Simulating averages of normally-distributed random variables in R - r

I'm trying to simulate some data in R to check my manual calculations of how variance changes in a simple model that involves a sequence of normally-distributed random variables being averaged. However, I find I'm getting results that are not only inconsistent with my manual calculations, but inconsistent with each other. Clearly I'm doing something wrong, but I'm having trouble isolating the problem(s).
Conceptually, the model involves two steps: First, storing a variable, and second, using the stored variable(s) to produce an output. The output is then stored as a new variable, contributing to future outputs, and so on. I assume storage is noisy (i.e., what's stored is a random variable rather than a constant), but that no further noise is added in output production, which simply involves averaging the existing stored variables. Thus, my model involves the following steps, where V_i is the variable stored at step i, and O_i is the ith output:
and so on.
I've tried simulating this in R in two ways: First,
nSamples <- 100000
o1 <- rnorm(nSamples) # First output
o2 <- rowMeans(cbind(rnorm(nSamples, mean=o1),rnorm(nSamples))) # Second output, averaged from first two stored variables.
o3 <- rowMeans(cbind(rnorm(nSamples, mean=o2),rnorm(nSamples, mean=o1),rnorm(nSamples))) # Third output, averaged from first three stored variables.
This gives me
var(o1) # Approximately 1, as per my manual calculations.
var(o2) # Approximately .75, as per my manual calculations.
var(o3) # Approximately .64, where my manual calculations give 19/36 or approximately .528.
Initially, I trusted the code and assumed my calculations were wrong. Then, I tried the following alternative code, which more explicitly follows the steps I used manually:
nSamples <- 100000
initialValue <- 0
v1 <- rnorm(nSamples, initialValue)
o1 <- v1
v2 <- rnorm(nSamples, o1)
o2 <- rowMeans(cbind(v1, v2))
v3 <- rnorm(nSamples, o2)
o3 <- rowMeans(cbind(v1, v2, v3))
This gives me
var(o1) # Approximately 1, as per my calculations.
var(o2) # Approximately 1.25, where my calculations give .75.
var(o3) # Approximately 1.36, where my calculations give approximately .528.
Thus, clearly I have done something wrong in using at least two of these three methods, but I'm having trouble isolating the source of the problems. What is it I'm missing that is leading my code to behave differently than what I'm expecting? And what is the difference between the two code examples that leads the variance to decrease in one and increase in the other?

Your correct calculation is the first one, where you are generating new realizations of the normal random variable when averaging, as opposed to using the realizations generated in the previous step.
In fact, the distribution of O2 assumes that the two normal random variables being averaged are mutually independent.
In your second calculation, this is not true, as you are averaging v1 and v2, which are not independent since both depend on o1. This is why you get larger variances in the second case.

Related

Simulation to find random sequences

With R I can try to find the probability that the Age vector below resulted from random sampling. I used the runs test (from randtests package) with resulted in p-value = 0.2892. Other colleagues used the rle functune (run length encoding in R) or others to simulate whether the probabilities of random allocation generating the observed sequences. Their result shows p < 0.00000001 that this sequence is the result of random sampling. I am trying to find the R code to replicate their findings. any help is highly appreciated on how to simulate to replicate their findings.
Update: I received advice from statistician that I can do this using non-parametric bootstrap. However, I still do not know how this can be done. I appreciate your help.
example:
Age <-c(68,71,72,69,80,78,80,81,84,82,67,73,65,68,66,70,69,72,74,73,68,75,70,72,75,73,69,75,74,79,80,78,80,81,79,82,69,73,67,66,70,72,69,72,75,80,68,69,71,77,70,73) ;
randtests::runs.test(Age);
X <- rle(Age);X$lengths
What was initially presented isn't the whole story. If one looks at the supplement where these numbers are from, the reported p-value is for comparing two vectors. OP only provides one, and hence the task is not well-defined.
The full assertion of the research article is that
group1 <- c(68,71,72,69,80,78,80,81,84,82,67,73,65,68,66,70,69,72,74,73,68,75,70,72,75,73)
group2 <- c(69,75,74,79,80,78,80,81,79,82,69,73,67,66,70,72,69,72,75,80,68,69,71,77,70,73)
being two independent random samples has a p-value < 0.00000001.
Even checking identity along position (10 entries in original) with permutations within a group, I'm seeing only 2 or 3 draws per million that have a similar number of identical values. I.e., something like:
set.seed(123)
mean(replicate(1e6, sum(sample(group1, length(group1)) == group2)) >= 10)
# 2e-06
Testing correlations and/or bootstrapping could easily be in the p-value range that is reported (nothing as extreme in 100 million simulations).

