I'm having trouble getting my nonparametric bootstrap working, since I'm clearly doing something off.
Say I have a vector with two columns (length of n) and I wish to find their correlation.
Let's call this variable data, so length(data[,1]) is equal to n.
Suppose I want N samples.
I start with
r=rep(NA,N)
r is referring to the bootstrap sample, as I'm trying to estimate correlation coefficient
then I construct a for loop:
for(i in 1:N){
column1=sample(data[,1], n, replace=T)
column2=sample(data[,2], n, replace=T)
sample.data=data.frame(column1,column2)
r[i]=corr(sample.data)
}
The thinking is, I want to resample each column individually, then take the correlation at the end, but I think theoretically that is incorrect. something is wrong because later results will not work.
If anyone could please lend any help I would appreciate.
The problem with your logic is that you're resampling as if the pairs were independent, when your assumption is clearly that they're not (since you're trying to estimate their correlation).
For a bootstrap, you need to resample the pairs together.
[If instead you were doing a permutation test for independence against an alternative that they're correlated, you'd treat them as independent under the null.]
Related
I want to generate 1000 random variables coming from different normal distributions. I use the function "rmvnorm" for that and in a small setting, it is easily done but I have no idea how to automate it, especially for the sigma matrix (I want no correlation between the Xs). I don't really care about their means or their standard deviation. I was thinking of using a loop (e.g. increase by A the mean and by B the variance) but I want something more random and have no idea how I can do that. Again, writing down a matrix of 1000 dimension is not smart (with the condition that the off-diag elements are 0).
I have searched online but I am probably not using the rights words so I apologize if it was already asked and answered.
Thanks!
You can pass equal-length vectors for the parameters of rnorm. The first value returned will be a random draw from a normal distribution with a mean equal to the first value in the mean vector and sd equal to the first value in the sd vector:
rnorm(1e3, 1:1e3, 1:1e3)
Not sure what is meant by "I want something more random", but you can use random values for the mean and sd vectors:
rnorm(1e3, runif(1e3)*1e3, 1/rgamma(1e3, 10, 20))
So, I'd like to test how precise is t-test for detecting a mean for various distributions. But I don't want to have to define the sampling distribution each time I run the function in the function. If I write function(data, mju) and then as data input rnorm(n) or any other random sample, I obviously get the same results when replicating the function, because I only have the one "data" sample, that was first inputted. To understand more clearly what I want, here is the code:
t_ci <- function(data,mju){
prod(t.test(data)$conf.int - mju)
}
set.seed(NULL)
prec_t <- function(data, n, N, mju){
sim <- replicate(N, t_ci(data, mju))
sim[sim<0]/N
}
The first function checks, whether the real theoretical parameter "mju" in in the confidence interval. The second one replicates the function t_ci N times, to see how precise the t test confidence intervals are for selected data. I'd like to have an option to just indicate the distribution and then it would generate n-sized samples N times and calculate the precision. But as far as my code goes, it only replicates the same data over and over. Maybe there is a solution for this problem?
Also, it seems that something is wrong with the function prec_t, because I'd like to have a count of times the t_ci produced negative outcome and then divide by N.
Any help would be greatly appreciated! Thanks in advance.
I have a txt file with numbers that looks like this(but with 100 numbers) -
[1] 7.1652348 5.6665965 4.4757553 4.8497086 15.2276296 -0.5730937
[7] 4.9798067 2.7396933 5.1468304 10.1221489 9.0165661 65.7118194
[13] 5.5205704 6.3067488 8.6777177 5.2528503 3.5039562 4.2477401
[19] 11.4137624 -48.1722034 -0.3764006 5.7647536 -27.3533138 4.0968204
I need to estimate MLE theta parameter from this distrubution -
[![this is my distrubution ][1]][1]
and I need to estimate theta from a sample of 1000 observations with replace, and save the sample, and do a hist.
How can I estimate theta from my sample? I have no information about normal distrubation.
I wrote something like this -
data<-read.table(file.choose(), header = TRUE, sep= "")
B <- 1000
sample.means <- numeric(data)
sample.sd <- numeric(data)
for (i in 1:B) {
MySample <- sample(data, length(data), replace = TRUE)
sample.means <- c(sample.means,mean(MySample))
sample.sd <- c(sample.sd,sd(MySample))
}
sd(sample.sd)
but it doesn't work..
