a  rate model  for 

 * the type 1 events  $\sim E(dN_1(t) | D > t)$
 * the terminal event $\sim E(dN_d(t) | D > t)$

and is defined as $\mu_1(t)=E(N_1^*(t))$ 
\begin{align}
   \int_0^t S(u) d R_1(u)	
\end{align}
where $S(t)=P(D \geq t)$ and $dR_1(t) = E(dN_1^*(t) | D > t)$ 

and can therefore be estimated by a 

 * Kaplan-Meier estimator, $\hat S(u)$ 
 * Nelson-Aalen estimator for $R_1(t)$

\begin{align}
  \hat R_1(t) & =   \sum_i \int_0^t  \frac{1}{Y_\bullet (s)}  dN_{1i}(s)
\end{align}
where $Y_{\bullet}(t)= \sum_i Y_i(t)$ such that the estimator is 
\begin{align}
  \hat \mu_1(t) & =    \int_0^t \hat S(u) d\hat R_1(u).
\end{align}

Cook & Lawless (1997), and developed further in Gosh & Lin (2000). 

The variance can be estimated based on the asymptotic expansion 
of $\hat \mu_1(t) - \mu_1(t)$
\begin{align*}
  & \sum_i \int_0^t \frac{S(s)}{\pi(s)} dM_{i1}  - \mu_1(t) \int_0^t  \frac{1}{\pi(s)} dM_i^d +  \int_0^t \frac{\mu_1(s) }{\pi(s)} dM_i^d,
\end{align*}

with mean-zero processes 

 * $M_i^d(t) = N_i^D(t)- \int_0^t Y_i(s) d \Lambda^D(s)$, 
 * $M_{i1}(t) = N_{i1}(t) - \int_0^t Y_{i}(s) dR_1(s)$. 

as in Gosh & Lin (2000)


Generating data
==================

We start by generating some data to illustrate the computation of the marginal mean 

```{r}
library(mets)
set.seed(1000) # to control output in simulatins for p-values below.
```


```{r}
data(base1cumhaz)
data(base4cumhaz)
data(drcumhaz)
ddr <- drcumhaz
base1 <- base1cumhaz
base4 <- base4cumhaz
rr <- simRecurrent(1000,base1,death.cumhaz=ddr)
rr$x <- rnorm(nrow(rr)) 
rr$strata <- floor((rr$id-0.01)/500)
dlist(rr,.~id| id %in% c(1,7,9))
```

The status variable keeps track of the recurrent evnts and their type, and death the timing of 
death.

To compute the marginal mean we simly estimate the two rates functions of the 
number of events of interest and death by using the phreg function 
(to start without covariates). Then the estimates are combined with standard 
error computation in the recurrentMarginal function

```{r}
#  to fit non-parametric models with just a baseline 
xr <- phreg(Surv(entry,time,status)~cluster(id),data=rr)
dr <- phreg(Surv(entry,time,death)~cluster(id),data=rr)
par(mfrow=c(1,3))
bplot(dr,se=TRUE)
title(main="death")
bplot(xr,se=TRUE)
# robust standard errors 
rxr <-   robust.phreg(xr,fixbeta=1)
bplot(rxr,se=TRUE,robust=TRUE,add=TRUE,col=4)

# marginal mean of expected number of recurrent events 
out <- recurrentMarginal(xr,dr)
bplot(out,se=TRUE,ylab="marginal mean",col=2)
```

We can also extract the estimate in different time-points 

```{r}
summary(out,times=c(1000,2000))
```

The marginal mean can also be estimated in a stratified case:


```{r}
xr <- phreg(Surv(entry,time,status)~strata(strata)+cluster(id),data=rr)
dr <- phreg(Surv(entry,time,death)~strata(strata)+cluster(id),data=rr)
par(mfrow=c(1,3))
bplot(dr,se=TRUE)
title(main="death")
bplot(xr,se=TRUE)
rxr <-   robust.phreg(xr,fixbeta=1)
bplot(rxr,se=TRUE,robust=TRUE,add=TRUE,col=1:2)

out <- recurrentMarginal(xr,dr)
bplot(out,se=TRUE,ylab="marginal mean",col=1:2)
```