R - linear model does not match experimental data

I am trying to perform a linear regression on experimental data consisting of replicate measures of the same condition (for several conditions) to check for the reliability of the experimental data. For each condition I have ~5k-10k observations stored in a data frame df:
[1] cond1 repA cond1 repB cond2 repA cond2 repB ...
[2] 4.158660e+06 4454400.703 ...
[3] 1.458585e+06 4454400.703 ...
[4] NA 887776.392 ...
...
[5024] 9571785.382 9.679092e+06 ...
I use the following code to plot scatterplot + lm + R^2 values (stored in rdata) for the different conditions:
for (i in seq(1,13,2)){
vec <- matrix(0, nrow = nrow(df), ncol = 2)
vec[,1] <- df[,i]
vec[,2] <- df[,i+1]
vec <- na.exclude(vec)
plot(log10(vec[,1]),log10(vec[,2]), xlab = 'rep A', ylab = 'rep B' ,col="#00000033")
abline(fit<-lm(log10(vec[,2])~log10(vec[,1])), col='red')
legend("topleft",bty="n",legend=paste("R2 is",rdata[1,((i+1)/2)] <- format(summary(fit)$adj.r.squared,digits=4)))
}
However, the lm seems to be shifted so that it does not fit the trend I see in the experimental data:
It consistently occurs for every condition. I unsuccesfully tried to find an explanation by looking up the scource code and browsing different forums and posts (this or here).
Would have like to simply comment/ask a few questions, but can't.
From what I've understood, both repA and repB are measured with error. Hence, you cannot fit your data using an ordinary least square procedure, which only takes into account the error in Y (some might argue a weighted OLS may work, however I'm not skilled enough to discuss that). Your question seem linked to this one.
What you can use is a total least square procedure: it takes into account the error in X and Y. In the example below, I've used a "normal" TLS assuming there is the same error in X and Y (thus error.ratio=1). If it is not, you can specify the error ratio by entering error.ratio=var(y1)/var(x1) (at least I think it's var(Y)/var(X): check on the documentation to ensure that).
library(mcr)
MCR_reg=mcreg(x1,y1,method.reg="Deming",error.ratio=1,method.ci="analytical")
MCR_intercept=getCoefficients(MCR_reg)[1,1]
MCR_slope=getCoefficients(MCR_reg)[2,1]
# CI for predicted values
x_to_predict=seq(0,35)
predicted_values=MCResultAnalytical.calcResponse(MCR_reg,x_to_predict,alpha=0.05)
CI_low=predicted_values[,4]
CI_up=predicted_values[,5]
Please note that, in Deming/TLS regressions, your x- and y-errors are supposed to follow normal distribution, as explained here. If it's not the case, go for a Passing-Bablok regressions (and the R code is here).
Also note that the R2 isn't defined for Deming nor Passing Bablok regressions (see here). A correlation coefficient is a good proxy, although it does not exactly provide the same information. Since you're studying a linear correlation between two factors, see Pearson's product moment correlation coefficient, and use e.g. the rcorrfunction.