This question incorporates multiple different ones, so let's tackle each step by step.
First, you will need to draw a random sample from your population (with replacement). Assuming your 100 population-observations sit in a vector named pop.
rs <- sample(pop, 1000, replace = True)
gives you your vector of random samples. If you wanna save it, you can write it to your disk in multiple formats, so I'll just suggest a few related questions (How to Export/Import Vectors in R?).
In a second step, you can use the mle()-function of the stats4-package (https://stat.ethz.ch/R-manual/R-devel/library/stats4/html/mle.html) and specify the objective function explicitly.
However, the second part of your question is more of a statistical/conceptual question than R related, IMO.
Try to understand what MLE actually does. You do not need normally distributed variables. The idea behind MLE is to choose theta in such a way, that under the resulting distribution the random sample is the most probable. Check https://en.wikipedia.org/wiki/Maximum_likelihood_estimation for more details or some youtube videos, if you'd like a more intuitive approach.
I assume, in the description of your task, it is stated that f(x|theta) is the conditional joint density function and that the observations x are iir?
What you wanna do in this case, is to select theta such that the squared difference between the observation x and the parameter theta is minimized.
For your statistical understanding, in such cases, it makes sense to perform log-linearization on the equation, instead of dealing with a non-linear function.
Minimizing the squared difference is equivalent to maximizing the log-transformed function since the sum is negative (<=> the product was in the denominator) and the log, as well as the +1 are solely linear transformations.
This leaves you with the maximization problem:
And the first-order condition:
Obviously, you would also have to check that you are actually dealing with a maximum via the second-order condition but I'll omit that at this stage for simplicity.
The algorithm in R does nothing else than solving this maximization problem.
Hope this helps for your understanding. Maybe some smarter people can give some additional input.
Here is what I want to do:
I have a time series data frame with let us say 100 time-series of length 600 - each in one column of the data frame.
I want to pick up 4 of the time-series randomly and then assign them random weights that sum up to one (ie 0.1, 0.5, 0.3, 0.1). Using those I want to compute the mean of the sum of the 4 weighted time series variables (e.g. convex combination).
I want to do this let us say 100k times and store each result in the form
ts1.name, ts2.name, ts3.name, ts4.name, weight1, weight2, weight3, weight4, mean
so that I get a 9*100k df.
I tried some things already but R is very bad with loops and I know vector oriented
solutions are better because of R design.
Here is what I did and I know it is horrible
The df is in the form
v1,v2,v2.....v100
1,5,6,.......9
2,4,6,.......10
3,5,8,.......6
2,2,8,.......2
etc
e=NULL
for (x in 1:100000)
{
s=sample(1:100,4)#pick 4 variables randomly
a=sample(seq(0,1,0.01),1)
b=sample(seq(0,1-a,0.01),1)
c=sample(seq(0,(1-a-b),0.01),1)
d=1-a-b-c
e=c(a,b,c,d)#4 random weights
average=mean(timeseries.df[,s]%*%t(e))
e=rbind(e,s,average)#in the end i get the 9*100k df
}
The procedure runs way to slow.
EDIT:
Thanks for the help i had,i am not used to think R and i am not very used to translate every problem into a matrix algebra equation which is what you need in R.
Then the problem becomes a little bit complex if i want to calculate the standard deviation.
i need the covariance matrix and i am not sure i can if/how i can pick random elements for each sample from the original timeseries.df covariance matrix then compute the sample variance
t(sampleweights)%*%sample_cov.mat%*%sampleweights
to get in the end the ts.weighted_standard_dev matrix
Last question what is the best way to proceed if i want to bootstrap the original df
x times and then apply the same computations to test the robustness of my datas
thanks
Ok, let me try to solve your problem. As a foreword: I can think of no application where it is sensible to do what you are doing. However, that is for you to judge (non the less I would be interested in the application...)