Furhter, if we adjust for covariates for the two rates we can still do
predictions of marginal mean, what can be plotted is the baseline marginal mean, 
that is for the covariates equal to 0 for both models. Predictions for specific 
covariates can also be obtained with the recmarg (recurren marginal mean used 
solely for predictions without standard error computation). 

```{r}
# cox case
xr <- phreg(Surv(entry,time,status)~x+cluster(id),data=rr)
dr <- phreg(Surv(entry,time,death)~x+cluster(id),data=rr)
par(mfrow=c(1,3))
bplot(dr,se=TRUE)
title(main="death")
bplot(xr,se=TRUE)
rxr <- robust.phreg(xr)
bplot(rxr,se=TRUE,robust=TRUE,add=TRUE,col=1:2)

out <- recurrentMarginal(xr,dr)
bplot(out,se=TRUE,ylab="marginal mean",col=1:2)

# predictions witout se's 
outX <- recmarg(xr,dr,Xr=1,Xd=1)
bplot(outX,add=TRUE,col=3)
```


Other marginal properties 
=========================

The mean is a useful summary measure but it is very easy and useful to look at other 
simple summary measures such as the probability of exceeding $k$ events 

 * $P(N_1^*(t) \ge k)$ 
   * cumulative incidence of $T_{k} = \inf \{ t: N_1^*(t)=k \}$ with competing $D$. 

that is thus equivalent to a certain cumulative incidence of $T_k$ occurring before $D$. We denote this
cumulative incidence as $\hat F_k(t)$. 

We note also that $N_1^*(t)^2$ can be written as
\begin{align*}
   \sum_{k=0}^K  \int_0^t I(D > s) I(N_1^*(s-)=k) f(k) dN_1^*(s)
\end{align*}
with $f(k)=(k+1)^2 - k^2$, such that its mean can be written as 
\begin{align*}
	\sum_{k=0}^K \int_0^t S(s) f(k) P(N_1^*(s-)= k  | D  \geq s) E( dN_1^*(s)  | N_1^*(s-)=k, D> s) 
\end{align*}
and estimated by
\begin{align*}
\tilde \mu_{1,2}(t) & = 
	\sum_{k=0}^K \int_0^t \hat S(s) f(k) 
	\frac{Y_{1\bullet}^k(s)}{Y_\bullet (s)} \frac{1}{Y_{1\bullet}^k(s)} d N_{1\bullet}^k(s)= \sum_{i=1}^n \int_0^t \hat S(s) f(N_{i1}(s-)) \frac{1}{Y_\bullet (s)} d N_{i1}(s),
\end{align*}
That is very similar to the "product-limit" estimator for $E( (N_1^*(t))^2 )$ 
\begin{align}
  \hat \mu_{1,2}(t) & =    \sum_{k=0}^K k^2 ( \hat F_{k}(t) - \hat F_{k+1}(t) ).
\end{align}

We use the esimator of the probabilty  of exceeding "k" events based on the fact that 
$I(N_1^*(t) \geq k)$ is  equivalent to 
\begin{align*}
	\int_0^t I(D > s) I(N_1^*(s-)=k-1) dN_1^*(s),
\end{align*}
suggesting that its mean can be computed as
\begin{align*}
\int_0^t S(s) P(N_1^*(s-)= k-1  | D  \geq s) E( dN_1^*(s)  | N_1^*(s-)=k-1, D> s) 
\end{align*}
and estimated by 
\begin{align*}
\tilde F_k(t) = \int_0^t \hat S(s)  \frac{Y_{1\bullet}^{k-1}(s)}{Y_\bullet (s)} 
          	\frac{1}{Y_{1\bullet}^{k-1}(s)} d N_{1\bullet}^{k-1}(s).
\end{align*}