Preventing a Gillespie SSA Stochastic Model From Running Negative

I have produce a stochastic model of infection (parasitic worm), using a Gillespie SSA. The model used the "GillespieSSA"package (https://cran.r-project.org/web/packages/GillespieSSA/index.html).
In short the code models a population of discrete compartments. Movement between compartments is dependent on user defined rate equations. The SSA algorithm acts to calculate the number of events produced by each rate equation for a given timestep (tau) and updates the population accordingly, process repeats up to a given time point. The problem is, the number of events is assumed Poisson distributed (Poisson(rate[i]*tau)), thus produces an error when the rate is negative, including when population numbers become negative.
# Parameter Values
sir.parms <- c(deltaHinfinity=0.00299, CHi=0.00586, deltaH0=0.0854, aH=0.5,
muH=0.02, SigmaW=0.1, SigmaM =0.8, SigmaL=104, phi=1.15, f = 0.6674,
deltaVo=0.0166, CVo=0.0205, alphaVo=0.5968, beta=52, mbeta=7300 ,muV=52, g=0.0096, N=100)
# Inital Population Values
sir.x0 <- c(W=20,M=10,L=0.02)
# Rate Equations
sir.a <- c("((deltaH0+deltaHinfinity*CHi*mbeta*L)/(1+CHi*mbeta*L))*mbeta*L*N"
,"SigmaW*W*N", "muH*W*N", "((1/2)*phi*f)*W*N", "SigmaM*M*N", "muH*M*N",
"(deltaVo/(1+CVo*M))*beta*M*N", "SigmaL*L*N", "muV*L*N", "alphaVo*M*L*N", "(aH/g)*L*N")
# Population change for even
sir.nu <- matrix(c(+0.01,0,0,
-0.01,0,0,
-0.01,0,0,
0,+0.01,0,
0,-0.01,0,
0,-0.01,0,
0,0,+0.01/230,
0,0,-0.01/230,
0,0,-0.01/230,
0,0,-0.01/230,
0,0,-0.01/32),nrow=3,ncol=11,byrow=FALSE)
runs <- 10
set.seed(1)
# Data Frame of output
sir.out <- data.frame(time=numeric(),W=numeric(),M=numeric(),L=numeric())
# Multiple runs and combining data and SSA methods
for(i in 1:runs){
sim <- ssa(sir.x0,sir.a,sir.nu,sir.parms, method="ETL", tau=1/12, tf=140, simName="SIR")
sim.out <- data.frame(time=sim$data[,1],W=sim$data[,2],M=sim$data[,3],L=sim$data[,4])
sim.out$run <- i
sir.out <- rbind(sir.out,sim.out)
}
Thus, rates are computed and the model updates the population values for each time step, with the data store in a data frame, then attached together with previous runs. However, when levels of the population get very low events can occur such that the number of events that occurs reducing a population is greater than the number in the compartment. One method is to make the time step very small, however this greatly increases the length of the simulation very long.
My question is there a way to augment the code so that as the data is created/ calculated at each time step any values of population numbers that are negative are converted to 0?
I have tried working on this problem, but only seem to be able to come up with methods that alter the values once the simulation is complete, with the negative values still causing issues in the runs themselves.
E.g.
if (sir.out$L < 0) sir.out$L == 0
Any help would be appreciated
I believe the problem is the method you set ("ETL") in the ssa function. The ETL method will eventually produce negative numbers. You can try the "OTL" method, based on Efficient step size selection for the tau-leaping simulation method- in which there are a few more parameters that you can tweak, but the basic command is:
ssa(sir.x0,sir.a,sir.nu,sir.parms, method="OTL", tf=140, simName="SIR")
Or the direct method, which will not produce negative number whatsoever:
ssa(sir.x0,sir.a,sir.nu,sir.parms, method="D", tf=140, simName="SIR")

Is it possible to specify a range for numbers randomly generated by mvrnorm( ) in R?