First, note that the mean of the weighted sums equals the weighted sum of the means, as:
Let's generate some sample data:
timeseries.df <- data.frame(matrix(runif(1000, 1, 10), ncol=40))
n <- 4 # number of items in the convex combination
replications <- 100 # number of replications
Thus, we may first compute the mean of all columns and do all further computations using this mean:
ts.means <- apply(timeseries.df, 2, mean)
Let's create some samples:
samples <- replicate(replications, sample(1:length(ts.means), n))
and the corresponding weights for those samples:
weights <- matrix(runif(replications*n), nrow=n)
# Now norm the weights so that each column sums up to 1:
weights <- weights / matrix(apply(weights, 2, sum), nrow=n, ncol=replications, byrow=T)
That part was a little bit tricky. Run the single functions on each own with a small number of replications to figure out what they are doing. Note that I took a different approach for generating the weights: First get uniformly distributed data and then norm them by their sum. The result should be identical to your approach, but with arbitrary resolution and much better performance.
Again a little bit trick: Get the means for each time series and multiply them with the weights just computed:
ts.weightedmeans <- matrix(ts.means[samples], nrow=n) * weights
# and sum them up:
weights.sum <- apply(ts.weightedmeans, 2, sum)
Now, we are basically done - all information are available and ready to use. The rest is just a matter of correctly formatting the data.frame.
result <- data.frame(t(matrix(names(ts.means)[samples], nrow=n)), t(weights), weights.sum)
# For perfectness, use better names:
colnames(result) <- c(paste("Sample", 1:n, sep=''), paste("Weight", 1:n, sep=''), "WeightedMean")
I would assume this approach to be rather fast - on my system the code took 1.25 seconds with the amount of repetitions you stated.
Final word: You were in luck that I was looking for something that kept me thinking for a while. Your question was not asked in a way to encourage users to think about your problem and give good answers. The next time you have a problem, I would suggest you to read www.whathaveyoutried.com before and try to break down the problem as far as you are able to. The more concrete your problem, the faster and of higher quality your answers will be.
Edit
You mentioned correctly that the weights generated above are not uniformly distributed over the whole range of values. (I still have to object that even (0.9, 0.05, 0.025, 0.025) is possible, but it is very unlikely).
Now we are playing in a different league, though. I am pretty sure that the approach you took is not uniformly distributed as well - the probability of the last value being 0.9 is far less than the probability of the first one being that large. Honestly I do not have a good idea ready for you concerning the generation of uniformly distributed random numbers on the unit sphere according to the L_1 distance. (Actually, it is not really a unit sphere, but both problems should be identical).
Thus, I have to give up on this.
I would suggest you to raise a new question at stats.stackexchange.com concerning the generation of those random vectors. It probably is fairly simple using the correct technique. However, I doubt that this question with that heading and a fairly long answer will attract a potential responder... (If you ask the question over there, I would appreciate a link, as I would like to know the solution ;)
Concerning the variance: I do not fully understand which standard deviation you want to compute. If you just want to compute the standard deviation of each time series, why do you not use the built-in function sd? In the computation above you could just replace mean by it.
Bootstrapping: That is a whole new question. Separate different topics by starting new questions.
Assume that I have sources of data X and Y that are indexable, say matrices. And I want to run a set of independent regressions and store the result. My initial approach would be
results = matrix(nrow=nrow(X), ncol=(2))
for(i in 1:ncol(X)) {
matrix[i,] = coefficients(lm(Y[i,] ~ X[i,])
}
But, loops are bad, so I could do it with lapply as
out <- lapply(1:nrow(X), function(i) { coefficients(lm(Y[i,] ~ X[i,])) } )
Is there a better way to do this?
You are certainly overoptimizing here. The overhead of a loop is negligible compared to the procedure of model fitting and therefore the simple answer is - use whatever way you find to be the most understandable. I'd go for the for-loop, but lapply is fine too.
I do this type of thing with plyr, but I agree that it's not a processing efficency issue as much as what you are comfortable reading and writing.
If you just want to perform straightforward multiple linear regression, then I would recommend not using lm(). There is lsfit(), but I'm not sure it would offer than much of a speed up (I have never performed a formal comparison). Instead I would recommend performing the (X'X)^{-1}X'y using qr() and qrcoef(). This will allow you to perform multivariate multiple linear regression; that is, treating the response variable as a matrix instead of a vector and applying the same regression to each row of observations.
Z # design matrix
Y # matrix of observations (each row is a vector of observations)
## Estimation via multivariate multiple linear regression
beta <- qr.coef(qr(Z), Y)
## Fitted values
Yhat <- Z %*% beta
## Residuals
u <- Y - Yhat
In your example, is there a different design matrix per vector of observations? If so, you may be able to modify Z in order to still accommodate this.