To compute these estimators we need to set up the data by computing the 
number of previous events of type "1" by the  count.history function


```{r}
###cor.mat <- corM <- rbind(c(1.0, 0.6, 0.9), c(0.6, 1.0, 0.5), c(0.9, 0.5, 1.0))
###rr <- simRecurrent(1000,base1,cumhaz2=base4,death.cumhaz=ddr)
rr <-  count.history(rr)
dtable(rr,~death+status)

oo <- prob.exceedRecurrent(rr,1)
bplot(oo)
```

We can also look at the mean and variance based on the estimators just described 

```{r}
par(mfrow=c(1,2))
with(oo,plot(time,mu,col=2,type="l"))
#
with(oo,plot(time,varN,type="l"))
```

We could also use the product-limit estimator to estimate the probability of 
exceeding "k" events, and then standard errors are also returned: 

```{r}
 oop <- prob.exceed.recurrent(rr,1)
 bplot(oo)
 matlines(oop$times,oop$prob,type="l")
 summaryTimeobject(oop$times,oop$prob,se.mu=oop$se.prob,times=1000)
```

We note from the plot that the estimates are quite similar. 

Finally, we make a plot with 95\% confidence intervals

```{r}
matplot(oop$times,oop$prob,type="l")
for (i in seq(ncol(oop$prob))) 
	plotConfRegion(oop$times,cbind(oop$se.lower[,i],oop$se.upper[,i]),col=i)
```


Multiple events 
================

We now generate recurrent events with two types of events. We start by
generating data as before where all events are independent. 

```{r}
rr <- simRecurrent(1000,base1,cumhaz2=base4,death.cumhaz=ddr)
rr <-  count.history(rr)
dtable(rr,~death+status)
```

Based on this we can estimate also the joint distribution function, that is
the probability that $(N_1(t) \geq k_1, N_2(t) \geq k_2)$

```{r}
# Bivariate probability of exceeding 
oo <- prob.exceedBiRecurrent(rr,1,2,exceed1=c(1,5,10),exceed2=c(1,2,3))
with(oo, matplot(time,pe1e2,type="s"))
nc <- ncol(oo$pe1e2)
legend("topleft",legend=colnames(oo$pe1e2),lty=1:nc,col=1:nc)
```

Dependence between events: Covariance 
--------------------------------------

The dependence can also be summarised in other ways. For example by computing the
covariance and comparing it to the covariance under the assumption of 
independence among survivors. 

Covariance among two types of events 
\begin{align}
\rho(t) &  = \frac{ E(N_1^*(t) N_2^*(t) )  - \mu_1(t) \mu_2(t) }{ \mbox{sd}(N_1^*(t)) \mbox{sd}(N_2^*(t)) }
\end{align}
where  $E(N_1^*(t) N_2^*(t))$  can be computed as 
\begin{align*}
  E(N_1^*(t) N_2^*(t)) &   = E( \int_0^t N_1^*(s-) dN_2^*(s) ) + E( \int_0^t N_2^*(s-) dN_1^*(s) ) 
\end{align*}

Recall that we might have a terminal event present such that we only see 
$N_1^*(t \wedge D)$  and $N_2^*(t \wedge D)$. 

To compute the covariance we thus compute 
\begin{align*}
E(\int_0^t N_1^*(s-) dN_2^*(s) ) & = \sum_k E( \int_0^t k I(N_1^*(s-)=k) I(D \geq s)  dN_2^*(s) ) 
\end{align*}
\begin{align*}
= \sum_k \int_0^t S(s) k P(N_1^*(s-)= k  | D  \geq s) E( dN_2^*(s)  | N_1^*(s-)=k, D \geq s) 
\end{align*}
estimated by 
\begin{align*}
  & \sum_k \int_0^t \hat S(s) k \frac{Y_1^k(s)}{Y_\bullet (s)} \frac{1}{Y_1^k(s)} d \tilde N_{2,k}(s),
\end{align*}
 * $Y_j^k(t) = \sum Y_i(t) I( N_{ji}^*(s-)=k)$ for $j=1,2$, 
 * $\tilde N_{j,k}(t) = \sum_i \int_0^t I(N_{ij^o}(s-)=k) dN_{ij}(s)$ 
 * $j^o$ gives the other type so that $1^o=2$ and $2^o=1$.