I am trying to generate a random set of numbers that exactly mirror a data set that I have (to test it). The dataset consists of 5 variables that are all correlated with different means and standard deviations as well as ranges (they are likert scales added together to form 1 variable). I have been able to get mvrnorm from the MASS package to create a dataset that replicated the correlation matrix with the observed number of observations (after 500,000+ iterations), and I can easily reassign means and std. dev. through z-score transformation, but I still have specific values within each variable vector that are far above or below the possible range of the scale whose score I wish to replicate.
Any suggestions how to fix the range appropriately?
Thank you for sharing your knowledge!
To generate a sample that does "exactly mirror" the original dataset, you need to make sure that the marginal distributions and the dependence structure of the sample matches those of the original dataset.
A simple way to achieve this is with resampling
my.data <- matrix(runif(1000, -1, 2), nrow = 200, ncol = 5) # Some dummy data
my.ind <- sample(1:nrow(my.data), nrow(my.data), replace = TRUE)
my.sample <- my.data[my.ind, ]
This will ensure that the margins and the dependence structure of the sample (closely) matches those of the original data.
An alternative is to use a parametric model for the margins and/or the dependence structure (copula). But as staded by #dickoa, this will require serious modeling effort.
Note that by using a multivariate normal distribution, you are (implicity) assuming that the dependence structure of the original data is the Gaussian copula. This is a strong assumption, and it would need to be validated beforehand.

