I'm brand new to R and trying to implement a simple model (which I will extend later) that deals with corporate bond defaults.
For starters, I'm using only two clients.
Parameters:
- two clients (which I name "A" and "B")
- a cash flow of $10,000 will be received from each client if they do not default within 10 years
- pulling together concepts using standard normal random variables, dependent uniform random variables and Gaussian copulas
- run some number of simulations
- store the sum of Client A cash flow plus Client B cash flow and store in a vector named "result"
- finally, take the average of the result vector
My code is:
# define variables
nSim <- 5 # of simulations
rho <- 0.3 # rho
lambda <- 0.01 # default intensity
T <- 10 # time to default
for (i in 1:nSim){
# Step 1: generate 2 independent standard normal random variables
z1 <- rnorm(1, mean=0, sd=1)
z2 <- rnorm(1, mean=0, sd=1)
# Step 2: map the normals into correlated normals
# by Cholesky composition of the correlation matrix
# w1 = z1
# w2 = rho(z1)+sqrt(1-(rho^2))*z2
w1 <- z1
w2 <- rho*z1 - sqrt(1-(rho^2))*z2
# Step 3: using the correlated normals, generate two dependent uniform variables
u <- runif(1, min=0, max=1)
v <- runif(1, min=0, max=1)
# Step 4: using the dependent uniforms, generate two dependent exponentials
tau.A <- (-1/lambda)*log(u)
tau.B <- (-1/lambda)*log(v)
payout.A <- if (tau.A > 10) {10000} else {0}
payout.B <- if (tau.B > 10) {10000} else {0}
result[i] = (payout.A[i] + payout.B[i])
}
# calculate expected value of portfolio
mean(result)
When I run this code, I'm getting an error of "NA" and can't figure out why (again, I'm brand new to R). I don't think each of the simulation values is being stored in the results vector, but don't know how to diagnose the problem.
Thanks in advance to anyone who can help!
--Sarah
Everything works until the results[i] <- (payout.A[i] + payout.B[i]) line. The problem is you never set results.
Before your for loop, add the line:
results <- vector('numeric', length = nSim)
This will create a vector of 0s with a length of nSim. In R is is best to preallocate the space instead of dynamically growing a vector using c().
No the problem is the presence of the [i] assignments in the results[i] <- (payout.A[i] + payout.B[i]) line.
The [i] assignment is okay for the results parameter but not the two payout parameters because each of these are being generated in each loop. So simply remove them to form the line:
results[i] <- (payout.A + payout.B)
will solve your issue. If you wish to keep each payout in its own vector then you need to assign it as such, but it seems that you don't.
Related
I have a stochastic SIR model that has the variables time T, contact rate beta, recovery rate gamma and number of susceptible n plus number of infectious m. And I would like to find a way to get a time varying value of beta.
I would like to have (at least) two different values of beta that change depending on the time T. My thought is to have beta == beta1 for t <= T/2 and then beta == beta2 for t > T, which means that I have one value for beta of the first half of the time and another value for the last half of the time. And I wonder if this can be implemented in any way? The following is my code for the model:
rSIR <- function(T, beta, gamma, n, m) {
t <- 0
x <- n # Susceptibles
y <- m # Infectious
# Possible events
eventLevels <- c("S->I","I->R")
# Initialize result
events <- data.frame(t=t,x=x,y=y,event=NA)
# Loop
while (t < T & (y>0)) {
# Draw
wait <- rexp(2,c("S->I"=beta*x*y,"I->R"=gamma*y))
# Which event occurs first
i <- which.min(wait)
# Advance time
t <- t+wait[i]
# Update population
if (eventLevels[i] == "S->I") {x <- x-1; y <- y+1}
if (eventLevels[i] == "I->R") {y <- y-1}
# Store results
events <- rbind(events,c(t,x,y,i))
}
events$event<- factor(eventLevels[events$event], levels=eventLevels)
return(events)
}
I have tried to do this by adding two if statements in the while loop where I repeat the code, one for beta1 which I define straight after the while loop is created, and another for beta2 where the if statements are the ones I mentioned in the beginning, i.e. if t <= T/2 and if t > T/2 but this does not seem to work. So I'm wondering if there is a more convenient way to specify this, maybe a statement at the beginning of the loop that directly specifies what the beta should be depending on where in time T we are?