We thus estimate  $ E(N_1^*(t) N_2^*(t))$  by 
\begin{align*}
  \sum_k \int_0^t \hat S(s) k \frac{Y_1^k(s)}{Y_\bullet (s)} \frac{1}{Y_1^k(s)} d \tilde N_{2,k}(s) +
   \sum_k \int_0^t \hat S(s) k \frac{Y_2^k(s)}{Y_\bullet (s)} \frac{1}{Y_2^k(s)} d \tilde N_{1,k}(s).
\end{align*}


 * Without terminating event covariance is a useful nonparametric measure.
 * With terminating event dependence can be generated  terminating event.
 * In reality what is of interest would be independence among survivors 
   that is if 
   * $N_1$ is not predicitive for $N_2$
   \begin{align}
      E( dN_2^*(t)  | N_1^*(t-)=k, D \geq t) =  E( dN_2^*(t)  | D \geq t) 
   \end{align}
   * $N_2$ is  not predicitive for $N_1$
   \begin{align}
      E( dN_1^*(t)  | N_2^*(t-)=k, D \geq t) =  E( dN_1^*(t)  | D \geq t) 
   \end{align}


If the two processes are independent among survivors then 
\begin{align}
 E( dN_2^*(t)  | N_1^*(t-)=k, D \geq t) =  E( dN_2^*(t)  | D \geq t) 
\end{align}
so 
\begin{align*}
  E( \int_0^t N_1^*(s-) dN_2^*(s) )  & =  \int_0^t S(s) E(N_1^*(s-) | D  \geq s) E( dN_2^*(s)  | D \geq s) 
\end{align*}
and 
\begin{align*}
  \int_0^t \hat S(s) \{  \sum_k k \frac{Y_1^k(s)}{Y_\bullet (s)} \} \frac{1}{Y_\bullet (s)} dN_{2\bullet}(s),
\end{align*}
where $N_{j\bullet}(t) = \sum_i \int_0^t dN_{j,i}(s)$.

Under the independence $E(N_1^*(t) N_2^*(t))$  is estimated 
\begin{align*}
  \int_0^t \hat S(s) \{  \sum_k k \frac{Y_1^k(s)}{Y_\bullet (s)} \} \frac{1}{Y_\bullet (s)} dN_{2\bullet}(s) 
    + \int_0^t \hat S(s) \{  \sum_k k \frac{Y_2^k(s)}{Y_\bullet (s)} \} \frac{1}{Y_\bullet (s)} dN_{1\bullet}(s).
\end{align*}


Both estimators, $\hat E(N_1^*(t) N_2^*(t))$ and $\hat E_I(N_1^*(t) N_2^*(t))$,  as well as 
$\hat E(N_1^*(t))$ and $\hat E(N_2^*(t))$, have asymptotic expansions that can be 
written as a sum of iid processes, similarly to the arguments 
of Ghosh & Lin 2000, $\sum_i \Psi_i(t)$.  
We here, however, use a simple block bootstrap to get standard errors. 

We can thus estimate the standard errors and of the estimators and their 
difference  $\hat E(N_1^*(t) N_2^*(t))- \hat E_I(N_1^*(t) N_2^*(t))$. 