R, cointegration, multivariate, co.ja(), johansen

I am new to R and cointegration so please have patience with me as I try to explain what it is that I am trying to do. I am trying to find cointegrated variables among 1500-2000 voltage variables in the west power system in Canada/US. THe frequency is hourly (common in power) and cointegrated combinations can be as few as N variables and a maximum of M variables.
I tried to use ca.jo but here are issues that I ran into:
1) ca.jo (Johansen) has a limit to the number of variables it can work with
2) ca.jo appears to force the first variable in the y(t) vector to be the dependent variable (see below).
Eigenvectors, normalised to first column: (These are the cointegration relations)
V1.l2 V2.l2 V3.l2
V1.l2 1.0000000 1.0000000 1.0000000
V2.l2 -0.2597057 -2.3888060 -0.4181294
V3.l2 -0.6443270 -0.6901678 0.5429844
As you can see ca.jo tries to find linear combinations of the 3 variables but by forcing the coefficient on the first variable (in this case V1) to be 1 (i.e. the dependent variable). My understanding was that ca.jo would try to find all combinations such that every variable is selected as a dependent variable. You can see the same treatment in the examples given in the documentation for ca.jo.
3) ca.jo does not appear to find linear combinations of fewer than the number of variables in the y(t) vector. So if there were 5 variables and 3 of them are cointegrated (i.e. V1 ~ V2 + V3) then ca.jo fails to find this combination. Perhaps I am not using ca.jo correctly but my expectation was that a cointegrated combination where V1 ~ V2 + V3 is the same as V1 ~ V2 + V3 + 0 x V4 + 0 x V5. In other words the coefficient of the variable that are NOT cointegrated should be zero and ca.jo should find this type of combination.
I would greatly appreciate some further insight as I am fairly new to R and cointegration and have spent the past 2 months teaching myself.
Thank you.
I have also posted on nabble:
http://r.789695.n4.nabble.com/ca-jo-cointegration-multivariate-case-tc3469210.html
I'm not an expert, but since no one is responding, I'm going to try to take a stab at this one.. EDIT: I noticed that I just answered to a 4 year old question. Hopefully it might still be useful to others in the future.
Your general understanding is correct. I'm not going to go in great detail about the whole procedure but will try to give some general insight. The first thing that the Johansen procedure does is create a VECM out of the VAR model that best corresponds to the data (This is why you need the lag length for the VAR as input to the procedure as well). The procedure will then investigate the non-lagged component matrix of the VECM by looking at its rank: If the variables are not cointegrated then the rank of the matrix will not be significantly different from 0. A more intuitive way of understanding the johansen VECM equations is to notice the comparibility with the ADF procedure for each distinct row of the model.
Furthermore, The rank of the matrix is equal to the number of its eigenvalues (characteristic roots) that are different from zero. Each eigenvalue is associated with a different cointegrating vector, which
is equal to its corresponding eigenvector. Hence, An eigenvalue significantly different
from zero indicates a significant cointegrating vector. Significance of the vectors can be tested with two distinct statistics: The max statistic or the trace statistic. The trace test tests the null hypothesis of less than or equal to r cointegrating vectors against the alternative of more than r cointegrating vectors. In contrast, The maximum eigenvalue test tests the null hypothesis of r cointegrating vectors against the alternative of r + 1 cointegrating vectors.
Now for an example,
# We fit data to a VAR to obtain the optimal VAR length. Use SC information criterion to find optimal model.
varest <- VAR(yourData,p=1,type="const",lag.max=24, ic="SC")
# obtain lag length of VAR that best fits the data
lagLength <- max(2,varest$p)
# Perform Johansen procedure for cointegration
# Allow intercepts in the cointegrating vector: data without zero mean
# Use trace statistic (null hypothesis: number of cointegrating vectors <= r)
res <- ca.jo(yourData,type="trace",ecdet="const",K=lagLength,spec="longrun")
testStatistics <- res#teststat
criticalValues <- res#criticalValues
# chi^2. If testStatic for r<= 0 is greater than the corresponding criticalValue, then r<=0 is rejected and we have at least one cointegrating vector
# We use 90% confidence level to make our decision
if(testStatistics[length(testStatistics)] >= criticalValues[dim(criticalValues)[1],1])
{
# Return eigenvector that has maximum eigenvalue. Note: we throw away the constant!!
return(res#V[1:ncol(yourData),which.max(res#lambda)])
}
This piece of code checks if there is at least one cointegrating vector (r<=0) and then returns the vector with the highest cointegrating properties or in other words, the vector with the highest eigenvalue (lamda).
Regarding your question: the procedure does not "force" anything. It checks all combinations, that is why you have your 3 different vectors. It is my understanding that the method just scales/normalizes the vector to the first variable.
Regarding your other question: The procedure will calculate the vectors for which the residual has the strongest mean reverting / stationarity properties. If one or more of your variables does not contribute further to these properties then the component for this variable in the vector will indeed be 0. However, if the component value is not 0 then it means that "stronger" cointegration was found by including the extra variable in the model.
Furthermore, you can test test significance of your components. Johansen allows a researcher to test a hypothesis about one or more
coefficients in the cointegrating relationship by viewing the hypothesis as
a restriction on the non-lagged component matrix in the VECM. If there exist r cointegrating vectors, only these linear combinations or linear transformations of them, or combinations of the cointegrating vectors, will be stationary. However, I'm not aware on how to perform these extra checks in R.
Probably, the best way for you to proceed is to first test the combinations that contain a smaller number of variables. You then have the option to not add extra variables to these cointegrating subsets if you don't want to. But as already mentioned, adding other variables can potentially increase the cointegrating properties / stationarity of your residuals. It will depend on your requirements whether or not this is the behaviour you want.
I've been searching for an answer to this and I think I found one so I'm sharing with you hoping it's the right solution.
By using the johansen test you test for the ranks (number of cointegration vectors), and it also returns the eigenvectors, and the alphas and betas do build said vectors.
In theory if you reject r=0 and accept r=1 (value of r=0 > critical value and r=1 < critical value) you would search for the highest eigenvalue and from that build your vector. On this case, if the highest eigenvalue was the first, it would be V1*1+V2*(-0.26)+V3*(-0.64).
This would generate the cointegration residuals for these variables.
Again, I'm not 100%, but preety sure the above is how it works.
Nonetheless, you can always use the cajools function from the urca package to create a VECM automatically. You only need to feed it a cajo object and define the number of ranks (https://cran.r-project.org/web/packages/urca/urca.pdf).
If someone could confirm / correct this, it would be appreciated.

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