I am trying to run some code on R based on this paper here through example 5.1. I want to simulate the following:
My background on R isn't great so I have the following code below, how can I generate a histogram and samples from this?
xseq<-seq(0, 100, 1)
n<-100
Z<- pnorm(xseq,0,1)
U<- pbern(xseq, 0.4, lower.tail = TRUE, log.p = FALSE)
Beta <- (-1)^U*(4*log(n)/(sqrt(n)) + abs(Z))
Some demonstrations of tools that will be of use:
rnorm(1) # generates one standard normal variable
rnorm(10) # generates 10 standard normal variables
rnorm(1, 5, 6) # generates 1 normal variable with mu = 5, sigma = 6
# not needed for this problem, but perhaps worth saying anyway
rbinom(5, 1, 0.4) # generates 5 Bernoulli variables that are 1 w/ prob. 0.4
So, to generate one instance of a beta:
n <- 100 # using the value you gave; I have no idea what n means here
u <- rbinom(1, 1, 0.4) # make one Bernoulli variable
z <- rnorm(1) # make one standard normal variable
beta <- (-1)^u * (4 * log(n) / sqrt(n) + abs(z))
But now, you'd like to do this many times for a Monte Carlo simulation. One way you might do this is by building a function, having beta be its output, and using the replicate() function, like this:
n <- 100 # putting this here because I assume it doesn't change
genbeta <- function(){ # output of this function will be one copy of beta
u <- rbinom(1, 1, 0.4)
z <- rnorm(1)
return((-1)^u * (4 * log(n) / sqrt(n) + abs(z)))
}
# note that we don't need to store beta anywhere directly;
# rather, it is just the return()ed value of the function we defined
betadraws <- replicate(5000, genbeta())
hist(betadraws)
This will have the effect of making 5000 copies of your beta variable and putting them in a histogram.
There are other ways to do this -- for instance, one might just make a big matrix of the random variables and work directly with it -- but I thought this would be the clearest approach for starting out.
EDIT: I realized that I ignored the second equation entirely, which you probably didn't want.
We've now made a vector of beta values, and you can control the length of the vector in the first parameter of the replicate() function above. I'll leave it as 5000 in my continued example below.
To get random samples of the Y vector, you could use something like:
x <- replicate(5000, rnorm(17))
# makes a 17 x 5000 matrix of independent standard normal variables
epsilon <- rnorm(17)
# vector of 17 standard normals
y <- x %*% betadraws + epsilon
# y is now a 17 x 1 matrix (morally equivalent to a vector of length 17)
and if you wanted to get many of these, you could wrap that inside another function and replicate() it.
Alternatively, if you didn't want the Y vector, but just a single Y_i component:
x <- rnorm(5000)
# x is a vector of 5000 iid standard normal variables
epsilon <- rnorm(1)
# epsilon_i is a single standard normal variable
y <- t(x) %*% betadraws + epsilon
# t() is the transpose function; y is now a 1 x 1 matrix
For a science project, I am looking for a way to generate random data in a certain range (e.g. min=0, max=100000) with a certain correlation with another variable which already exists in R. The goal is to enrich the dataset a little so I can produce some more meaningful graphs (no worries, I am working with fictional data).
For example, I want to generate random values correlating with r=-.78 with the following data:
var1 <- rnorm(100, 50, 10)
I already came across some pretty good solutions (i.e. https://stats.stackexchange.com/questions/15011/generate-a-random-variable-with-a-defined-correlation-to-an-existing-variable), but only get very small values, which I cannot transform so the make sense in the context of the other, original values.