Note that we have terms for  whether 
 * $N_1$ is predicitive for $N_2$
   * N1 -> N2 : $E( \int_0^t N_1^*(s-) dN_2^*(s) )$
   * this is equivalent to a weighted log-rank test 
 * $N_2$ is  predicitive for $N_1$
   * N2 -> N1 : $E( \int_0^t N_2^*(s-) dN_1^*(s) )$
   * this is equivalent to a weighted log-rank test 


```{r}
rr$strata <- 1
dtable(rr,~death+status)

covrp <- covarianceRecurrent(rr,1,2,status="status",death="death",
                        start="entry",stop="time",id="id",names.count="Count")
par(mfrow=c(1,3)) 
plot(covrp)

# with strata, each strata in matrix column, provides basis for fast Bootstrap
covrpS <- covarianceRecurrentS(rr,1,2,status="status",death="death",
        start="entry",stop="time",strata="strata",id="id",names.count="Count")
```

Bootstrap standard errors for terms
------------------------------------

First fitting the model again to get our estimates of interst, and then 
computing them for some specific time-points  

```{r}
times <- seq(500,5000,500)

coo1 <- covarianceRecurrent(rr,1,2,status="status",start="entry",stop="time")
#
mug <- Cpred(cbind(coo1$time,coo1$EN1N2),times)[,2]
mui <- Cpred(cbind(coo1$time,coo1$EIN1N2),times)[,2]
mu2.1 <- Cpred(cbind(coo1$time,coo1$mu2.1),times)[,2]
mu2.i <- Cpred(cbind(coo1$time,coo1$mu2.i),times)[,2]
mu1.2 <- Cpred(cbind(coo1$time,coo1$mu1.2),times)[,2]
mu1.i <- Cpred(cbind(coo1$time,coo1$mu1.i),times)[,2]
cbind(times,mu2.1,mu2.i)
cbind(times,mu1.2,mu1.i)
```

To get the bootstrap standard errors there is a quick memory demanding function 
(with S for speed and strata)  BootcovariancerecurrenceS
and slow function that goes through the loops in R Bootcovariancerecurrence. 

```{r}
bt1 <- BootcovariancerecurrenceS(rr,1,2,status="status",start="entry",stop="time",K=100,times=times)
#bt1 <- Bootcovariancerecurrence(rr,1,2,status="status",start="entry",stop="time",K=K,times=times)

output <- list(bt1=bt1,mug=mug,mui=mui,
               bse.mug=bt1$se.mug,bse.mui=bt1$se.mui,
               dmugi=mug-mui,
	bse.dmugi=apply(bt1$EN1N2-bt1$EIN1N2,1,sd),
	mu2.1 = mu2.1 , mu2.i = mu2.i , dmu2.i=mu2.1-mu2.i,
	mu1.2 = mu1.2 , mu1.i = mu1.i , dmu1.i=mu1.2-mu1.i,
	bse.mu2.1=apply(bt1$mu2.i,1,sd), bse.mu2.1=apply(bt1$mu2.1,1,sd),
	bse.dmu2.i=apply(bt1$mu2.1-bt1$mu2.i,1,sd),
	bse.mu1.2=apply(bt1$mu1.2,1,sd), bse.mu1.i=apply(bt1$mu1.i,1,sd),
	bse.dmu1.i=apply(bt1$mu1.2-bt1$mu1.i,1,sd)
	)
```

We then look at the test for overall dependence  in the different time-points. 
We here have no suggestion of dependence.

```{r}
tt  <- output$dmugi/output$bse.dmugi
cbind(times,2*(1-pnorm(abs(tt))))
```

We can also take out the specific components for whether $N_1$ is predictive for 
$N_2$ and vice versa.  We here have no suggestion of dependence.

```{r}
t21  <- output$dmu1.i/output$bse.dmu1.i 
t12  <- output$dmu2.i/output$bse.dmu2.i 
cbind(times,2*(1-pnorm(abs(t21))),2*(1-pnorm(abs(t12))))
```

We finally plot the boostrap samples 

```{r}
par(mfrow=c(1,2))
matplot(bt1$time,bt1$EN1N2,type="l",lwd=0.3)
matplot(bt1$time,bt1$EIN1N2,type="l",lwd=0.3)
```

Looking at other simulations with dependence 
=============================================


Using the normally distributed random effects we plot 4 different settings. We have variance $0.5$ for all
random effects and change the correlation. We let the correlation between the random effect associated with
$N_1$ and $N_2$ be denoted $\rho_{12}$ and the correlation between the random effects 
associated between $N_j$ and $D$ the terminal event be denoted as $\rho_{j3}$, and organize all correlation
in a vector $\rho=(\rho_{12},\rho_{13},\rho_{23})$.

