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))
Related
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.
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.]
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.
I'm trying to generate random numbers with a multivariate skew normal distribution using the rmsn command from the sn package in R. I would like, ideally, to be able to get three columns of numbers with a specified variances and covariances, while having one column strongly skewed. But I'm struggling to achieve both goals simultaneously.
The post at skew normal distribution was related and useful (and the source of some of the code below), but hasn't completely clarified the issue for me.
I've been trying:
a <- c(5, 0, 0) # set shape parameter
s <- diag(3) # create variance-covariance matrix
w <- sqrt(1/(1-((2*(a^2)/(1 + a^2))/pi))) # determine scale parameter to get sd of 1
xi <- w*a/sqrt(1 + a^2)*sqrt(2/pi) # determine location parameter to get mean of 0
apply(rmsn(n=1000, xi=c(xi), Omega=s, alpha=a), 2, sd)
colMeans(rmsn(n=1000, xi=c(xi), Omega=s, alpha=a))
The columns means and SDs are correct for the second and third columns (which have no skew) but not the first (which does). Can anyone clarify where my code above, or my thinking, has gone wrong? I may be misunderstanding how to use rmsn, or the output. Any assistance would be appreciated.
The location is not the mean (except when there is no skew). From the documentation:
Notice that the location vector ‘xi’ does not represent the mean
vector of the distribution (which in fact may not even exist if ‘df <=
1’), and similarly ‘Omega’ is not the covariance matrix of the
distribution
And you may want to replace Omega=s with Omega=w.
And this is supposed to be a variance matrix: there should be no square root.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
how to generate pseudo-random positive definite matrix with constraints on the off-diagonal elements?
The user wants to impose a unique, non-trivial, upper/lower bound on the correlation between every pair of variable in a var/covar matrix.
For example: I want a variance matrix in which all variables have 0.9 > |rho(x_i,x_j)| > 0.6, rho(x_i,x_j) being the correlation between variables x_i and x_j.
Thanks.
There are MANY issues here.
First of all, are the pseudo-random deviates assumed to be normally distributed? I'll assume they are, as any discussion of correlation matrices gets nasty if we diverge into non-normal distributions.
Next, it is rather simple to generate pseudo-random normal deviates, given a covariance matrix. Generate standard normal (independent) deviates, and then transform by multiplying by the Cholesky factor of the covariance matrix. Add in the mean at the end if the mean was not zero.
And, a covariance matrix is also rather simple to generate given a correlation matrix. Just pre and post multiply the correlation matrix by a diagonal matrix composed of the standard deviations. This scales a correlation matrix into a covariance matrix.
I'm still not sure where the problem lies in this question, since it would seem easy enough to generate a "random" correlation matrix, with elements uniformly distributed in the desired range.
So all of the above is rather trivial by any reasonable standards, and there are many tools out there to generate pseudo-random normal deviates given the above information.
Perhaps the issue is the user insists that the resulting random matrix of deviates must have correlations in the specified range. You must recognize that a set of random numbers will only have the desired distribution parameters in an asymptotic sense. Thus, as the sample size goes to infinity, you should expect to see the specified distribution parameters. But any small sample set will not necessarily have the desired parameters, in the desired ranges.
For example, (in MATLAB) here is a simple positive definite 3x3 matrix. As such, it makes a very nice covariance matrix.
S = randn(3);
S = S'*S
S =
0.78863 0.01123 -0.27879
0.01123 4.9316 3.5732
-0.27879 3.5732 2.7872
I'll convert S into a correlation matrix.
s = sqrt(diag(S));
C = diag(1./s)*S*diag(1./s)
C =
1 0.0056945 -0.18804
0.0056945 1 0.96377
-0.18804 0.96377 1
Now, I can sample from a normal distribution using the statistics toolbox (mvnrnd should do the trick.) As easy is to use a Cholesky factor.
L = chol(S)
L =
0.88805 0.012646 -0.31394
0 2.2207 1.6108
0 0 0.30643
Now, generate pseudo-random deviates, then transform them as desired.
X = randn(20,3)*L;
cov(X)
ans =
0.79069 -0.14297 -0.45032
-0.14297 6.0607 4.5459
-0.45032 4.5459 3.6549
corr(X)
ans =
1 -0.06531 -0.2649
-0.06531 1 0.96587
-0.2649 0.96587 1
If your desire was that the correlations must ALWAYS be greater than -0.188, then this sampling technique has failed, since the numbers are pseudo-random. In fact, that goal will be a difficult one to achieve unless your sample size is large enough.
You might employ a simple rejection scheme, whereby you do the sampling, then redo it repeatedly until the sample has the desired properties, with the correlations in the desired ranges. This may get tiring.
An approach that might work (but one that I've not totally thought out at this point) is to use the standard scheme as above to generate a random sample. Compute the correlations. I they fail to lie in the proper ranges, then identify the perturbation one would need to make to the actual (measured) covariance matrix of your data, so that the correlations would be as desired. Now, find a zero mean random perturbation to your sampled data that would move the sample covariance matrix in the desired direction.
This might work, but unless I knew that this is actually the question at hand, I won't bother to go any more deeply into it. (Edit: I've thought some more about this problem, and it appears to be a quadratic programming problem, with quadratic constraints, to find the smallest perturbation to a matrix X, such that the resulting covariance (or correlation) matrix has the desired properties.)
This is not a complete answer, but a suggestion of a possible constructive method:
Looking at the characterizations of the positive definite matrices (http://en.wikipedia.org/wiki/Positive-definite_matrix) I think one of the most affordable approaches could be using the Sylvester criterion.
You can start with a trivial 1x1 random matrix with positive determinant and expand it in one row and column step by step while ensuring that the new matrix has also a positive determinant (how to achieve that is up to you ^_^).
Woodship,
"First of all, are the pseudo-random deviates assumed to be normally distributed?"
yes.
"Perhaps the issue is the user insists that the resulting random matrix of deviates must have correlations in the specified range."
Yes, that's the whole difficulty
"You must recognize that a set of random numbers will only have the desired distribution parameters in an asymptotic sense."
True, but this is not the problem here: your strategy works for p=2, but fails for p>2, regardless of sample size.
"If your desire was that the correlations must ALWAYS be greater than -0.188, then this sampling technique has failed, since the numbers are pseudo-random. In fact, that goal will be a difficult one to achieve unless your sample size is large enough."
It is not a sample size issue b/c with p>2 you do not even observe convergence to the right range for the correlations, as sample size growths: i tried the technique you suggest before posting here, it obviously is flawed.
"You might employ a simple rejection scheme, whereby you do the sampling, then redo it repeatedly until the sample has the desired properties, with the correlations in the desired ranges. This may get tiring."
Not an option, for p large (say larger than 10) this option is intractable.
"Compute the correlations. I they fail to lie in the proper ranges, then identify the perturbation one would need to make to the actual (measured) covariance matrix of your data, so that the correlations would be as desired."
Ditto
As for the QP, i understand the constraints, but i'm not sure about the way you define the objective function; by using the "smallest perturbation" off some initial matrix, you will always end up getting the same (solution) matrix: all the off diagonal entries will be exactly equal to either one of the two bounds (e.g. not pseudo random); plus it is kind of an overkill isn't it ?
Come on people, there must be something simpler