Following the example:
var1 <- rnorm(100, 50, 10)
n <- length(var1)
rho <- -0.78
theta <- acos(rho)
x1 <- var1
x2 <- rnorm(n, 50, 50)
X <- cbind(x1, x2)
Xctr <- scale(X, center=TRUE, scale=FALSE)
Id <- diag(n)
Q <- qr.Q(qr(Xctr[ , 1, drop=FALSE]))
P <- tcrossprod(Q) # = Q Q'
x2o <- (Id-P) %*% Xctr[ , 2]
Xc2 <- cbind(Xctr[ , 1], x2o)
Y <- Xc2 %*% diag(1/sqrt(colSums(Xc2^2)))
var2 <- Y[ , 2] + (1 / tan(theta)) * Y[ , 1]
cor(var1, var2)
What I get for var2 are values ranging between -0.5 and 0.5. with a mean of 0. I would like to have much more distributed data, so I could simply transform it by adding 50 and have a quite simililar range compared to my first variable.
Does anyone of you know a way to generate this kind of - more or less -meaningful data?
Thanks a lot in advance!
Starting with var1, renamed to A, and using 10,000 points:
set.seed(1)
A <- rnorm(10000,50,10) # Mean of 50
First convert values in A to have the new desired mean 50,000 and have an inverse relationship (ie subtract):
B <- 1e5 - (A*1e3) # Note that { mean(A) * 1000 = 50,000 }
This only results in r = -1. Add some noise to achieve the desired r:
B <- B + rnorm(10000,0,8.15e3) # Note this noise has mean = 0
# the amount of noise, 8.15e3, was found through parameter-search
This has your desired correlation:
cor(A,B)
[1] -0.7805972
View with:
plot(A,B)
Caution
Your B values might fall outside your range 0 100,000. You might need to filter for values outside your range if you use a different seed or generate more numbers.
That said, the current range is fine:
range(B)
[1] 1668.733 95604.457
If you're happy with the correlation and the marginal distribution (ie, shape) of the generated values, multiply the values (that fall between (-.5, +.5) by 100,000 and add 50,000.
> c(-0.5, 0.5) * 100000 + 50000
[1] 0e+00 1e+05
edit: this approach, or any thing else where 100,000 & 50,000 are exchanged for different numbers, will be an example of a 'linear transformation' recommended by #gregor-de-cillia.
I've been working with MCMC for population genetics and I have some doubts.
I'm not experienced in statistics and because of that I have difficulty.
I have code to run MCMC, 1000 iterations. I start by creating a matrix with 0's (50 columns = 50 individuals and 1000 lines for 1000 iterations).
Then I create a random vector to substitute the first line of the matrix. This vector has 1's and 2's, representing population 1 or population 2.
I also have genotype frequencies and the genotypes of the 50 individuals.
What I want is to, according to the genotype frequencies and genotypes, determine to what population an individual belongs.
Then, I'll keep changing the population assigned to a random individual and checking if the new value should be accepted.
niter <- 1000
z <- matrix(0,nrow=niter,ncol=ncol(targetinds))
z[1,] <- sample(1:2, size=ncol(z), replace=T)
lhood <- numeric(niter)
lhood[1] <- compute_lhood_K2(targetinds, z[1,], freqPops)
accepted <- 0
priorz <- c(1e-6, 0.999999)
for(i in 2:niter) {
z[i,] <- z[i-1,]
# propose new vector z, by selecting a random individual, proposing a new zi value
selind <- sample(1:nind, size=1)
# proposal probability of selecting individual at random
proposal_ratio_ind <- log(1/nind)-log(1/nind)
# propose a new index for the selected individual
if(z[i,selind]==1) {
z[i,selind] <- 2
} else {
z[i,selind] <- 1
}
# proposal probability of changing the index of individual is 1/2
proposal_ratio_cluster <- log(1/2)-log(1/2)
propratio <- proposal_ratio_ind+proposal_ratio_cluster
# compute f(x_i|z_i*, p)
# the probability of the selected individual given the two clusters
probindcluster <- compute_lhood_ind_K2(targetinds[,selind],freqPops)
# likelihood ratio f(x_i|z_i*,p)/f(x_i|z_i, p)
lhoodratio <- probindcluster[z[i,selind]]-probindcluster[z[i-1,selind]]
# prior ratio pi(z_i*)/pi(z_i)
priorratio <- log(priorz[z[i,selind]])-log(priorz[z[i-1,selind]])
# accept new value according to the MH ratio
mh <- lhoodratio+propratio+priorratio
# reject if the random value is larger than the MH ratio
if(runif(1)>exp(mh)) {
z[i,] <- z[i-1,] # keep the same z
lhood[i] <- lhood[i-1] # keep the same likelihood
} else { # if accepted
lhood[i] <- lhood[i-1]+lhoodratio # update the likelihood
accepted <- accepted+1 # increase the number of accepted
}
}
It is asked that I have to change the proposal probability so that the new proposed values are proportional to the likelihood. This leads to a Gibbs sampling MCMC algorithm, supposedly.