 * Scenario I $\rho=(0,0.0,0.0)$ Independence among all efects.

```{r}
  data(base1cumhaz)
  data(base4cumhaz)
  data(drcumhaz)
  dr <- drcumhaz
  base1 <- base1cumhaz
  base4 <- base4cumhaz

  par(mfrow=c(1,3))
  var.z <- c(0.5,0.5,0.5)
  # death related to  both causes in same way 
  cor.mat <- corM <- rbind(c(1.0, 0.0, 0.0), c(0.0, 1.0, 0.0), c(0.0, 0.0, 1.0))
  rr <- simRecurrentII(3000,base1,base4,death.cumhaz=dr,var.z=var.z,cor.mat=cor.mat,dependence=2)
  rr <- count.history(rr,types=1:2)
  cor(attr(rr,"z"))
  coo <- covarianceRecurrent(rr,1,2,status="status",start="entry",stop="time")
  plot(coo,main ="Scenario I")
```
 * Scenario II $\rho=(0,0.5,0.5)$ Independence among survivors but dependence on terminal event 

```{r}
  var.z <- c(0.5,0.5,0.5)
  # death related to  both causes in same way 
  cor.mat <- corM <- rbind(c(1.0, 0.0, 0.5), c(0.0, 1.0, 0.5), c(0.5, 0.5, 1.0))
  rr <- simRecurrentII(3000,base1,base4,death.cumhaz=dr,var.z=var.z,cor.mat=cor.mat,dependence=2)
  rr <- count.history(rr,types=1:2)
  coo <- covarianceRecurrent(rr,1,2,status="status",start="entry",stop="time")
  par(mfrow=c(1,3))
  plot(coo,main ="Scenario II")
```

 * Scenario III $\rho=(0.5,0.5,0.5)$ Positive dependence among survivors and dependence on terminal event 

```{r}
  var.z <- c(0.5,0.5,0.5)
  # positive dependence for N1 and N2 all related in same way
  cor.mat <- corM <- rbind(c(1.0, 0.5, 0.5), c(0.5, 1.0, 0.5), c(0.5, 0.5, 1.0))
  rr <- simRecurrentII(3000,base1,base4,death.cumhaz=dr,var.z=var.z,cor.mat=cor.mat,dependence=2)
  rr <- count.history(rr,types=1:2)
  coo <- covarianceRecurrent(rr,1,2,status="status",start="entry",stop="time")
  par(mfrow=c(1,3))
  plot(coo,main="Scenario III")
```

 * Scenario IV $\rho=(-0.4,0.5,0.5)$ Negative dependence among survivors and positive dependence on terminal event 

```{r}
  var.z <- c(0.5,0.5,0.5)
  # negative dependence for N1 and N2 all related in same way
  cor.mat <- corM <- rbind(c(1.0, -0.4, 0.5), c(-0.4, 1.0, 0.5), c(0.5, 0.5, 1.0))
  rr <- simRecurrentII(3000,base1,base4,death.cumhaz=dr,var.z=var.z,cor.mat=cor.mat,dependence=2)
  rr <- count.history(rr,types=1:2)
  coo <- covarianceRecurrent(rr,1,2,status="status",start="entry",stop="time")
  par(mfrow=c(1,3))
  plot(coo,main="Scenario IV")
```



SessionInfo
============


```{r}
sessionInfo()
```