I don't know what to change in the code to do this. I also don't understand very well the concept of proposal probability and how to chose the prior.
Grateful if someone knows how to clarify my doubts.
Your current proposal is done here:
# propose a new index for the selected individual
if(z[i,selind]==1) {
z[i,selind] <- 2
} else {
z[i,selind] <- 1
}
if the individual is assigned to cluster 1, then you propose to switch assignment deterministically by assigning them to cluster 2 (and vice versa).
You didn't show us what freqPops is, but if you want to propose according to freqPops then I believe the above code has to be replaced by
z[i,selind] <- sample(c(1,2),size=1,prob=freqPops)
(at least that is what I understand when you say you want to propose based on the likelihood - however, that statement of yours is unclear).
For this now to be a valid mcmc gibbs sampling algorithm you also need to change the next line of code:
proposal_ratio_cluster <- log(freqPops[z[i-1,selind]])-log(fregPops[z[i,selind]])
I want to generate 2 continuous random variables Q1, Q2 (quantitative traits, each are normal) and 2 binary random variables Z1, Z2 (binary traits) with given pairwise correlations between all possible pairs of them.
Say
(Q1,Q2):0.23
(Q1,Z1):0.55
(Q1,Z2):0.45
(Q2,Z1):0.4
(Q2,Z2):0.5
(Z1,Z2):0.47
Please help me generate such data in R.
This is crude but might get you started in the right direction.
library(copula)
options(digits=3)
probs <- c(0.5,0.5)
corrs <- c(0.23,0.55,0.45,0.4,0.5,0.47) ## lower triangle
Simulate correlated values (first two quantitative, last two transformed to binary)
sim <- function(n,probs,corrs) {
tmp <- normalCopula( corrs, dim=4 , "un")
getSigma(tmp) ## test
x <- rCopula(1000, tmp)
x2 <- x
x2[,3:4] <- qbinom(x[,3:4],size=1,prob=rep(probs,each=nrow(x)))
x2
}
Test SSQ distance between observed and target correlations:
objfun <- function(corrs,targetcorrs,probs,n=1000) {
cc <- try(cor(sim(n,probs,corrs)),silent=TRUE)
if (is(cc,"try-error")) return(NA)
sum((cc[lower.tri(cc)]-targetcorrs)^2)
}
See how bad things are when input corrs=target:
cc0 <- cor(sim(1000,probs=probs,corrs=corrs))
cc0[lower.tri(cc0)]
corrs
objfun(corrs,corrs,probs=probs) ## 0.112
Now try to optimize.
opt1 <- optim(fn=objfun,
par=corrs,
targetcorrs=corrs,probs=c(0.5,0.5))
opt1$value ## 0.0208
Stops after 501 iterations with "max iterations exceeded". This will never work really well because we're trying to use a deterministic hill-climbing algorithm on a stochastic objective function ...
cc1 <- cor(sim(1000,probs=c(0.5,0.5),corrs=opt1$par))
cc1[lower.tri(cc1)]
corrs
Maybe try simulated annealing?
opt2 <- optim(fn=objfun,
par=corrs,
targetcorrs=corrs,probs=c(0.5,0.5),
method="SANN")
It doesn't seem to do much better than the previous value. Two possible problems (left as an exercise for the reader are) (1) we have specified a set of correlations that are not feasible with the marginal distributions we have chosen, or (2) the error in the objective function surface is getting in the way -- to do better we would have to average over more replicates (i.e. increase n).